Quantum Physics

Color-enhanced scanning electronmicrograph of the storage miteLepidoglyphus destructor. Thesecommon mites grow to 0.75 mm andfeed on molds, flour, and rice. Theythrive at 25°C and high humidity andcan trigger allergies.CHAPTER27O U T L I N E27.1 Blackbody Radiation andPlanck’s Hypothesis27.2 The Photoelectric Effectand the Particle Theory ofLight27.3 X-Rays27.4 Diffraction of X-Rays byCrystals27.5 The Compton Effect27.6 The Dual Nature of Lightand Matter27.7 The Wave Function27.8 The Uncertainty Principle27.9 The Scanning TunnelingMicroscope© Eye of Science/Science Source/Photo Researchers, Inc.Quantum PhysicsAlthough many problems were resolved by the theory of relativity in the early part of the 20thcentury, many other problems remained unsolved. Attempts to explain the behavior of matteron the atomic level with the laws of classical physics were consistently unsuccessful. Variousphenomena, such as the electromagnetic radiation emitted by a heated object (blackbodyradiation), the emission of electrons by illuminated metals (the photoelectric effect), and theemission of sharp spectral lines by gas atoms in an electric discharge tube, couldn’t be understoodwithin the framework of classical physics. Between 1900 and 1930, however, a modernversion of mechanics called quantum mechanics or wave mechanics was highly successful inexplaining the behavior of atoms, molecules, and nuclei.The earliest ideas of quantum theory were introduced by Planck, and most of the subsequentmathematical developments, interpretations, and improvements were made by a numberof distinguished physicists, including Einstein, Bohr, Schrödinger, de Broglie, Heisenberg,Born, and Dirac. In this chapter we introduce the underlying ideas of quantum theory and thewave–particle nature of matter, and discuss some simple applications of quantum theory,including the photoelectric effect, the Compton effect, and x-rays.27.1 BLACKBODY RADIATION ANDPLANCK’S HYPOTHESISAn object at any temperature emits electromagnetic radiation, called thermalradiation. Stefan’s law, discussed in Section 11.5, describes the total power radiated.The spectrum of the radiation depends on the temperature and propertiesof the object. At low temperatures, the wavelengths of the thermal radiation aremainly in the infrared region and hence not observable by the eye. As the temperatureof an object increases, the object eventually begins to glow red. At sufficientlyhigh temperatures, it appears to be white, as in the glow of the hot tungsten filamentof a lightbulb. A careful study of thermal radiation shows that it consists of a874

27.1 Blackbody Radiation and Planck’s Hypothesis 875continuous distribution of wavelengths from the infrared, visible, and ultravioletportions of the spectrum.From a classical viewpoint, thermal radiation originates from acceleratedcharged particles near the surface of an object; such charges emit radiation, muchas small antennas do. The thermally agitated charges can have a distribution of frequencies,which accounts for the continuous spectrum of radiation emitted by theobject. By the end of the 19th century, it had become apparent that the classicaltheory of thermal radiation was inadequate. The basic problem was in understandingthe observed distribution energy as a function of wavelength in the radiationemitted by a blackbody. By definition, a blackbody is an ideal system that absorbsall radiation incident on it. A good approximation of a blackbody is a small holeleading to the inside of a hollow object, as shown in Figure 27.1. The nature ofthe radiation emitted through the small hole leading to the cavity depends only onthe temperature of the cavity walls, and not at all on the material composition of theobject, its shape, or other factors.Experimental data for the distribution of energy in blackbody radiation at threetemperatures are shown in Active Figure 27.2 (page 876). The radiated energyvaries with wavelength and temperature. As the temperature of the blackbody increases,the total amount of energy (area under the curve) it emits increases. Also,with increasing temperature, the peak of the distribution shifts to shorter wavelengths.This shift obeys the following relationship, called Wien’s displacement law, max T 0.2898 10 2 m K [27.1]where max is the wavelength at which the curve peaks and T is the absolute temperatureof the object emitting the radiation.Figure 27.1 An opening in the cavityof a body is a good approximationof a blackbody. As light enters thecavity through the small opening, partis reflected and part is absorbed oneach reflection from the interior walls.After many reflections, essentially allof the incident energy is absorbed.TIP 27.1 Expect to BeConfusedYour life experiences take place inthe macroscopic world, where quantumeffects are not evident. Quantumeffects can be even more bizarre thanrelativistic effects, but don’t despair:confusion is normal and expected. Asthe Nobel prize-winning physicistRichard Feynman once said, “Nobodyunderstands quantum mechanics.”Applying Physics 27.1Star ColorsIf you look carefully at stars in the night sky, you candistinguish three main colors: red, white, and blue.What causes these particular colors?Explanation These colors result from the differentsurface temperatures of stars. A relatively cool star, witha surface temperature of 3 000 K, has a radiation curvelike the middle curve in Active Figure 27.2 (page 876).The peak in this curve is above the visible wavelengths,0.4 m–0.7m, beyond the wavelength of red light, sosignificantly more radiation is emitted within the visiblerange at the red end than the blue end of the spectrum.Consequently, the star appears reddish in color, similarto the red glow from the burner of an electric stove.A hotter star has a radiation curve more like the upper curve in Active Figure 27.2.In this case, the star emits significant radiation throughout the visible range, andthe combination of all colors causes the star to look white. This is the case with ourown Sun, with a surface temperature of 5 800 K. For very hot stars, the peak canbe shifted so far below the visible range that significantly more blue radiation isemitted than red, so the star appears bluish in color.EXAMPLE 27.1 Thermal Radiation from the Human BodyGoal Apply Wien’s law.Problem The temperature of the skin is approximately 35.0°C. At what wavelength does the radiation emitted fromthe skin reach its peak?Strategy This is a matter of substitution into Wien’s law, Equation 27.1.SolutionApply Wien’s displacement law: max T 0.289 8 10 2 m K

876 Chapter 27 Quantum PhysicsSolve for max , noting that 35.0°C corresponds to anabsolute temperature of 308 K:max 0.289 8 102 mK308 K9.41 mRemarkThis radiation is in the infrared region of the spectrum.Exercise 27.1(a) Find the wavelength corresponding to the peak of the radiation curve for the heating element of an electric ovenat a temperature of 1.20 10 3 K. (Note that although this radiation peak lies in the infrared, there is enough visibleradiation at this temperature to give the element a red glow.) (b) The peak in the radiation curve of the Sun is510 nm. Calculate the temperature of the surface of the Sun.Answers (a) 2.42 m; (b) 5 700 KIntensity0124 000 K3 000 K2 000 KWavelength ( µ m)3 4ACTIVE FIGURE 27.2Intensity of blackbody radiation versuswavelength at three different temperatures.Note that the total radiationemitted (the area under a curve)increases with increasing temperature.Log into PhysicsNow at www.cp7e.comand go to Active Figure 27.2, whereyou can adjust the temperature of theblackbody and study the radiationemitted from it.IntensityClassicaltheoryWavelengthExperimentaldataACTIVE FIGURE 27.3Comparison of experimental datawith the classical theory of blackbodyradiation. Planck’s theory matchesthe experimental data perfectly.Log into PhysicsNow at www.cp7e.comand go to Active Figure 27.3, whereyou can investigate the discreteenergies emitted in the Planck model.Attempts to use classical ideas to explain the shapes of the curves shown inActive Figure 27.2 failed. Active Figure 27.3 shows an experimental plot of theblackbody radiation spectrum (red curve), together with the theoretical picture ofwhat this curve should look like based on classical theories (blue curve). At longwavelengths, classical theory is in good agreement with the experimental data. Atshort wavelengths, however, major disagreement exists between classical theoryand experiment. As approaches zero, classical theory predicts that the amount ofenergy being radiated should increase. In fact, the theory erroneously predictsthat the intensity should be infinite, when the experimental data shows it shouldapproach zero. This contradiction is called the ultraviolet catastrophe, becausetheory and experiment disagree strongly in the short-wavelength, ultravioletregion of the spectrum.In 1900 Planck developed a formula for blackbody radiation that wasin complete agreement with experiments at all wavelengths, leading to acurve shown by the red line in Active Figure 27.3. Planck hypothesized thatblackbody radiation was produced by submicroscopic charged oscillators, whichhe called resonators. He assumed that the walls of a glowing cavity werecomposed of billions of these resonators, although their exact nature wasunknown. The resonators were allowed to have only certain discrete energiesE n , given byE n nhf[27.2]where n is a positive integer called a quantum number, f is the frequency of vibrationof the resonator, and h is a constant known as Planck’s constant, which has thevalueh 6.626 10 34 Js[27.3]Because the energy of each resonator can have only discrete values given byEquation 27.2, we say the energy is quantized. Each discrete energy value representsa different quantum state, with each value of n representing a specific quantum state.(When the resonator is in the n 1 quantum state, its energy is hf ; when it is inthe n 2 quantum state, its energy is 2 hf ; and so on.)The key point in Planck’s theory is the assumption of quantized energy states.This is a radical departure from classical physics, the “quantum leap” that led to atotally new understanding of nature. It’s shocking: it’s like saying a pitched baseballcan have only a fixed number of different speeds, and no speeds in betweenthose fixed values. When Planck presented his theory, most scientists (includingPlanck!) didn’t consider the quantum concept to be realistic; however, subsequentdevelopments showed that a theory based on the quantum concept (rather thanon classical concepts) had to be used to explain a number of other phenomena atthe atomic level.

878 Chapter 27 Quantum PhysicsACTIVE FIGURE 27.4A circuit diagram for studying thephotoelectric effect. When lightstrikes plate E (the emitter), photoelectronsare ejected from the plate.Electrons moving from plate E toplate C (the collector) create acurrent in the circuit, registered atthe ammeter, A.LightCPhotoelectronsE© Bettmann/CORBISLog into PhysicsNow at www.cp7e.comand go to Active Figure 27.4, whereyou can observe the motion ofelectrons for various frequencies andvoltages.VAMAX PLANCK, German Physicist,(1858 – 1947)Planck introduced the concept of a “quantumof action” (Planck’s constant h) in anattempt to explain the spectral distributionof blackbody radiation, which laid the foundationsfor quantum theory. In 1918, hewas awarded the Nobel Prize for this discoveryof the quantized nature of energy.Current–V sHigh intensityLow intensityApplied voltageACTIVE FIGURE 27.5Photoelectric current versus appliedpotential difference for two lightintensities. The current increases withintensity, but reaches a saturationlevel for large values of V. At voltagesequal to or less than V s ,where V s is the stopping potential,the current is zero.Log into PhysicsNow at www.cp7e.comand go to Active Figure 27.5, whereyou can sweep through the voltagerange and observe the current curvefor different intensities of radiation.Variable powersupplyActive Figure 27.4 is a schematic diagram of a photoelectric effect apparatus.An evacuated glass tube known as a photocell contains a metal plate E (the emitter)connected to the negative terminal of a variable power supply. Another metalplate, C (the collector), is maintained at a positive potential by the power supply.When the tube is kept in the dark, the ammeter reads zero, indicating that there isno current in the circuit. However, when plate E is illuminated by light having awavelength shorter than some particular wavelength that depends on the materialused to make plate E, a current is detected by the ammeter, indicating a flow ofcharges across the gap between E and C. This current arises from photoelectronsemitted from the negative plate E and collected at the positive plate C.Active Figure 27.5 is a plot of the photoelectric current versus the potential differenceV between E and C for two light intensities. At large values of V, the currentreaches a maximum value. In addition, the current increases as the incident light intensityincreases, as you might expect. Finally, when V is negative—that is, when thepower supply in the circuit is reversed to make E positive and C negative—the currentdrops to a low value because most of the emitted photoelectrons are repelled by thenow negative plate C. In this situation, only those electrons having a kinetic energygreater than the magnitude of eV reach C, where e is the charge on the electron.When V is equal to or more negative than V s , the stopping potential, noelectrons reach C and the current is zero. The stopping potential is independent ofthe radiation intensity. The maximum kinetic energy of the photoelectrons isrelated to the stopping potential through the relationshipKE max eV s[27.4]Several features of the photoelectric effect can’t be explained with classicalphysics or with the wave theory of light:• No electrons are emitted if the incident light frequency falls below some cutofffrequency f c , which is characteristic of the material being illuminated. This isinconsistent with the wave theory, which predicts that the photoelectric effectshould occur at any frequency, provided the light intensity is sufficiently high.• The maximum kinetic energy of the photoelectrons is independent of lightintensity. According to wave theory, light of higher intensity should carry moreenergy into the metal per unit time and therefore eject photoelectrons havinghigher kinetic energies.• The maximum kinetic energy of the photoelectrons increases with increasinglight frequency. The wave theory predicts no relationship between photoelectronenergy and incident light frequency.• Electrons are emitted from the surface almost instantaneously (less than 10 9 safter the surface is illuminated), even at low light intensities. Classically, weexpect the photoelectrons to require some time to absorb the incident radiationbefore they acquire enough kinetic energy to escape from the metal.

27.2 The Photoelectric Effect and the Particle Theory of Light 879A successful explanation of the photoelectric effect was given by Einstein in1905, the same year he published his special theory of relativity. As part of a generalpaper on electromagnetic radiation, for which he received the Nobel Prize in1921, Einstein extended Planck’s concept of quantization to electromagneticwaves. He suggested that a tiny packet of light energy or photon would be emittedwhen a quantized oscillator made a jump from an energy state E n nhf to thenext lower state E n1 (n 1)hf. Conservation of energy would require thedecrease in oscillator energy, hf, to be equal to the photon’s energy E, so thatE hf [27.5]where h is Planck’s constant and f is the frequency of the light, which is equal tothe frequency of Planck’s oscillator.The key point here is that the light energy lost by the emitter, hf, stays sharplylocalized in a tiny packet or particle called a photon. In Einstein’s model, a photonis so localized that it can give all its energy hf to a single electron in the metal.According to Einstein, the maximum kinetic energy for these liberated photoelectronsis Energy of a photonKE max hf [27.6] Photoelectric effect equationwhere is called the work function of the metal. The work function, which representsthe minimum energy with which an electron is bound in the metal, is on theorder of a few electron volts. Table 27.1 lists work functions for various metals.With the photon theory of light, we can explain the previously mentionedfeatures of the photoelectric effect that cannot be understood using concepts ofclassical physics:• Photoelectrons are created by absorption of a single photon, so the energy ofthat photon must be greater than or equal to the work function, else no photoelectronswill be produced. This explains the cutoff frequency.• From Equation 27.6, KE max depends only on the frequency of the light and thevalue of the work function. Light intensity is immaterial, because absorption ofa single photon is responsible for the electron’s change in kinetic energy.• Equation 27.6 is linear in the frequency, so KE max increases with increasingfrequency.• Electrons are emitted almost instantaneously, regardless of intensity, becausethe light energy is concentrated in packets rather than spread out in waves. Ifthe frequency is high enough, no time is needed for the electron to graduallyacquire sufficient energy to escape the metal.Experimentally, a linear relationship is observed between f and KE max , as sketchedin Figure 27.6. The intercept on the horizontal axis, corresponding to KE max 0,gives the cutoff frequency below which no photoelectrons are emitted, regardless oflight intensity. The cutoff wavelength c can be derived from Equation 27.6:KE max hf c 0 : hc hc[27.7]where c is the speed of light. Wavelengths greater than c incident on a materialwith work function don’t result in the emission of photoelectrons.cc 0TABLE 27.1Work Functions ofSelected MetalsMetal (eV)Na 2.46Al 4.08Cu 4.70Zn 4.31Ag 4.73Pt 6.35Pb 4.14Fe 4.50KE maxf c fFigure 27.6 A sketch of KE max versusthe frequency of incident light forphotoelectrons in a typical photoelectriceffect experiment. Photons withfrequency less than f c don’t havesufficient energy to eject an electronfrom the metal.INTERACTIVE EXAMPLE 27.3GoalPhotoelectrons from SodiumUnderstand the quantization of light and its role in the photoelectric effect.Problem A sodium surface is illuminated with light of wavelength 0.300 m. The work function for sodium is 2.46 eV.(a) Calculate the energy of each photon in electron volts, (b) the maximum kinetic energy of the ejected photoelectrons,and (c) the cutoff wavelength for sodium.

880 Chapter 27 Quantum PhysicsStrategyParts (a), (b), and (c) require substitution of values into Equations 27.5, 27.6, and 27.7, respectively.Solution(a) Calculate the energy of each photon.Obtain the frequency from the wavelength:Use Equation 27.5 to calculate the photon’s energy:c f : f cf 1.00 10 15 Hz 6.63 10 19 J (6.63 10 19 J) 3.00 108 m/s0.300 10 6 mE hf (6.63 10 34 J s)(1.00 10 15 Hz)1.00 eV1.60 10 19 J 4.14 eV(b) Find the maximum kinetic energy of thephotoelectrons.Substitute into Equation 27.6:KE max hf 4.14 eV 2.46 eV 1.68 eV(c) Compute the cutoff wavelength.Convert from electron volts to joules:Find the cutoff wavelength using Equation 27.7. 2.46 eV (2.46 eV)(1.60 10 19 J/eV) 3.94 10 19 Jc hc (6.63 1034 Js)(3.00 10 8 m/s)3.94 10 19 J 5.05 10 7 m 505 nmRemarkThe cutoff wavelength is in the yellow-green region of the visible spectrum.Exercise 27.3(a) What minimum-frequency light will eject photoelectrons from a copper surface? (b) If this frequency is tripled,find the maximum kinetic energy (in eV) of the resulting photoelectrons. (Answer in eV.)Answers (a) 1.13 10 15 Hz (b) 9.40 eVInvestigate the photoelectric effect for different materials and different wavelengths of light by logginginto PhysicsNow at www.cp7e.com and going to Interactive Example 27.3.APPLICATIONPhotocellsPhotocellsThe photoelectric effect has many interesting applications using a device calledthe photocell. The photocell shown in Active Figure 27.4 produces a current in thecircuit when light of sufficiently high frequency falls on the cell, but it doesn’tallow a current in the dark. This device is used in streetlights: a photoelectric controlunit in the base of the light activates a switch that turns off the streetlightwhen ambient light strikes it. Many garage-door systems and elevators use a lightbeam and a photocell as a safety feature in their design. When the light beamstrikes the photocell, the electric current generated is sufficiently large to maintaina closed circuit. When an object or a person blocks the light beam, the currentis interrupted, which signals the door to open.27.3 X-RAYSIn 1895 at the University of Wurzburg, Wilhelm Roentgen (1845–1923) was studyingelectrical discharges in low-pressure gases when he noticed that a fluorescentscreen glowed even when placed several meters from the gas discharge tube and

27.3 X-Rays 881even when black cardboard was placed between the tube and the screen. He concludedthat the effect was caused by a mysterious type of radiation, which he calledx-rays because of their unknown nature. Subsequent study showed that these raystraveled at or near the speed of light and that they couldn’t be deflected by eitherelectric or magnetic fields. This last fact indicated that x-rays did not consist ofbeams of charged particles, although the possibility that they were beams ofuncharged particles remained.In 1912 Max von Laue (1879–1960) suggested that if x-rays were electromagneticwaves with very short wavelengths, it should be possible to diffract themby using the regular atomic spacings of a crystal lattice as a diffraction grating,just as visible light is diffracted by a ruled grating. Shortly thereafter, researchersdemonstrated that such a diffraction pattern could be observed, similar to thatshown in Figure 27.7 for NaCl. The wavelengths of the x-rays were then determinedfrom the diffraction data and the known values of the spacing betweenatoms in the crystal. X-ray diffraction has proved to be an invaluable technique forunderstanding the structure of matter (as discussed in more detail in the nextsection).Typical x-ray wavelengths are about 0.1 nm, which is on the order of the atomicspacing in a solid. We now know that x-rays are a part of the electromagnetic spectrum,characterized by frequencies higher than those of ultraviolet radiation andhaving the ability to penetrate most materials with relative ease.X-rays are produced when high-speed electrons are suddenly slowed down—for example, when a metal target is struck by electrons that have been acceleratedthrough a potential difference of several thousand volts. Figure 27.8a showsa schematic diagram of an x-ray tube. A current in the filament causeselectrons to be emitted, and these freed electrons are accelerated toward a densemetal target, such as tungsten, which is held at a higher potential than thefilament.Figure 27.9 represents a plot of x-ray intensity versus wavelength for the spectrumof radiation emitted by an x-ray tube. Note that the spectrum has are twodistinct components. One component is a continuous broad spectrum thatdepends on the voltage applied to the tube. Superimposed on this componentis a series of sharp, intense lines that depend on the nature of the targetmaterial. The accelerating voltage must exceed a certain value, called thethreshold voltage, in order to observe these sharp lines, which representradiation emitted by the target atoms as their electrons undergo rearrangements.We will discuss this further in Chapter 28. The continuous radiation issometimes called bremsstrahlung, a German word meaning “braking radiation,”because electrons emit radiation when they undergo an acceleration inside thetarget.Figure 27.10 (page 882) illustrates how x-rays are produced when an electronpasses near a charged target nucleus. As the electron passes close to a positivelyIntensityK bK aCourtesy of GE Medical SystemsFigure 27.7 X-ray diffractionpattern of NaCl.FilamentvoltageFilamentHigh voltage– +(a)Tungstentargetelectronsx-rays(b)VacuumFigure 27.8 (a) Diagram of anx-ray tube. (b) Photograph of an x-raytube.CopperrodFigure 27.9 The x-ray spectrum of a metal target consists of abroad continuous spectrum plus a number of sharp lines, whichare due to characteristic x-rays. The data shown were obtainedwhen 35-keV electrons bombarded a molybdenum target. Notethat 1 pm 10 12 m 10 3 nm.3040 50 60 70 80 90l, pm

882 Chapter 27 Quantum Physics–IncomingelectronNucleus oftarget atom+EmittedphotonDeflectedlower energyelectronFigure 27.10 An electron passingnear a charged target atom experiencesan acceleration, and a photonis emitted in the process.–hfAPPLICATIONUsing X-Rays to Study theWork of Master Painterscharged nucleus contained in the target material, it is deflected from its path becauseof its electrical attraction to the nucleus; hence, it undergoes an acceleration.An analysis from classical physics shows that any charged particle will emit electromagneticradiation when it is accelerated. (An example of this phenomenon is theproduction of electromagnetic waves by accelerated charges in a radio antenna, asdescribed in Chapter 21.) According to quantum theory, this radiation must appearin the form of photons. Because the radiated photon shown in Figure 27.10 carriesenergy, the electron must lose kinetic energy because of its encounter with thetarget nucleus. An extreme example would consist of the electron losing allof its energy in a single collision. In this case, the initial energy of the electron(eV ) is transformed completely into the energy of the photon (hf max ). Inequation form,eV hf max [27.8]where eV is the energy of the electron after it has been accelerated through apotential difference of V volts and e is the charge on the electron. This says thatthe shortest wavelength radiation that can be produced is min hc[27.9]eVThe reason that not all the radiation produced has this particular wavelength isbecause many of the electrons aren’t stopped in a single collision. This results inthe production of the continuous spectrum of wavelengths.Interesting insights into the process of painting and revising a masterpiece arebeing revealed by x-rays. Long wavelength x-rays are absorbed in varying degreesby some paints, such as those having lead, cadmium, chromium, or cobalt as abase. The x-ray interactions with the paints give contrast, because the differentelements in the paints have different electron densities. Also, thicker layers willabsorb more than thin layers. To examine a painting by an old master, a film isplaced behind it while it is x-rayed from the front. Ghost outlines of earlier paintingsand earlier forms of the final masterpiece are sometimes revealed when thefilm is developed.hc minEXAMPLE 27.4 An X-Ray TubeGoal Calculate the minimum x-ray wavelength due to accelerated electrons.Problem Medical x-ray machines typically operate at a potential difference of 1.00 10 5 V. Calculate the minimumwavelength their x-ray tubes produce when electrons are accelerated through this potential difference.Strategy The minimum wavelength corresponds to the most energetic photons. Substitute the given potential differenceinto Equation 27.9.SolutionSubstitute into Equation 27.9: min hceV (6.63 1034 Js)(3.00 10 8 m/s)(1.60 10 19 C)(1.00 10 5 V) 1.24 10 11 mRemarks X-ray tubes generally operate with half the voltage with respect to Earth, 50 000 V, applied to theanode, and the other half, 50 000 V, applied to the cathode. This lengthens tube lifetime by reducing the probabilityof voltage breakthroughs.

27.4 Diffraction of X-Rays by Crystals 883Exercise 27.4What potential difference would be necessary to produce gamma rays with wavelength 1.00 10 15 m? This wavelengthis about the same size as the diameter of a proton.Solution1.24 10 9 V27.4 DIFFRACTION OF X-RAYS BY CRYSTALSIn Chapter 24 we described how a diffraction grating could be used to measurethe wavelength of light. In principle, the wavelength of any electromagnetic wavecan be measured if a grating having a suitable line spacing can be found. Thespacing between lines must be approximately equal to the wavelength of the radiationto be measured. X-rays are electromagnetic waves with wavelengths on theorder of 0.1 nm. It would be impossible to construct a grating with such a smallspacing. However, as noted in the previous section, Max von Laue suggested thatthe regular array of atoms in a crystal could act as a three-dimensional grating forobserving the diffraction of x-rays.One experimental arrangement for observing x-ray diffraction is shown inFigure 27.11. A narrow beam of x-rays with a continuous wavelength range is incidenton a crystal such as sodium chloride. The diffracted radiation is very intensein certain directions, corresponding to constructive interference from wavesreflected from layers of atoms in the crystal. The diffracted radiation is detected bya photographic film and forms an array of spots known as a Laue pattern. The crystalstructure is determined by analyzing the positions and intensities of the variousspots in the pattern.The arrangement of atoms in a crystal of NaCl is shown in Figure 27.12. Thesmaller red spheres represent Na ions, and the larger blue spheres represent Cl ions. The spacing between successive Na (or Cl ) ions in this cubic structure,denoted by the symbol a in Figure 27.12, is approximately 0.563 nm.A careful examination of the NaCl structure shows that the ions lie in variousplanes. The shaded areas in Figure 27.12 represent one example, in which theatoms lie in equally spaced planes. Now suppose an x-ray beam is incident atgrazing angle on one of the planes, as in Figure 27.13. The beam can bereflected from both the upper and lower plane of atoms. However, thegeometric construction in Figure 27.13 shows that the beam reflected fromthe lower surface travels farther than the beam reflected from the upper surfaceby a distance of 2d sin . The two portions of the reflected beam will combineto produce constructive interference when this path difference equals some integralmultiple of the wavelength . The condition for constructive interference isgiven byX-raytubeX-raysCollimatorCrystalPhotographicfilmFigure 27.11 Schematic diagramof the technique used to observe thediffraction of x-rays by a single crystal.The array of spots formed on the filmby the diffracted beams is called aLaue pattern. (See Fig. 27.7.)Figure 27.12 A model of the cubiccrystalline structure of sodium chloride.The blue spheres represent theCl ions, and the red spheres representthe Na ions. The length of thecube edge is a 0.563 nm.a2d sin m (m 1, 2, 3, . . .)[27.10] Bragg’s lawIncidentbeamReflectedbeamUpper planeLower planeud sin uuudFigure 27.13 A two-dimensionaldepiction of the reflection of an x-raybeam from two parallel crystallineplanes separated by a distance d. Thebeam reflected from the lower planetravels farther than the one reflectedfrom the upper plane by an amountequal to 2d sin .

884 Chapter 27 Quantum PhysicsFigure 27.14 An x-ray diffractionphotograph of DNA taken by RosalindFranklin. The cross pattern of spots wasa clue that DNA has a helical structure.Science Source/Photo Researchers, Inc.This condition is known as Bragg’s law, after W. L. Bragg (1890–1971), whofirst derived the relationship. If the wavelength and diffraction angle aremeasured, Equation 27.10 can be used to calculate the spacing between atomicplanes.The method of x-ray diffraction to determine crystalline structures was thoroughlydeveloped in England by W. H. Bragg and his son W. L. Bragg, who shareda Nobel prize in 1915 for their work. Since then, thousands of crystallinestructures have been investigated. Most importantly, the technique of x-ray diffractionhas been used to determine the atomic arrangement of complex organic moleculessuch as proteins. Proteins are large molecules containing thousands ofatoms that help to regulate and speed up chemical life processes in cells. Someproteins are amazing catalysts, speeding up the slow room temperature reactionsin cells by 17 orders of magnitude. In order to understand this incrediblebiochemical reactivity, it is important to determine the structure of these intricatemolecules.The main technique used to determine the molecular structure of proteins,DNA, and RNA is x-ray diffraction using x-rays of wavelength of about 1.0 A. Thistechnique allows the experimenter to “see” individual atoms that are separated byabout this distance in molecules. Since the biochemical x-ray diffraction sample isprepared in crystal form, the geometry (position of the bright spots in space) of thediffraction pattern is determined by the regular three-dimensional crystal latticearrangement of molecules in the sample. The intensities of the bright diffractionspots are determined by the atoms and their electronic distributions in the fundamentalbuilding block of the crystal: the unit cell. Using complicated computationaltechniques, investigators can essentially deduce the molecular structure bymatching the observed intensities of diffracted beams with a series of assumedatomic positions that determine the atomic structure and electron density of themolecule. Figure 27.14 shows a classic x-ray diffraction image of DNA made byRosalind Franklin in 1952.This and similar x-ray diffraction photos played an important role in thedetermination of the double-helix structure of DNA by F. H. C. Crick andJ. D. Watson in 1953. A model of the famous DNA double helix is shown inFigure 27.15.2 nmFigure 27.15 The double-helixstructure of DNA.EXAMPLE 27.5 X-Ray Diffraction from CalciteGoal Understand Bragg’s law and apply it to a crystal.Problem If the spacing between certain planes in a crystal of calcite (CaCO 3 ) is 0.314 nm, find the grazing angles atwhich first- and third-order interference will occur for x-rays of wavelength 0.070 0 nm.StrategySolve Bragg’s law for sin and substitute, using the inverse-sine function to obtain the angle.SolutionFind the grazing angle corresponding to m 1, for firstorderinterference:sin m (0.070 0 nm)2d 2(0.314 nm) 0.111 sin 1 (0.111) 6.37Repeat the calculation for third-order interference(m 3):sin m 3(0.070 0 nm)2d 2(0.314 nm) sin 1 (0.334) 19.5 0.334RemarkNotice there is little difference between this kind of problem and a Young’s slit experiment.

27.5 The Compton Effect 885Exercise 27.5X-rays of wavelength 0.060 0 nm are scattered from a crystal with a grazing angle of 11.7°. Assume m 1 for thisprocess. Calculate the spacing between the crystal planes.Answer0.148 nm27.5 THE COMPTON EFFECTFurther justification for the photon nature of light came from an experimentconducted by Arthur H. Compton in 1923. In his experiment, Compton directed anx-ray beam of wavelength 0 toward a block of graphite. He found that the scatteredx-rays had a slightly longer wavelength than the incident x-rays, and hence the energiesof the scattered rays were lower. The amount of energy reduction dependedon the angle at which the x-rays were scattered. The change in wavelength between a scattered x-ray and an incident x-ray is called the Compton shift.In order to explain this effect, Compton assumed that if a photon behaves like aparticle, its collision with other particles is similar to a collision between two billiardballs. Hence, the x-ray photon carries both measurable energy and momentum,and these two quantities must be conserved in a collision. If the incident photoncollides with an electron initially at rest, as in Figure 27.16, the photon transferssome of its energy and momentum to the electron. As a consequence, the energyand frequency of the scattered photon are lowered and its wavelength increases.Applying relativistic energy and momentum conservation to the collision describedin Figure 27.16, the shift in wavelength of the scattered photon is given by 0 hm e c (1 cos )[27.11]where m e is the mass of the electron and is the angle between the directions ofthe scattered and incident photons. The quantity h/m e c is called the Comptonwavelength and has a value of 0.002 43 nm. The Compton wavelength is very smallrelative to the wavelengths of visible light, so the shift in wavelength would bedifficult to detect if visible light were used. Further, note that the Compton shiftdepends on the scattering angle and not on the wavelength. Experimentalresults for x-rays scattered from various targets obey Equation 27.11 and stronglysupport the photon concept.Quick Quiz 27.1An x-ray photon is scattered by an electron. The frequency of the scattered photonrelative to that of the incident photon (a) increases, (b) decreases, or (c) remainsthe same.Recoiling electronCourtesy of AIP Niels Bohr Library The Compton shift formulaARTHUR HOLLY COMPTON,American Physicist (1892 – 1962)Compton was born in Wooster, Ohio, andhe attended Wooster College and PrincetonUniversity. He became director of the laboratoryat the University of Chicago, whereexperimental work concerned with sustainedchain reactions was conducted. Thiswork was of central importance to theconstruction of the first atomic bomb. Hisdiscovery of the Compton effect and hiswork with cosmic rays led to his sharing the1927 Nobel Prize in physics with CharlesWilson.φf 0 , λ 0θf ′, λ′ λFigure 27.16 Diagram representingCompton scattering of a photonby an electron. The scattered photonhas less energy (or a longer wavelength)than the incident photon.

886 Chapter 27 Quantum PhysicsQuick Quiz 27.2A photon of energy E 0 strikes a free electron, with the scattered photon of energyE moving in the direction opposite that of the incident photon. In this Comptoneffect interaction, the resulting kinetic energy of the electron is(a) E 0 (b) E (c) E 0 E (d) E 0 E (e) None of the aboveApplying Physics 27.2The Compton effect involves a change in wavelengthas photons are scattered through different angles.Suppose we illuminate a piece of material with a beamof light and then view the material from different anglesrelative to the beam of light. Will we see a colorchange corresponding to the change in wavelength ofthe scattered light?Explanation There will be a wavelength change for visiblelight scattered by the material, but the change willColor Changes through the Compton Effectbe far too small to detect as a color change. The largestpossible wavelength change, at 180° scattering, will betwice the Compton wavelength, about 0.005 nm. Thisrepresents a change of less than 0.001% of the wavelengthof red light. The Compton effect is only detectablefor wavelengths that are very short to beginwith, so that the Compton wavelength is an appreciablefraction of the incident wavelength. As a result, theusual radiation for observing the Compton effect is inthe x-ray range of the electromagnetic spectrum.INTERACTIVE EXAMPLE 27.6GoalScattering X-RaysUnderstand Compton scattering and its effect on the photon’s energy.Problem X-rays of wavelength 0 0.200 000 nm are scattered from a block of material. The scattered x-rays areobserved at an angle of 45.0° to the incident beam. (a) Calculate the wavelength of the x-rays scattered at this angle.(b) Compute the fractional change in the energy of a photon in the collision.Solution(a) Calculate the wavelength of the x-rays.Substitute into Equation 27.11 to obtain the wavelengthshift:Add this shift to the original wavelength to obtain thewavelength of the scattered photon: hm e c (1 cos )6.63 10 34 Js(9.11 10 31 kg)(3.00 10 8 m/s) 7.11 10 13 m 0.000 711 nm 0 0.200 711 nm(1 cos 45.0)(b) Find the fraction of energy lost by the photon in thecollision.Rewrite the energy E in terms of wavelength, using c f :E hf hcCompute E/E using this expression:EE E f E iE i hc/f hc/ihc/iCancel hc and rearrange terms:EE 1/f 1/i1/iif 1 i ff fSubstitute values from part (a):EE711 nm0.0000.200 711 nm 3.54 10 3Remarks It is also possible to find this answer by substituting into the energy expression at an earlier stage, but thealgebraic derivation is more elegant and instructive.

27.6 The Dual Nature of Light and Matter 887Exercise 27.6Repeat the exercise for a photon with wavelength 3.00 10 2 nm that scatters at an angle of 60.0°.Answers (a) 3.12 10 2 nm (b) E/ E 3.88 10 2Study Compton scattering for different angles by logging into PhysicsNow at www.cp7e.com andgoing to Interactive Example 27.6.27.6 THE DUAL NATURE OF LIGHT AND MATTERLight and Electromagnetic RadiationPhenomena such as the photoelectric effect and the Compton effect offer evidencethat when light (or other forms of electromagnetic radiation) and matterinteract, the light behaves as if it were composed of particles having energy hf andmomentum h/. In other contexts, however, light acts like a wave, exhibiting interferenceand diffraction effects. Is light a wave or a particle?The answer depends on the phenomenon being observed. Some experimentscan be better explained with the photon concept, whereas others are bestdescribed with a wave model. The end result is that both models are needed.Light has a dual nature, exhibiting both wave and particle characteristics.To understand why photons are compatible with electromagnetic waves, consider2.5-MHz radio waves as an example. The energy of a photon having thisfrequency is only about 10 8 eV, too small to allow the photon to be detected. Asensitive radio receiver might require as many as 10 10 of these photons to producea detectable signal. Such a large number of photons would appear, on the average,as a continuous wave. With so many photons reaching the detector every second,we wouldn’t be able to detect the individual photons striking the antenna.Now consider what happens as we go to higher frequencies. In the visibleregion, it’s possible to observe both the particle characteristics and the wave characteristicsof light. As we mentioned earlier, a light beam shows interference phenomena(thus, it is a wave) and at the same time can produce photoelectrons(thus, it is a particle). At even higher frequencies, the momentum and energy ofthe photons increase. Consequently, the particle nature of light becomes moreevident than its wave nature. For example, the absorption of an x-ray photon iseasily detected as a single event, but wave effects are difficult to observe.The Wave Properties of ParticlesIn his doctoral dissertation in 1924, Louis de Broglie postulated that, becausephotons have wave and particle characteristics, perhaps all forms of matter haveboth properties. This was a highly revolutionary idea with no experimental confirmationat that time. According to de Broglie, electrons, just like light, have a dualparticle–wave nature.In Chapter 26 we found that the relationship between energy and momentumfor a photon, which has a rest energy of zero, is p E/c. We also know from Equation27.5 that the energy of a photon isAIP Niels Bohr LibraryLOUIS DE BROGLIE, FrenchPhysicist, (1892 – 1987)De Broglie was born in Dieppe, France. Atthe Sorbonne in Paris, he studied history inpreparation for what he hoped to be acareer in the diplomatic service. The worldof science is lucky that he changed hiscareer path to become a theoretical physicist.De Broglie was awarded the NobelPrize in 1929 for his discovery of the wavenature of electrons.E hf hc[27.12]Consequently, the momentum of a photon can be expressed asp E c hc[27.13]From this equation, we see that the photon wavelength can be specified by itsmomentum, or h/p. De Broglie suggested that all material particles withmomentum p should have a characteristic wavelength h/p. Because thec h Momentum of a photon

888 Chapter 27 Quantum Physicsmomentum of a particle of mass m and speed v is mv p, the de Broglie wavelengthof a particle isde Broglie’s hypothesis h p hmv[27.14]Further, de Broglie postulated that the frequencies of matter waves (waves associatedwith particles having nonzero rest energy) obey the Einstein relationshipfor photons, E hf, so thatFrequency of matter waves f E h[27.15]The dual nature of matter is quite apparent in Equations 27.14 and 27.15, becauseeach contains both particle concepts (mv and E ) and wave concepts ( and f ).The fact that these relationships had been established experimentally for photonsmade the de Broglie hypothesis that much easier to accept.The Davisson–Germer ExperimentDe Broglie’s proposal in 1923 that matter exhibits both wave and particle propertieswas first regarded as pure speculation. If particles such as electrons had wavelikeproperties, then, under the correct conditions, they should exhibit diffractioneffects. In 1927, three years after de Broglie published his work, C. J. Davisson(1881–1958) and L. H. Germer (1896–1971) of the United States succeeded inmeasuring the wavelength of electrons. Their important discovery provided thefirst experimental confirmation of the matter waves proposed by de Broglie.The intent of the initial Davisson–Germer experiment was not to confirm the deBroglie hypothesis. In fact, their discovery was made by accident (as is often thecase). The experiment involved the scattering of low-energy electrons (about 54 eV)from a nickel target in a vacuum. During one experiment, the nickel surface wasbadly oxidized because of an accidental break in the vacuum system. After the nickeltarget was heated in a flowing stream of hydrogen to remove the oxide coating, electronsscattered by it exhibited intensity maxima and minima at specific angles. Theexperimenters finally realized that the nickel had formed large crystalline regionsupon heating and that the regularly spaced planes of atoms in the crystalline regionsserved as a diffraction grating for electron matter waves. (See Section 27.5.)Shortly thereafter, Davisson and Germer performed more extensive diffractionmeasurements on electrons scattered from single-crystal targets. Their resultsshowed conclusively the wave nature of electrons and confirmed the de Broglierelation h/p. In the same year, G. P. Thomson (1892–1975) of Scotland alsoobserved electron diffraction patterns by passing electrons through very thin goldfoils. Diffraction patterns have since been observed for helium atoms, hydrogenatoms, and neutrons. Hence, the universal nature of matter waves has been establishedin various ways.Quick Quiz 27.3A nonrelativistic electron and a nonrelativistic proton are moving and have thesame de Broglie wavelength. Which of the following are also the same for the twoparticles?(a) speed (b) kinetic energy (c) momentum (d) frequencyQuick Quiz 27.4We have seen two wavelengths assigned to the electron: the Compton wavelengthand the de Broglie wavelength. Which is an actual physical wavelength associatedwith the electron? (a) the Compton wavelength (b) the de Broglie wavelength(c) both wavelengths (d) neither wavelength

27.6 The Dual Nature of Light and Matter 889EXAMPLE 27.7GoalThe Electron versus the BaseballApply the de Broglie hypothesis to a quantum and a classical object.Problem (a) Compare the de Broglie wavelength for an electron (m e 9.11 10 31 kg) moving at a speed of1.00 10 7 m/s with that of a baseball of mass 0.145 kg pitched at 45.0 m/s. (b) Compare these wavelengths with thatof an electron traveling at 0.999c.Strategy This is a matter of substitution into Equation 27.14 for the de Broglie wavelength. In part (b), the relativisticmomentum must be used.Solution(a) Compare the de Broglie wavelengths of theelectron and the baseball.Substitute data for the electron into Equation 27.14:Repeat the calculation with the baseball data:(b) Find the wavelength for an electron traveling at0.999c.Replace the momentum in Equation 27.14 with the relativisticmomentum:Substitute:e hm e v 7.28 10 11 mb hm b v e 6.63 10 34 Js(9.11 10 31 kg)(1.00 10 7 m/s)6.63 1034 Js(0.145 kg)(45.0 m/s) hm e v/√1 v 2 /c h √1 v 2 /c 22 m e v1.02 10 34 m(6.63 10 34 Js)√1 (0.999c) 2 /c 2(9.11 10 31 kg)(0.9993.00 10 8 m/s) 1.09 10 13 me Remarks The electron wavelength corresponds to that of x-rays in the electromagnetic spectrum. The baseball,by contrast, has a wavelength much smaller than any aperture through which the baseball could possibly pass,so we couldn’t observe any of its diffraction effects. It is generally true that the wave properties of large-scale objectscan’t be observed. Notice that even at extreme relativistic speeds, the electron wavelength is still far larger than thebaseball’s.Exercise 27.7Find the de Broglie wavelength of a proton (m p 1.67 10 27 kg) moving with a speed of 1.00 10 7 m/s.Answer3.97 10 14 mApplication: The Electron MicroscopeA practical device that relies on the wave characteristics of electrons is the electronmicroscope. A transmission electron microscope, used for viewing flat, thin samples,is shown in Figure 27.17 (page 890). In many respects, it is similar to an opticalmicroscope, but the electron microscope has a much greater resolving powerbecause it can accelerate electrons to very high kinetic energies, giving them veryshort wavelengths. No microscope can resolve details that are significantly smallerthan the wavelength of the radiation used to illuminate the object. Typically, thewavelengths of electrons are about 100 times smaller than those of the visible lightused in optical microscopes. (Radiation of the same wavelength as the electrons inan electron microscope is in the x-ray region of the spectrum.)APPLICATIONElectron Microscopes

890 Chapter 27 Quantum PhysicsElectron gunCathodeVacuumAnodeElectromagneticlensElectromagneticcondenserlensCoreCoilElectronbeamSpecimengoeshereSpecimenchamberdoorScreenVisualtransmission(a)ProjectorlensPhotochamberFigure 27.17 (a) Diagram of a transmission electron microscope for viewing a thin sectioned sample.The “lenses” that control the electron beam are magnetic deflection coils. (b) An electron microscope.(b)© David Parker/Photo Researchers, Inc.The electron beam in an electron microscope is controlled by electrostatic ormagnetic deflection, which acts on the electrons to focus the beam to an image.Rather than examining the image through an eyepiece as in an optical microscope,the viewer looks at an image formed on a fluorescent screen. (The viewing screenmust be fluorescent because otherwise the image produced wouldn’t be visible.)Applying Physics 27.3X-Ray Microscopes?Electron microscopes (Fig. 27.17) take advantageof the wave nature of particles. Electrons are acceleratedto high speeds, giving them a short de Brogliewavelength. Imagine an electron microscope usingelectrons with a de Broglie wavelength of 0.2 nm.Why don’t we design a microscope using 0.2-nmphotons to do the same thing?Explanation Because electrons are charged particles,they interact electrically with the sample in the microscopeand scatter according to the shape and densityof various portions of the sample, providing a meansof viewing the sample. Photons of wavelength 0.2 nmare uncharged and in the x-ray region of the spectrum.They tend to simply pass through the thin samplewithout interacting.27.7 THE WAVE FUNCTIONDe Broglie’s revolutionary idea that particles should have a wave nature soonmoved out of the realm of skepticism to the point where it was viewed as a necessaryconcept in understanding the subatomic world. In 1926, the Austrian–Germanphysicist Erwin Schrödinger proposed a wave equation that described how matter

27.8 The Uncertainty Principle 891waves change in space and time. The Schrödinger wave equation represents a keyelement in the theory of quantum mechanics. It’s as important in quantum mechanicsas Newton’s laws in classical mechanics. Schrödinger’s equation has beensuccessfully applied to the hydrogen atom and to many other microscopic systems.Solving Schrödinger’s equation (beyond the level of this course) determines aquantity called the wave function. Each particle is represented by a wave function that depends both on position and on time. Once is found, 2 gives usinformation on the probability (per unit volume) of finding the particle in anygiven region. To understand this, we return to Young’s experiment involving coherentlight passing through a double slit.First, recall from Chapter 21 that the intensity of a light beam is proportional tothe square of the electric field strength E associated with the beam: I E 2 . Accordingto the wave model of light, there are certain points on the viewing screenwhere the net electric field is zero as a result of destructive interference of wavesfrom the two slits. Because E is zero at these points, the intensity is also zero, andthe screen is dark there. Likewise, at points on the screen at which constructiveinterference occurs, E is large, as is the intensity; hence, these locations are bright.Now consider the same experiment when light is viewed as having a particle nature.The number of photons reaching a point on the screen per second increasesas the intensity (brightness) increases. Consequently, the number of photons thatstrike a unit area on the screen each second is proportional to the square of theelectric field, or N E 2 . From a probabilistic point of view, a photon has a highprobability of striking the screen at a point at which the intensity (and E 2 ) is highand a low probability of striking the screen where the intensity is low.When describing particles rather than photons, rather than E plays the roleof the amplitude. Using an analogy with the description of light, we make the followinginterpretation of for particles: If is a wave function used to describe asingle particle, the value of 2 at some location at a given time is proportional tothe probability per unit volume of finding the particle at that location at that time.Adding up all the values of 2 in a given region gives the probability of finding theparticle in that region.AIP Emilio Segré Visual ArchivesERWIN SCHRÖDINGER, AustrianTheoretical Physicist (1887–1961)Schrödinger is best known as the creator ofwave mechanics, a less cumbersome theorythan the equivalent matrix mechanicsdeveloped by Werner Heisenberg. In 1933Schrödinger left Germany and eventuallysettled at the Dublin Institute of AdvancedStudy, where he spent 17 happy, creativeyears working on problems in general relativity,cosmology, and the application ofquantum physics to biology. In 1956, he returnedhome to Austria and his belovedTirolean mountains, where he died in 1961.27.8 THE UNCERTAINTY PRINCIPLEIf you were to measure the position and speed of a particle at any instant, you wouldalways be faced with experimental uncertainties in your measurements. According toclassical mechanics, no fundamental barrier to an ultimate refinement of the apparatusor experimental procedures exists. In other words, it’s possible, in principle, tomake such measurements with arbitrarily small uncertainty. Quantum theory predicts,however, that such a barrier does exist. In 1927, Werner Heisenberg (1901–1976) introducedthis notion, which is now known as the uncertainty principle:If a measurement of the position of a particle is made with precision x and asimultaneous measurement of linear momentum is made with precision p x ,then the product of the two uncertainties can never be smaller than h/4:x p x [27.16]In other words, it is physically impossible to measure simultaneously the exactposition and exact linear momentum of a particle. If x is very small, then p x islarge, and vice versa.To understand the physical origin of the uncertainty principle, consider the followingthought experiment introduced by Heisenberg. Suppose you wish to measurethe position and linear momentum of an electron as accurately as possible. Youmight be able to do this by viewing the electron with a powerful light microscope.For you to see the electron and determine its location, at least one photon of lightmust bounce off the electron, as shown in Figure 27.18a, and pass through theh4Courtesy of the University of HamburgWERNER HEISENBERG, GermanTheoretical Physicist (1901 – 1976)Heisenberg obtained his Ph.D. in 1923 at theUniversity of Munich, where he studied underArnold Sommerfeld. While physicists such asde Broglie and Schrödinger tried to developphysical models of the atom, Heisenbergdeveloped an abstract mathematical modelcalled matrix mechanics to explain the wavelengthsof spectral lines. Heisenberg mademany other significant contributions tophysics, including his famous uncertaintyprinciple, for which he received the NobelPrize in 1932; the prediction of two forms ofmolecular hydrogen; and theoretical modelsof the nucleus of an atom.

892 Chapter 27 Quantum PhysicsFigure 27.18 A thought experimentfor viewing an electron with apowerful microscope. (a) The electronis viewed before colliding with thephoton. (b) The electron recoils(is disturbed) as the result of thecollision with the photon.IncidentphotonBeforecollisionScatteredphotonAftercollisionElectronRecoilingelectron(a)(b)microscope into your eye, as shown in Figure 27.18b. When it strikes the electron,however, the photon transfers some unknown amount of its momentum to the electron.Thus, in the process of locating the electron very accurately (that is, by makingx very small), the light that enables you to succeed in your measurement changesthe electron’s momentum to some undeterminable extent (making p x very large).The incoming photon has momentum h/. As a result of the collision, the photontransfers part or all of its momentum along the x-axis to the electron. Therefore,the uncertainty in the electron’s momentum after the collision is as great asthe momentum of the incoming photon: p x h/. Further, because the photonalso has wave properties, we expect to be able to determine the electron’s positionto within one wavelength of the light being used to view it, so x . Multiplyingthese two uncertainties givesx p x hThe value h represents the minimum in the product of the uncertainties. Becausethe uncertainty can always be greater than this minimum, we havex p x hApart from the numerical factor 1/4 introduced by Heisenberg’s more preciseanalysis, this inequality agrees with Equation 27.16.Another form of the uncertainty relationship sets a limit on the accuracy withwhich the energy E of a system can be measured in a finite time interval t :E t h[27.17]It can be inferred from this relationship that the energy of a particle cannot bemeasured with complete precision in a very short interval of time. Thus, when anelectron is viewed as a particle, the uncertainty principle tells us that (a) its positionand velocity cannot both be known precisely at the same time and (b) itsenergy can be uncertain for a period given by t h/(4 E ).h4Applying Physics 27.4A common, but erroneous, description of the absolutezero of temperature is “that temperature at which allmolecular motion ceases.” How can the uncertaintyprinciple be used to argue against this description?Motion at Absolute ZeroExplanation Imagine a particular molecule in a piece ofmaterial. The molecule is confined within the material,so there is a fixed uncertainty x in its position alongone axis, corresponding to the size of that piece of material.If all molecular motion ceased at absolute zero, thegiven molecule’s velocity, in particular, would be exactlyzero, so its uncertainty in velocity would be v 0,meaning its uncertainty in momentum would also bezero, since p mv. The product of zero uncertainty inmomentum and a nonzero uncertainty in position iszero, violating the uncertainty principle. So according tothe uncertainty principle, there must be some molecularmotion even at absolute zero.

27.8 The Uncertainty Principle 893EXAMPLE 27.8 Locating an ElectronGoal Apply Heisenberg’s position–momentum uncertainty principle.Problem The speed of an electron is measured to be 5.00 10 3 m/s to an accuracy of 0.003 00%. Find the minimumuncertainty in determining the position of this electron.Strategy After computing the momentum and its uncertainty, substitute into Heisenberg’s uncertainty principle,Equation 27.16.SolutionCalculate the momentum of the electron:The uncertainty in p is 0.003 00% of this value:Now calculate the uncertainty in position using thisvalue of p x and Equation 27.17:p x m e v (9.11 10 31 kg)(5.00 10 3 m/s) 4.56 10 27 kg m/sp x 0.000 030 0p (0.000 030 0)(4.56 10 27 kg m/s) 1.37 10 31 kg m/sxp x x h4p x6.626 10 34 Js4(1.37 10 31 kgm/s) 0.384 103 m4: x h 0.384 mmRemarks Notice that this isn’t an exact calculation: the uncertainty in position can take any value, as long as it’sgreater than or equal to the value given by the uncertainty principle.Exercise 27.8Suppose an electron is found somewhere in an atom of diameter 1.25 10 10 m. Estimate the uncertainty in theelectron’s momentum (in one dimension).Answerp 4.22 10 25 kg m/sEXAMPLE 27.9GoalExcited States of AtomsApply the energy–time form of the uncertainty relation.Problem As we’ll see in the next chapter, electrons in atoms can be found in certain high states of energy calledexcited states for short periods of time. If the average time that an electron exists in one of these states is 1.00 10 8 s,what is the minimum uncertainty in energy of the excited state?StrategySubstitute values into Equation 27.17, the energy–time form of Heisenberg’s uncertainty relation.SolutionUse Equation 27.17 to obtain the minimum uncertaintyin the energy:E h4t 6.63 1034 Js4(1.00 10 8 s) 5.28 1027 J 3.30 10 8 eVRemarksThis is again an imprecise calculation, giving only a lower bound on the uncertainty.Exercise 27.9A muon may be considered to be an excited state of an electron, to which it decays in an average of 2.2 10 6 s.What’s the minimum uncertainty in the muon’s (rest) energy, according to the uncertainty principle?Answer2.40 10 29 J

894 Chapter 27 Quantum PhysicsAPPLICATIONScanning TunnelingMicroscopes27.9 THE SCANNING TUNNELING MICROSCOPE 1One of the basic phenomena of quantum mechanics—tunneling—is at the heartof a very practical device—the scanning tunneling microscope, or STM—whichenables us to get highly detailed images of surfaces with a resolution comparableto the size of a single atom.Figure 27.19 shows an image of a ring of 48 iron atoms located on a coppersurface. Note the high quality of the STM image and the recognizable ring of ironatoms. What makes this image so remarkable is that its resolution—the size of thesmallest detail that can be discerned—is about 0.2 nm. For an ordinarymicroscope, the resolution is limited by the wavelength of the waves used tomake the image. An optical microscope has a resolution no better than 200 nm,about half the wavelength of visible light, and so could never show thedetail displayed in Figure 27.19. Electron microscopes can have a resolutionof 0.2 nm by using electron waves of that wavelength, given by the de Broglieformula h/p. The electron momentum p required to give this wavelengthis 10 000 eV/c, corresponding to an electron speed of 2% of the speed oflight. Electrons traveling at this speed would penetrate into the interior of thesample in Figure 27.20 and so could not give us information about individualsurface atoms.The STM achieves its very fine resolution by using the basic idea shown inFigure 27.20. A conducting probe with a sharp tip is brought near the surface tobe studied. Because it is attracted to the positive ions in the surface, an electron inthe surface has a lower total energy than an electron in the empty space betweensurface and tip. The same thing is true for an electron in the probe tip, which isattracted to the positive ions in the tip. In Newtonian mechanics, this means thatelectrons cannot move between surface and tip because they lack the energy toescape either material. Because the electrons obey quantum mechanics, however,they can “tunnel” across the barrier of empty space. By applying a voltagebetween surface and tip, the electrons can be made to tunnel preferentially fromsurface to tip. In this way, the tip samples the distribution of electrons just abovethe surface.Because of the nature of tunneling, the STM is very sensitive to the distance zfrom tip to surface. The reason is that in the empty space between tip and surface,the electron wave function falls off exponentially with a decay length on the orderof 0.1 nm; that is, the wave function decreases by 1/e over that distance. For dis-Figure 27.19 This is a photographof a “quantum corral” consisting of aring of 48 iron atoms located on acopper surface. The diameter of thering is 143 nm. The photograph wasobtained with a low-temperaturescanning tunneling microscope(STM).IBM Corporation Research Division1 This section was written by Roger A. Freedman, University of California, Santa Barbara.

27.9 The Scanning Tunneling Microscope 895x piezoz piezoIy piezoBased on a drawing from P. K. Hansma, V. B. Elings, O. Marti, andC. Bracker, Science 242:209, 1988, Copyright 1988 by the AAAS.Figure 27.20 A schematic view of an STM.The tip, shown as a rounded cone, is mountedon a piezoelectric x, y, z scanner. A scan of thetip over the sample can reveal contours of thesurface down to the atomic level. An STM imageis composed of a series of scans displacedlaterally from each other.tances z greater than 1 nm (that is, beyond a few atomic diameters), essentially notunneling takes place. This exponential behavior causes the current of electronstunneling from surface to tip to depend very strongly on z. This sensitivity is thebasis of the operation of the STM: by monitoring the tunneling current as the tipis scanned over the surface, scientists obtain a sensitive measure of the topographyof the electron distribution on the surface. The result of this scan is used tomake images like that in Figure 27.20. In this way the STM can measure theheight of surface features to within 0.001 nm, approximately 1/100 of an atomicdiameter!The STM has, however, one serious limitation: it depends on electrical conductivityof the sample and the tip. Unfortunately, the surfaces of most materials arenot electrically conductive. Even metals such as aluminum are covered with nonconductiveoxides. A newer microscope—the atomic force microscope, or AFM—overcomes this limitation. It measures the force between a tip and the sample,rather than an electrical current. This force depends strongly on the tip–sampleseparation just as the electron tunneling current does for the STM. The AFM hascomparable sensitivity for measuring topography and has become widely used fortechnological applications.Perhaps the most remarkable thing about the STM is that its operation is basedon a quantum mechanical phenomenon—tunneling—that was well understoodin the 1920s, even though the first STM was not built until the 1980s. What otherapplications of quantum mechanics may yet be waiting to be discovered?

896 Chapter 27 Quantum PhysicsSUMMARYTake a practice test by logging intoPhysicsNow at www.cp7e.com and clicking on the Pre-Testlink for this chapter.27.1 Blackbody Radiation and Planck’sHypothesisThe characteristics of blackbody radiation can’t beexplained with classical concepts. The peak of a blackbodyradiation curve is given by Wien’s displacementlaw; max T 0.289 8 10 2 m K [27.1]where max is the wavelength at which the curve peaksand T is the absolute temperature of the object emittingthe radiation.Planck first introduced the quantum concept whenhe assumed that the subatomic oscillators responsiblefor blackbody radiation could have only discreteamounts of energy given byE n nhf [27.2]where n is a positive integer called a quantum numberand f is the frequency of vibration of the resonator.27.2 The Photoelectric Effect and theParticle Theory of LightThe photoelectric effect is a process whereby electronsare ejected from a metal surface when light is incidenton that surface. Einstein provided a successful explanationof this effect by extending Planck’s quantum hypothesisto electromagnetic waves. In this model, light isviewed as a stream of particles called photons, each withenergy E hf, where f is the light frequency and h isPlanck’s constant. The maximum kinetic energy of theejected photoelectrons isKE max hf [27.6]where is the work function of the metal.27.3 X-Rays27.4 Diffraction of X-Rays byCrystalsX-rays are produced when high-speed electrons aresuddenly decelerated. When electrons have been acceleratedthrough a voltage V, the shortest-wavelength radiationthat can be produced is[27.9]eVThe regular array of atoms in a crystal can act as a diffractiongrating for x-rays and for electrons. The conditionfor constructive interference of the diffracted raysis given by Bragg’s law:2d sin m (m 1, 2, 3, . . .) [27.10]Bragg’s law bears a similarity to the equation for the diffractionpattern of a double slit.27.5 The Compton EffectX-rays from an incident beam are scattered at variousangles by electrons in a target such as carbon. In such ascattering event, a shift in wavelength is observed for thescattered x-rays. This phenomenon is known as theCompton shift. Conservation of momentum and energyapplied to a photon–electron collision yields thefollowing expression for the shift in wavelength of thescattered x-rays: min hc0 hm e c[27.11]Here, m e is the mass of the electron, c is the speed oflight, and is the scattering angle.27.6 The Dual Nature of Lightand MatterLight exhibits both a particle and a wave nature. DeBroglie proposed that all matter has both a particle anda wave nature. The de Broglie wavelength of any particleof mass m and speed v is h p h mv[27.14]De Broglie also proposed that the frequencies of thewaves associated with particles obey the Einstein relationshipE hf.27.7 The Wave Function(1 cos)In the theory of quantum mechanics, each particle isdescribed by a quantity called the wave function.

Problems 897The probability per unit volume of finding the particleat a particular point at some instant is proportional to 2 . Quantum mechanics has been highly successfulin describing the behavior of atomic and molecularsystems.27.8 The Uncertainty PrincipleAccording to Heisenberg’s uncertainty principle, it isimpossible to measure simultaneously the exact positionand exact momentum of a particle. If x is theuncertainty in the measured position and p x theuncertainty in the momentum, the product x p x isgiven byAlso,E t [27.17]where E is the uncertainty in the energy of the particleand t is the uncertainty in the time it takes to measurethe energy.h4x p x h4[27.16]CONCEPTUAL QUESTIONS1. If you observe objects inside a very hot kiln why is itdifficult to discern the shapes of the objects?2. Why is an electron microscope more suitable thanan optical microscope for “seeing” objects of atomicsize?3. Are blackbodies black?4. Why is it impossible to simultaneously measure theposition and velocity of a particle with infinite accuracy?5. All objects radiate energy. Why, then, are we not ableto see all objects in a dark room?6. Is light a wave or a particle? Support your answer by citingspecific experimental evidence.7. A student claims that he is going to eject electronsfrom a piece of metal by placing a radio transmitterantenna adjacent to the metal and sending a strongAM radio signal into the antenna. The work functionof a metal is typically a few electron volts. Will thiswork?8. Light acts sometimes like a wave and sometimes like aparticle. For the following situations, which best describesthe behavior of light? Defend your answers.(a) The photoelectric effect. (b) The Compton effect.(c) Young’s double-slit experiment.9. In the photoelectric effect, explain why the stoppingpotential depends on the frequency of the light butnot on the intensity.10. Which has more energy, a photon of ultraviolet radiationor a photon of yellow light?11. Why does the existence of a cutoff frequency in thephotoelectric effect favor a particle theory of lightrather than a wave theory?12. What effect, if any, would you expect the temperatureof a material to have on the ease with whichelectrons can be ejected from it via the photoelectriceffect?13. The cutoff frequency of a material is f 0 . Are electronsemitted from the material when (a) light of frequencygreater than f 0 is incident on the material? (b) Less thanf 0 ?14. The brightest star in the constellation Lyra is thebluish star Vega, whereas the brightest star in Boötes isthe reddish star Arcturus. How do you account for thedifference in color of the two stars?15. If the photoelectric effect is observed in one metal,can you conclude that the effect will also be observedin another metal under the same conditions?Explain.16. A beam of blue light and a beam of red light carry thesame total amount of energy. Which beam contains thelarger number of photons?17. An x-ray photon is scattered by an electron which isinitially at rest. What happens to the frequency ofthe scattered photon relative to that of the incidentphoton?

898 Chapter 27 Quantum PhysicsPROBLEMS1, 2, 3 straightforward, intermediate, challenging full solution available in Student Solutions Manual/Study Guide coached problem with hints available at www.cp7e.com biomedical applicationSection 27.1 Blackbody Radiation and Planck’sHypothesis1.What is the surface temperature of(a) Betelgeuse, a red giant star in the constellation ofOrion, which radiates with a peak wavelength of about970 nm? (b) Rigel, a bluish-white star in Orion, radiateswith a peak wavelength of 145 nm? Find the temperatureof Rigel’s surface.7.2. (a) Lightning produces a maximum air temperatureon the order of 10 4 K, while (b) a nuclear explosionproduces a temperature on the order of 10 7 K. UseWien’s displacement law to find the order of magnitudeof the wavelength of the thermally produced photonsradiated with greatest intensity by each of thesesources. Name the part of the electromagnetic spectrumwhere you would expect each to radiate moststrongly.3. If the surface temperature of the Sun is 5 800 K, findthe wavelength that corresponds to the maximum rateof energy emission from the Sun.4. A beam of blue light and a beam of red light eachcarry a total energy of 2 500 eV. If the wavelength ofthe red light is 690 nm and the wavelength of theblue light is 420 nm, find the number of photons ineach beam.5. Calculate the energy in electron volts of a photon havinga wavelength (a) in the microwave range, 5.00 cm,(b) in the visible light range, 500 nm, and (c) in thex-ray range, 5.00 nm.6. A certain light source is found to emit radiation whosepeak value has a frequency of 1.00 10 15 Hz. Find thetemperature of the source assuming that it is a blackbodyradiator.An FM radio transmitter has a power output of150 kW and operates at a frequency of 99.7 MHz.How many photons per second does the transmitteremit?8. The threshold of dark-adapted (scotopic) vision is4.0 10 11 W/m 2 at a central wavelength of 500 nm.If light with this intensity and wavelength enters theeye when the pupil is open to its maximum diameterof 8.5 mm, how many photons per second enter theeye?9. A 1.5-kg mass vibrates at an amplitude of 3.0 cm on theend of a spring of spring constant 20.0 N/m. (a) If theenergy of the spring is quantized, find its quantum num-ber. (b) If n changes by 1, find the fractional change inenergy of the spring.10. A 70.0-kg jungle hero swings at the end of a vine at afrequency of 0.50 Hz at 2.0 m/s as he moves throughthe lowest point on his arc. (a) Assume the energyis quantized and find the quantum number n for thissystem. (b) Find the energy carried away in a onequantumchange in the jungle hero’s energy.Section 27.2 The Photoelectric Effect andthe Particle Theory of Light11.When light of wavelength 350 nmfalls on a potassium surface, electrons having a maximumkinetic energy of 1.31 eV are emitted. Find (a) thework function of potassium, (b) the cutoff wavelength,and (c) the frequency corresponding to the cutoffwavelength.12. When a certain metal is illuminated with light of frequency3.0 10 15 Hz, a stopping potential of 7.0 V isrequired to stop the most energetic ejected electrons.What is the work function of this metal?13. What wavelength of light would have to fall on sodium(with a work function of 2.46 eV) if it is to emit electronswith a maximum speed of 1.0 10 6 m/s?14. Lithium, beryllium, and mercury have work functionsof 2.30 eV, 3.90 eV, and 4.50 eV, respectively. If 400-nmlight is incident on each of these metals, determine(a) which metals exhibit the photoelectric effect and(b) the maximum kinetic energy of the photoelectronsin each case.15. From the scattering of sunlight, Thomson calculatedthat the classical radius of the electron has a valueof 2.82 10 15 m. If sunlight having an intensityof 500 W/m 2 falls on a disk with this radius, estimatethe time required to accumulate 1.00 eV of energy.Assume that light is a classical wave and that thelight striking the disk is completely absorbed. Howdoes your estimate compare with the observationthat photoelectrons are promptly (within 10 9 s)emitted?16. An isolated copper sphere of radius 5.00 cm, initiallyuncharged, is illuminated by ultraviolet light of wavelength200 nm. What charge will the photoelectric effectinduce on the sphere? The work function for copperis 4.70 eV.

Problems 89917. When light of wavelength 254 nm falls on cesium,the required stopping potential is 3.00 V. If light ofwavelength 436 nm is used, the stopping potentialis 0.900 V. Use this information to plot a graph likethat shown in Figure 27.6, and from the graph determinethe cutoff frequency for cesium and its workfunction.18. Ultraviolet light is incident normally on the surface ofa certain substance. The binding energy of theelectrons in this substance is 3.44 eV. The incidentlight has an intensity of 0.055 W/m 2 . The electronsare photoelectrically emitted with a maximum speed of4.2 10 5 m/s. How many electrons are emitted froma square centimeter of the surface each second?Assume that the absorption of every photon ejects anelectron.Section 27.3 X-Rays19. The extremes of the x-ray portion of the electromagneticspectrum range from approximately 1.0 10 8m to 1.0 10 13 m. Find the minimum acceleratingvoltages required to produce wavelengths at thesetwo extremes.20. Calculate the minimum-wavelength x-ray that can beproduced when a target is struck by an electron thathas been accelerated through a potential difference of(a) 15.0 kV and (b) 100 kV.21. What minimum accelerating voltage would berequired to produce an x-ray with a wavelength of0.030 0 nm?Section 27.4 Diffraction of X-Rays byCrystals22. A monochromatic x-ray beam is incident on a NaClcrystal surface with d 0.353 nm. The second-ordermaximum in the reflected beam is found when theangle between the incident beam and the surface is20.5°. Determine the wavelength of the x-rays.23. Potassium iodide has an interplanar spacing ofd 0.296 nm. A monochromatic x-ray beam shows afirst-order diffraction maximum when the grazingangle is 7.6°. Calculate the x-ray wavelength.24. The spacing between certain planes in a crystal isknown to be 0.30 nm. Find the smallest angle of incidenceat which constructive interference will occur forwavelength 0.070 nm.25. X-rays of wavelength 0.140 nm are reflected from a certaincrystal, and the first-order maximum occurs at anangle of 14.4°. What value does this give for the interplanarspacing of the crystal?Section 27.5 The Compton Effect26. X-rays are scattered from electrons in a carbon target.The measured wavelength shift is 1.50 10 3 nm.Calculate the scattering angle.27. Calculate the energy and momentum of a photon ofwavelength 700 nm.28. A beam of 0.68-nm photons undergoes Compton scatteringfrom free electrons. What are the energy andmomentum of the photons that emerge at a 45° anglewith respect to the incident beam?29. A 0.001 6-nm photon scatters from a free electron. Forwhat (photon) scattering angle will the recoiling electronand scattered photon have the same kineticenergy?30. X-rays with an energy of 300 keV undergo Comptonscattering from a target. If the scattered rays aredeflected at 37.0° relative to the direction of theincident rays, find (a) the Compton shift at this angle,(b) the energy of the scattered x-ray, and (c) the kineticenergy of the recoiling electron.31.A 0.110-nm photon collides with astationary electron. After the collision, theelectron moves forward and the photon recoils backwards.Find the momentum and kinetic energy of theelectron.32. After a 0.800-nm x-ray photon scatters from a freeelectron, the electron recoils with a speed equal to1.40 10 6 m/s. (a) What was the Compton shiftin the photon’s wavelength? (b) Through what anglewas the photon scattered?33. A 0.45-nm x-ray photon is deflected through a 23°angle after scattering from a free electron. (a) What isthe kinetic energy of the recoiling electron? (b) Whatis its speed?Section 27.6 The Dual Nature of Light and Matter34. Calculate the de Broglie wavelength of a proton movingat (a) 2.00 10 4 m/s; (b) 2.00 10 7 m/s.35.(a) If the wavelength of an electron is 5.00 10 7 m,how fast is it moving? (b) If the electron has a speed of1.00 10 7 m/s, what is its wavelength?36. A 0.200-kg ball is released from rest at the top of a50.0-m tall building. Find the de Broglie wavelength ofthe ball just before it strikes the Earth.37. The nucleus of an atom is on the order of 10 14 min diameter. For an electron to be confined to anucleus, its de Broglie wavelength would have to be ofthat order of magnitude or smaller. (a) What wouldbe the kinetic energy of an electron confined to

900 Chapter 27 Quantum Physicsthis region? (b) On the basis of your result in part(a), would you expect to find an electron in a nucleus?Explain.38. After learning about de Broglie’s hypothesis thatparticles of momentum p have wave characteristicswith wavelength h/p, an 80.0-kg student hasgrown concerned about being diffracted when passingthrough a 75.0-cm-wide doorway. Assume thatsignificant diffraction occurs when the width of thediffraction aperture is less than 10.0 times thewavelength of the wave being diffracted. (a) Determinethe maximum speed at which the studentcan pass through the doorway in order to be significantlydiffracted. (b) With that speed, how longwill it take the student to pass through the doorwayif it is 15.0 cm thick? Compare your result withthe currently accepted age of the Universe, which is4.00 10 17 s. (c) Should this student worry aboutbeing diffracted?39. De Broglie postulated that the relationship h/p isvalid for relativistic particles. What is the de Brogliewavelength for a (relativistic) electron whose kineticenergy is 3.00 MeV?40. A monoenergetic beam of electrons is incident on asingle slit of width 0.500 nm. A diffraction pattern isformed on a screen 20.0 cm from the slit. If thedistance between successive minima of the diffractionpattern is 2.10 cm, what is the energy of the incidentelectrons?41. The resolving power of a microscope is proportionalto the wavelength used. A resolution of 1.0 10 11 m (0.010 nm) would be required in orderto “see” an atom. (a) If electrons were used (electronmicroscope), what minimum kinetic energy would berequired of the electrons? (b) If photons were used,what minimum photon energy would be needed toobtain 1.0 10 11 m resolution?45.and is initially known to be within a pond 1.00 mwide. (a) What is the minimum uncertainty in hisspeed? (b) Assuming this uncertainty in speed to prevailfor 5.00 s, determine the uncertainty in Fuzzy’s positionafter this time.Suppose optical radiation ( 5.00 10 7 m) is used to determine the position of anelectron to within the wavelength of the light. Whatwill be the resulting uncertainty in the electron’svelocity?46. (a) Show that the kinetic energy of a nonrelativisticparticle can be written in terms of its momentum asKE p 2 /2m. (b) Use the results of (a) to find theminimum kinetic energy of a proton confined within anucleus having a diameter of 1.0 10 15 m.ADDITIONAL PROBLEMS47. Figure P27.47 shows the spectrum of light emitted by afirefly. Determine the temperature of a blackbody thatwould emit radiation peaked at the same frequency.Based on your result, would you say firefly radiation isblackbody radiation?Relative intensity1.00.80.60.40.2400 500 600Wavelength (nm)Figure P27.47Section 27.7 The Wave FunctionSection 27.8 The Uncertainty Principle42. A 50.0-g ball moves at 30.0 m/s. If its speed is measuredto an accuracy of 0.10%, what is the minimumuncertainty in its position?43. In the ground state of hydrogen, the uncertaintyin the position of the electron is roughly 0.10 nm.If the speed of the electron is on the order of theuncertainty in its speed, how fast is the electronmoving?44. Suppose Fuzzy, a quantum mechanical duck, lives in aworld in which h 2 J s. Fuzzy has a mass of 2.00 kg48. An x-ray tube is operated at 50 000 V. (a) Find the minimumwavelength of the radiation emitted by this tube.(b) If the radiation is directed at a crystal, the firstordermaximum in the reflected radiation occurswhen the grazing angle is 2.5°. What is the spacingbetween reflecting planes in the crystal?49. The spacing between planes of nickel atoms in anickel crystal is 0.352 nm. At what angle does a secondorderBragg reflection occur in nickel for 11.3-keVx-rays?50. Johnny Jumper’s favorite trick is to step out of his16th-story window and fall 50.0 m into a pool. Anews reporter takes a picture of 75.0-kg Johnnyjust before he makes a splash, using an exposure

Problems 90151.time of 5.00 ms. Find (a) Johnny’s de Broglie wavelengthat this moment, (b) the uncertainty of hiskinetic energy measurement during such a period oftime, and (c) the percent error caused by such anuncertainty.Photons of wavelength 450 nm are incident on ametal. The most energetic electrons ejected from themetal are bent into a circular arc of radius 20.0 cm by amagnetic field with a magnitude of 2.00 10 5 T.What is the work function of the metal?52. A 200-MeV photon is scattered at 40.0° by a free protonthat is initially at rest. Find the energy (in MeV) ofthe scattered photon.53. A light source of wavelength illuminates a metal andejects photoelectrons with a maximum kineticenergy of 1.00 eV. A second light source of wavelength/2 ejects photoelectrons with a maximum kineticenergy of 4.00 eV. What is the work function of themetal?54. Red light of wavelength 670 nm produces photoelectronsfrom a certain photoemissive material. Greenlight of wavelength 520 nm produces photoelectronsfrom the same material with 1.50 times the maximumkinetic energy. What is the material’s workfunction?55.How fast must an electron be moving if all its kineticenergy is lost to a single x-ray photon (a) at the highend of the x-ray electromagnetic spectrum with awavelength of 1.00 10 8 m; (b) at the low end ofthe x-ray electromagnetic spectrum with a wavelengthof 1.00 10 13 m?56. Show that if an electron were confined inside anatomic nucleus of diameter 2.0 10 15 m, it wouldhave to be moving relativistically, while a proton confinedto the same nucleus can be moving at less thanone-tenth the speed of light.57. A photon strikes a metal with a work function andproduces a photoelectron with a de Broglie wavelengthequal to the wavelength of the original photon.(a) Show that the energy of this photon must havebeen given byE (m e c 2 /2)m e c 2 where m e is the mass of the electron. [Hint : Beginwith the conservation of energy, E m e c 2 √(pc)2 (m e c 2)2.] (b) If one of these photons strikesplatinum ( 6.35 eV), determine the resultingmaximum speed of the photoelectron that isemitted.58. In a Compton scattering event, the scattered photonhas an energy of 120.0 keV and the recoilingelectron has a kinetic energy of 40.0 keV. Find(a) the wavelength of the incident photon, (b) theangle at which the photon is scattered, and (c) therecoil angle of the electron. [Hint : Conserve bothmass–energy and relativistic momentum.]59. A woman on a ladder drops small pellets toward apoint target on the floor. (a) Show that, according tothe uncertainty principle, the average distance bywhich she misses the target must be at leastwhere H is the initial height of each pellet abovethe floor and m is the mass of each pellet. Assumethat the spread in impact points is given by x f x i (v x )t. (b) If H 2.00 m and m 0.500 g, whatis x f ?60. Show that the speed of a particle having de Brogliewavelength and Compton wavelength C h/(mc) is61. (a) Find the mass of a solid iron sphere 2.00 cm inradius. (b) Assume that it is at 20°C and has emissivity0.860. Find the power with which it is radiatingelectromagnetic waves. (c) If this sphere were alonein the Universe, at what rate would its temperaturebe changing? (d) Assume Wien’s law describes thesphere. Find the wavelength max of electromagneticradiation it emits most strongly. Although it emits aspectrum of waves having all different wavelengths,model its whole power output as carried by photonsof wavelength max . Find (e) the energy of one photonand (f) the number of photons it emits eachsecond. When the sphere is at thermal equilibriumwith its surroundings, it emits and also absorbs photonsat this rate.ACTIVITIESx f 2 m 1/2 2H g 1/4v 1. Use a black marker or pieces of dark electrical tapeto make a very dark area on the outside of a shoebox.Poke a hole in the center of the dark area with apencil. Now put a lid on the box, and compare theblackness of the hole with the blackness of the surroundingdark area. Based on your observation,explain why the radiation emitted from the hole islike that emitted from a black body.c√1 (/C) 2

902 Chapter 27 Quantum Physics2. On a clear night, go outdoors far from city lightsand find the constellation Orion. Your instructorshould be able to furnish you with a star chart toassist you in locating this grouping of stars.Look very carefully at the color of the two starsBetelgeuse and Rigel. (See Fig. A27.2.) Can youtell which star is hotter? Orion is visible only fromNovember through April in the evening sky, soif Orion is not visible when you go out, comparetwo of the brightest stars you can see, such asVega in the constellation Lyra and Arcturus inBoötes.John Chumack/Photo ResearchersBetelgeuseRigelFigure A27.2

“Neon lights,” commonly used inadvertising signs, consist of thin glasstubes filled with various gases, suchas neon and helium. The gas atomsare excited to higher energy levels byelectric discharge through the tube.When the electrons in these excitedlevels return to lower energy levels,the atoms emit light having awavelength (color) that depends onthe type of gas in the tube. Forexample, a tube filled with neonproduces a red-orange color, whilehelium produces pink.Atomic PhysicsA large portion of this chapter concerns the hydrogen atom. Although the hydrogen atom isthe simplest atomic system, it’s especially important for several reasons:• The quantum numbers used to characterize the allowed states of hydrogen can also beused to describe (approximately) the allowed states of more complex atoms. This enablesus to understand the periodic table of the elements, one of the greatest triumphs of quantummechanics.• The hydrogen atom is an ideal system for performing precise comparisons of theory withexperiment and for improving our overall understanding of atomic structure.• Much of what is learned about the hydrogen atom with its single electron can be extendedto such single-electron ions as He and Li 2 .In this chapter we first discuss the Bohr model of hydrogen, which helps us understand manyfeatures of that element but fails to explain finer details of atomic structure. Next we examinethe hydrogen atom from the viewpoint of quantum mechanics and the quantum numbersused to characterize various atomic states. Quantum numbers aren’t mere mathematical abstractions:they have physical significance, such as the role they play in the effect of a magneticfield on certain quantum states. The fact that no two electrons in an atom can have thesame set of quantum numbers—the Pauli exclusion principle—is extremely important in understandingthe properties of complex atoms and the arrangement of elements in the periodictable. Finally, we apply our knowledge of atomic structure to describe the mechanismsinvolved in the production of x-rays, the operation of a laser, and the behavior of solid-statedevices such as diodes and transistors.Dembinsky Photo AssociatesCHAPTER28O U T L I N E28.1 Early Models of the Atom28.2 Atomic Spectra28.3 The Bohr Theoryof Hydrogen28.4 Modification of the BohrTheory28.5 De Broglie Waves and theHydrogen Atom28.6 Quantum Mechanics andthe Hydrogen Atom28.7 The Spin MagneticQuantum Number28.8 Electron Clouds28.9 The Exclusion Principleand the Periodic Table28.10 Characteristic X-Rays28.11 Atomic Transitions28.12 Lasers and Holography28.13 Energy Bands in Solids28.14 Semiconductor Devices28.1 EARLY MODELS OF THE ATOMThe model of the atom in the days of Newton was a tiny, hard, indestructiblesphere. Although this model was a good basis for the kinetic theory of gases, newmodels had to be devised when later experiments revealed the electronic nature of903

904 Chapter 28 Atomic PhysicsElectronFigure 28.1 Thomson’s modelof the atom, with the electronsembedded inside the positive chargelike seeds in a watermelon.SIR JOSEPH JOHN THOMSON,English Physicist (1856 – 1940)Thomson, usually considered the discovererof the electron, opened up the field ofsubatomic particle physics with his extensivework on the deflection of cathoderays (electrons) in an electric field. Hereceived the 1906 Nobel prize for hisdiscovery of the electron.Stock Montage, Inc.atoms. J. J. Thomson (1856–1940) suggested a model of the atom as a volume ofpositive charge with electrons embedded throughout the volume, much like theseeds in a watermelon (Fig. 28.1).In 1911 Ernest Rutherford (1871–1937) and his students Hans Geiger andErnest Marsden performed a critical experiment showing that Thomson’s modelcouldn’t be correct. In this experiment, a beam of positively charged alpha particleswas projected against a thin metal foil, as in Figure 28.2a. The results of theexperiment were astounding. Most of the alpha particles passed through the foil asif it were empty space, but a few particles deflected from their original direction oftravel were scattered through large angles. Some particles were even deflectedbackwards, reversing their direction of travel. When Geiger informed Rutherfordof these results, Rutherford wrote, “It was quite the most incredible event that hasever happened to me in my life. It was almost as incredible as if you fired a 15-inchshell at a piece of tissue paper and it came back and hit you.”Such large deflections were not expected on the basis of Thomson’s model. Accordingto that model, a positively charged alpha particle would never come closeenough to a large positive charge to cause any large-angle deflections. Rutherfordexplained these astounding results by assuming that the positive charge in an atomwas concentrated in a region that was small relative to the size of the atom. Hecalled this concentration of positive charge the nucleus of the atom. Any electronsbelonging to the atom were assumed to be in the relatively large volume outsidethe nucleus. In order to explain why electrons in this outer region of the atomwere not pulled into the nucleus, Rutherford viewed them as moving in orbitsabout the positively charged nucleus in the same way that planets orbit the Sun, asshown in Figure 28.2b. Alpha particles themselves were later identified as the nucleiof helium atoms.There are two basic difficulties with Rutherford’s planetary model. First, an atomemits certain discrete characteristic frequencies of electromagnetic radiation and noothers; the Rutherford model is unable to explain this phenomenon. Second, theelectrons in Rutherford’s model undergo a centripetal acceleration. According toMaxwell’s theory of electromagnetism, centripetally accelerated charges revolvingwith frequency f should radiate electromagnetic waves of the same frequency.Unfortunately, this classical model leads to disaster when applied to the atom. As theelectron radiates energy, the radius of its orbit steadily decreases and its frequency ofrevolution increases. This leads to an ever-increasing frequency of emitted radiationand a rapid collapse of the atom as the electron spirals into the nucleus.28.2 ATOMIC SPECTRAThe hydrogen atom is the simplest atomic system and an especially important oneto understand. Much of what we know about the hydrogen atom (which consists ofone proton and one electron) can be extended directly to other single-electronViewingscreen–Figure 28.2 (a) Geiger andMarsden’s technique for observingthe scattering of alpha particlesfrom a thin foil target. The sourceis a naturally occurring radioactivesubstance, such as radium.(b) Rutherford’s planetary modelof the atom.SourceLeadscreen(a)Target–+(b)

28.2 Atomic Spectra 905l(nm) 400 500 600 700(a)HHgNeHl(nm) 400 500 600 700(b)Figure 28.3 Visible spectra. (a) Line spectra produced by emission in the visible range for the elementshydrogen, mercury, and neon. (b) The absorption spectrum for hydrogen. The dark absorptionlines occur at the same wavelengths as the emission lines for hydrogen shown in (a).K. W. Whitten, R. E. Davis, M. L. Peck, and G. G. Stanley, General Chemistry,7th ed., Belmont, CA, Brooks/Cole, 2004.ions such as He and Li 2 . Further, a thorough understanding of the physics underlyingthe hydrogen atom can then be used to describe more complex atomsand the periodic table of the elements.Suppose an evacuated glass tube is filled with hydrogen (or some other gas) atlow pressure. If a voltage applied between metal electrodes in the tube is greatenough to produce an electric current in the gas, the tube emits light having acolor that depends on the gas inside. (This is how a neon sign works.) When theemitted light is analyzed with a spectrometer, discrete bright lines are observed,each having a different wavelength, or color. Such a series of spectral lines is commonlycalled an emission spectrum. The wavelengths contained in such a spectrumare characteristic of the element emitting the light (Fig. 28.3). Because notwo elements emit the same line spectrum, this phenomenon represents a marvelousand reliable technique for identifying elements in a gaseous substance.The emission spectrum of hydrogen shown in Figure 28.4 includes four prominentlines that occur at wavelengths of 656.3 nm, 486.1 nm, 434.1 nm, and410.2 nm, respectively. In 1885 Johann Balmer (1825–1898) found that the wavelengthsof these and less prominent lines can be described by the simple empiricalequation1 R H 1 [28.1]2 2 1 2 nwhere n may have integral values of 3, 4, 5, . . . , and R H is a constant, called theRydberg constant. If the wavelength is in meters, R H has the valueR H 1.097 373 2 10 7 m 1 [28.2]The first line in the Balmer series, at 656.3 nm, corresponds to n 3 in Equation28.1, the line at 486.1 nm corresponds to n 4, and so on. In addition to theBalmer series of spectral lines, a Lyman series was subsequently discovered in thefar ultraviolet, with the radiated wavelengths described by a similar equation.In addition to emitting light at specific wavelengths, an element can absorblight at specific wavelengths. The spectral lines corresponding to this process formwhat is known as an absorption spectrum. An absorption spectrum can be obtainedby passing a continuous radiation spectrum (one containing all wavelengths)through a vapor of the element being analyzed. The absorption spectrumconsists of a series of dark lines superimposed on the otherwise bright continuousspectrum. Each line in the absorption spectrum of a given element coincides witha line in the emission spectrum of the element. This means that if hydrogen is theλ(nm)364.6410.2 434.1486.1 656.3Figure 28.4 The Balmer series ofspectral lines for atomic hydrogen,with several lines marked with thewavelength in nanometers. The linelabeled 346.6 is the shortest-wavelengthline and is in the ultravioletregion of the electromagneticspectrum. The other labeled lines arein the visible region. Balmer series Rydberg constant

906 Chapter 28 Atomic PhysicsAPPLICATIONDiscovery of Heliumabsorbing vapor, dark lines will appear at the visible wavelengths 656.3 nm,486.1 nm, 434.1 nm, and 410.2 nm, as shown in Figures 28.3b and 28.4.The absorption spectrum of an element has many practical applications. For example,the continuous spectrum of radiation emitted by the Sun must passthrough the cooler gases of the solar atmosphere before reaching the Earth. Thevarious absorption lines observed in the solar spectrum have been used to identifyelements in the solar atmosphere, including one that was previously unknown.When the solar spectrum was first being studied, some lines were found that didn’tcorrespond to any known element. A new element had been discovered! Becausethe Greek word for Sun is helios, the new element was named helium. It was lateridentified in underground gases on Earth. Scientists are able to examine the lightfrom stars other than our Sun in this way, but elements other than those presenton Earth have never been detected.Applying Physics 28.1Thermal or Spectral?On observing a yellow candle flame, your laboratorypartner claims that the light from the flame originatesfrom excited sodium atoms in the flame. You disagree,stating that because the candle flame is hot, the radiationmust be thermal in origin. Before the disagreementleads to fisticuffs, how could you determine whois correct?Explanation A simple determination could be madeby observing the light from the candle flame througha spectrometer, which is a slit and diffraction gratingcombination discussed in Chapter 25. If the spectrumof the light is continuous, then it’s probably thermalin origin. If the spectrum shows discrete lines, it’satomic in origin. The results of the experiment showthat the light is indeed thermal in origin and originatesfrom random molecular motion in the candleflame.Applying Physics 28.2AurorasAt extreme northern latitudes, the aurora borealisprovides a beautiful and colorful display in thenight sky. A similar display occurs near thesouthern polar region and is called the auroraaustralis. What’s the origin of the various colorsseen in the auroras?Explanation The aurora is due to high speed particlesinteracting with the Earth’s magnetic field andentering the atmosphere. When these particles collidewith molecules in the atmosphere, they excite themolecules in a way similar to the voltage in thespectrum tubes discussed earlier in this section. Inresponse, the molecules emit colors of light accordingto the characteristic spectrum of their atomicconstituents. For our atmosphere, the primaryconstituents are nitrogen and oxygen, which providethe red, blue, and green colors of the aurora.+ er– em eFigure 28.5 Diagram representingBohr’s model of the hydrogen atom.The orbiting electron is allowed onlyin specific orbits of discrete radii.Fv28.3 THE BOHR THEORY OF HYDROGENAt the beginning of the 20th century, scientists were perplexed by the failure ofclassical physics to explain the characteristics of spectra. Why did atoms of a givenelement emit only certain lines? Further, why did the atoms absorb only those wavelengthsthat they emitted? In 1913 Bohr provided an explanation of atomic spectrathat includes some features of the currently accepted theory. Using the simplestatom, hydrogen, Bohr developed a model of what he thought must be the atom’sstructure in an attempt to explain why the atom was stable. His model of the hydrogenatom contains some classical features, as well as some revolutionary postulatesthat could not be justified within the framework of classical physics. The basic assumptionsof the Bohr theory as it applies to the hydrogen atom are as follows:1. The electron moves in circular orbits about the proton under the influence ofthe Coulomb force of attraction, as in Figure 28.5. The Coulomb force producesthe electron’s centripetal acceleration.

28.3 The Bohr Theory of Hydrogen 9072. Only certain electron orbits are stable. These are orbits in which the hydrogenatom doesn’t emit energy in the form of electromagnetic radiation. Hence, thetotal energy of the atom remains constant, and classical mechanics can be usedto describe the electron’s motion.3. Radiation is emitted by the hydrogen atom when the electron “jumps” from amore energetic initial state to a less energetic state. The “jump” can’t be visualizedor treated classically. In particular, the frequency f of the radiation emitted in thejump is related to the change in the atom’s energy and is independent of the frequencyof the electron’s orbital motion. The frequency of the emitted radiation is given byE i E f hf [28.3]where E i is the energy of the initial state, E f is the energy of the final state, h isPlanck’s constant, and E i E f .4. The size of the allowed electron orbits is determined by a condition imposedon the electron’s orbital angular momentum: the allowed orbits are those forwhich the electron’s orbital angular momentum about the nucleus is an integralmultiple of (pronounced “h bar”), where :m e vr nn 1, 2, 3, . . . [28.4]With these four assumptions, we can calculate the allowed energies and emissionwavelengths of the hydrogen atom. We use the model pictured in Figure 28.5,in which the electron travels in a circular orbit of radius r with an orbital speed v.The electrical potential energy of the atom isPE k eq 1 q 2r k e(e)(e)rh/2k ewhere k e is the Coulomb constant. Assuming the nucleus is at rest, the total energyE of the atom is the sum of the kinetic and potential energy:E KE PE 1 2 m ev 2 e 2 k e[28.5]rWe apply Newton’s second law to the electron. We know that the electric forceof attraction on the electron, k e e 2 /r 2 , must equal m e a r , where a r v 2 /r is the centripetalacceleration of the electron. Thus,e 2k e[28.6]r 2 m v 2erFrom this equation, we see that the kinetic energy of the electron is12 m ev 2 k ee 2[28.7]2rWe can combine this result with Equation 28.5 and express the energy of the atom asE k ee 2[28.8]2rwhere the negative value of the energy indicates that the electron is bound to theproton.An expression for r is obtained by solving Equations 28.4 and 28.6 for v andequating the results:v 2 n2 m 2 e r 2 k ee 2m e rr n n2 2n 1, 2, 3, . . . [28.9]m e k e e 2This equation is based on the assumption that the electron can exist only in certainallowed orbits determined by the integer n.e 2rNIELS BOHR, Danish Physicist(1885 – 1962)Bohr was an active participant in the earlydevelopment of quantum mechanics andprovided much of its philosophical framework.During the 1920s and 1930s, heheaded the Institute for Advanced Studiesin Copenhagen. The institute was a magnetfor many of the world’s best physicistsand provided a forum for the exchange ofideas. When Bohr visited the United Statesin 1939 to attend a scientific conference,he brought news that the fission of uraniumhad been observed by Hahn andStrassman in Berlin. The results were thefoundations of the atomic bomb developedin the United States during WorldWar II. Bohr was awarded the 1922 NobelPrize for his investigation of the structureof atoms and of the radiation emanatingfrom them. Energy of the hydrogen atom The radii of the Bohr orbitsare quantizedPrinceton University/Courtesy of AIP Emilio Segre Visual Archives

908 Chapter 28 Atomic PhysicsThe orbit with the smallest radius, called the Bohr radius, a 0 , corresponds ton 1 and has the value4a 0– ea 0+ e9a 0ACTIVE FIGURE 28.6The first three circular orbitspredicted by the Bohr model ofthe hydrogen atom.Log into PhysicsNow atwww.cp7e.com and go to ActiveFigure 28.6, where you can choosethe initial and final states of thehydrogen atom and observe thetransition.ENERGYn∞54321BalmerseriesLymanseriesE (eV)0.00–0.5442–0.8504–1.512Paschenseries–3.401–13.606ACTIVE FIGURE 28.7An energy level diagram for hydrogen.Quantum numbers are given on theleft and energies (in electron volts)are given on the right. Vertical arrowsrepresent the four lowest-energytransitions for each of the spectralseries shown. The colored arrows forthe Balmer series indicate that thisseries results in visible light.Log into PhysicsNow atwww.cp7e.com and go to ActiveFigure 28.7, where you can choosethe initial and final states of thehydrogen atom and observe thetransition.a 0 [28.10]mk e e2 0.052 9 nmA general expression for the radius of any orbit in the hydrogen atom is obtainedby substituting Equation 28.10 into Equation 28.9:r n n 2 a 0 n 2 (0.052 9 nm) [28.11]The first three Bohr orbits for hydrogen are shown in Active Figure 28.6.Equation 28.9 may be substituted into Equation 28.8 to give the following expressionfor the energies of the quantum states:E n m ek 2 e e 4 1 2 n 1, 2, 3, . . . [28.12]2 2 nIf we insert numerical values into Equation 28.12, we obtainE n 13.6[28.13]n 2 eVThe lowest energy state, or ground state, corresponds to n 1 and has an energyE 1 m e k 2 e e 4 /2 2 13.6 eV. The next state, corresponding to n 2, has anenergy E 2 E 1 /4 3.40 eV, and so on. An energy level diagram showing theenergies of these stationary states and the corresponding quantum numbers isgiven in Active Figure 28.7. The uppermost level shown, corresponding to E 0and n : , represents the state for which the electron is completely removedfrom the atom. In this state, the electron’s KE and PE are both zero, which meansthat the electron is at rest infinitely far away from the proton. The minimum energyrequired to ionize the atom—that is, to completely remove the electron—iscalled the ionization energy. The ionization energy for hydrogen is 13.6 eV.Equations 28.3 and 28.12 and the third Bohr postulate show that if the electronjumps from one orbit with quantum number n i to a second orbit with quantumnumber, n f , it emits a photon of frequency f given byf E i E f m ek 2 e e 42 [28.14]hn 1f n i2where n f n i .Finally, to compare this result with the empirical formulas for the various spectralseries, we use Equation 28.14 and the fact that for light, f c, to get1 f c m ek e 2 e 44c 3 1n f2 1n i2[28.15]A comparison of this result with Equation 28.1 gives the following expression forthe Rydberg constant:R H m ek 2 e e 4[28.16]4c 3If we insert the known values of m e , k e , e, c, and into this expression, the resultingtheoretical value for R H is found to be in excellent agreement with the value determinedexperimentally for the Rydberg constant. When Bohr demonstrated thisagreement, it was recognized as a major accomplishment of his theory.In order to compare Equation 28.15 with spectroscopic data, it is convenient toexpress it in the form124 3 1 R H 1n f2 1n i2[28.17]

28.3 Th Bohr Theory of Hydrogen 909We can use this expression to evaluate the wavelengths for the various series in thehydrogen spectrum. For example, in the Balmer series, n f 2 and n i 3, 4, 5, . . .(Eq. 28.1). For the Lyman series, we take n f 1 and n i 2, 3, 4, . . . . The energylevel diagram for hydrogen shown in Active Figure 28.7 indicates the origin of thespectral lines described previously. The transitions between levels are representedby vertical arrows. Note that whenever a transition occurs between a state designatedby n i to one designated by n f (where n i n f ), a photon with a frequency(E i E f )/h is emitted. This can be interpreted as follows: the lines in the visiblepart of the hydrogen spectrum arise when the electron jumps from the third,fourth, or even higher orbit to the second orbit. Likewise, the lines of the Lymanseries (in the ultraviolet) arise when the electron jumps from the second, third, oreven higher orbit to the innermost (n f 1) orbit. Hence, the Bohr theory successfullypredicts the wavelengths of all the observed spectral lines of hydrogen.TIP 28.1 Energy DependsOn n Only for HydrogenAccording to Equation 28.13, the energydepends only on the quantumnumber n. Note that this is only truefor the hydrogen atom. For morecomplicated atoms, the energy levelsdepend primarily on n, but also onother quantum numbers.INTERACTIVE EXAMPLE 28.1The Balmer Series for HydrogenGoal Calculate the wavelength, frequency, and energy of a photon emitted during anelectron transition in an atom.Problem The Balmer series for the hydrogen atom corresponds to electronic transitionsthat terminate in the state with quantum number n 2, as shown in Figure 28.8.(a) Find the longest-wavelength photon emitted in the Balmer series and determine itsfrequency and energy. (b) Find the shortest-wavelength photon emitted in the sameseries.Strategy This is a matter of substituting values into Equation 28.17. The frequency canthen be obtained from c f and the energy from E hf. The longest wavelengthphoton corresponds to the one that is emitted when the electron jumps from the n i 3state to the n f 2 state. The shortest wavelength photon corresponds to the one that isemitted when the electron jumps from n i to the state n f 2.n∞6 –0.385–0.544–0.853–1.512BalmerseriesE (eV)0.00–3.40Figure 28.8 (Example 28.1)Transitions responsible for theBalmer series for the hydrogenatom. All transitions terminateat the n 2 level.Solution(a) Find the longest wavelength photon emitted in theBalmer series, and determine its energy.Substitute into Equation 28.17, with n i 3 and n f 2:Take the reciprocal and substitute, finding the wavelength:1 R H 1n 2 f 365R H 656.3 nm 1ni 2 R H 12 2 1 2 3 5R H36365(1.097 10 7 m 1 ) 6.563 107 mNow use c f to obtain the frequency:f c 2.998 108 m/s6.563 10 7 m4.568 10 14 HzCalculate the photon’s energy by substituting into Equation27.5:(b) Find the shortest wavelength photon emitted in theBalmer series.Substitute into Equation 28.17, with n i and n f 2.Take the reciprocal and substitute, finding the wavelength:E hf (6.626 10 34 J s)(4.568 10 14 Hz) 3.027 10 19 J 1.892 eV1 R H 1n 2 f 4R H 364.6 nm 1n i 2 R H 12 2 1 R H44(1.097 10 7 m 1 ) 3.646 107 m

910 Chapter 28 Atomic PhysicsRemarks The first wavelength is in the red region of the visible spectrum. We could also obtain the energy of thephoton by using Equation 28.3 in the form hf E 3 E 2 , where E 2 and E 3 are the energy levels of the hydrogenatom, calculated from Equation 28.13. Note that this is the lowest energy photon in the Balmer series, because it involvesthe smallest energy change. The second photon, the most energetic, is in the ultraviolet region.Exercise 28.1(a) Calculate the energy of the shortest wavelength photon emitted in the Balmer series for hydrogen. (b) Calculatethe wavelength of a transition from n 4 to n 2.Answers(a) 3.40 eV (b) 486 nmInvestigate transitions between various states by logging into PhysicsNow at www.cp7e.com and goingto Interactive Example 28.1.Bohr’s Correspondence PrincipleIn our study of relativity in Chapter 26, we found that Newtonian mechanics cannotbe used to describe phenomena that occur at speeds approaching the speed oflight. Newtonian mechanics is a special case of relativistic mechanics and appliesonly when v is much smaller than c. Similarly, quantum mechanics is in agreementwith classical physics when the energy differences between quantized levels arevery small. This principle, first set forth by Bohr, is called the correspondenceprinciple.For example, consider the hydrogen atom with n 10 000. For such large valuesof n, the energy differences between adjacent levels approach zero and the levelsare nearly continuous, as Equation 28.13 shows. As a consequence, the classicalmodel is reasonably accurate in describing the system for large values of n. Accordingto the classical model, the frequency of the light emitted by the atom is equalto the frequency of revolution of the electron in its orbit about the nucleus. Calculationsshow that for n 10 000, this frequency is different from that predicted byquantum mechanics by less than 0.015%.28.4 MODIFICATION OF THE BOHR THEORYThe Bohr theory of the hydrogen atom was a tremendous success in certain areasbecause it explained several features of the hydrogen spectrum that had previouslydefied explanation. It accounted for the Balmer series and other series; it predicteda value for the Rydberg constant that is in excellent agreement with the experimentalvalue; it gave an expression for the radius of the atom; and it predictedthe energy levels of hydrogen. Although these successes were important to scientists,it is perhaps even more significant that the Bohr theory gave us a model ofwhat the atom looks like and how it behaves. Once a basic model is constructed,refinements and modifications can be made to enlarge on the concept and to explainfiner details.The analysis used in the Bohr theory is also successful when applied to hydrogenlikeatoms. An atom is said to be hydrogen-like when it contains only one electron.Examples are singly ionized helium, doubly ionized lithium, triply ionized beryllium,and so forth. The results of the Bohr theory for hydrogen can be extended to hydrogen-likeatoms by substituting Ze 2 for e 2 in the hydrogen equations, where Z is theatomic number of the element. For example, Equations 28.12 and 28.15 becomeandE n m ek e 2 Z 2 e 42 21 1 n 2 m ek e 2 Z 2 e 44c 3 1n f2 1n i2n 1, 2, 3, . . . [28.18][28.19]

28.4 Modification of the Bohr Theory 911Although many attempts were made to extend the Bohr theory to more complex,multi-electron atoms, the results were unsuccessful. Even today, only approximatemethods are available for treating multi-electron atoms.Quick Quiz 28.1Consider a hydrogen atom and a singly-ionized helium atom. Which atom has thelower ground state energy? (a) hydrogen (b) helium (c) the ground state energy isthe same for bothQuick Quiz 28.2Consider once again a singly-ionized helium atom. Suppose the remaining electronjumps from a higher to a lower energy level, resulting in the emission of photon,which we’ll call photon-He. An electron in a hydrogen atom then jumpsbetween the same two levels, resulting in an emitted photon-H. Which photon hasthe shorter wavelength? (a) photon-He (b) photon-H (c) The wavelengths are thesame.EXAMPLE 28.2GoalSingly Ionized HeliumApply the modified Bohr theory to a hydrogen-like atom.Problem Singly ionized helium, He , a hydrogen-like system, has one electron in the 1s orbit when the atom is inits ground state. Find (a) the energy of the system in the ground state in electron volts, and (b) the radius of theground-state orbit.Strategy Part (a) requires substitution into the modified Bohr model, Equation 28.18. In part (b), modify Equation28.9 for the radius of the Bohr orbits by replacing e 2 by Ze 2 , where Z is the number of protons in the nucleus.Solution(a) Find the energy of the system in the ground state.Write Equation 28.18 for the energies of a hydrogen-likesystem:E n m ek e 2 Z 2 e 42 2 1 n 2Substitute the constants and convert to electron volts: E n Z 2 (13.6)n 2 eVSubstitute Z 2 (the atomic number of helium) andn 1 to obtain the ground state energy:(b) Find the radius of the ground state.Generalize Equation 28.9 to a hydrogen-like atom bysubstituting Ze 2 for e 2 :For our case, n 1 and Z 2:E 1 4(13.6) eV 54.4 eVr n n2 2m e k e Ze 2 n2Z (a 0) n2(0.052 9 nm)Zr 1 0.026 5 nmRemarks Notice that for higher Z the energy of a hydrogen-like atom is lower, which means that the electron ismore tightly bound than in hydrogen. This results in a smaller atom, as seen in part (b).Exercise 28.2Repeat the problem for the first excited state of doubly-ionized lithium (Z 3, n 2).Answers (a) E 2 30.6 eV (b) r 2 0.070 5 nm

912 Chapter 28 Atomic PhysicsTABLE 28.1Shell and Subshell NotationShellSubshelln Symbol Symbol1 K 0 s2 L 1 p3 M 2 d4 N 3 f5 O 4 g6 P 5 h. . . . . .Figure 28.9 A single line (A) cansplit into three separate lines (B) in amagnetic field.ABWithin a few months following the publication of Bohr’s paper, Arnold Sommerfeld(1868–1951) extended the Bohr model to include elliptical orbits. We examinehis model briefly because much of the nomenclature used in this treatment is still inuse today. Bohr’s concept of quantization of angular momentum led to the principalquantum number n, which determines the energy of the allowed states of hydrogen.Sommerfeld’s theory retained n, but also introduced a new quantum number called the orbital quantum number, where the value of ranges from 0 to n 1 in integersteps. According to this model, an electron in any one of the allowed energystates of a hydrogen atom may move in any one of a number of orbits correspondingto different values. For each value of n, there are n possible orbits corresponding todifferent values. Because n 1 and 0 for the first energy level (ground state),there is only one possible orbit for this state. The second energy level, with n 2, hastwo possible orbits, corresponding to 0 and 1. The third energy level, withn 3, has three possible orbits, corresponding to 0, 1, and 2.For historical reasons, all states with the same principal quantum number n aresaid to form a shell. Shells are identified by the letters K, L, M, . . . , which designatethe states for which n 1, 2, 3, . . . . Likewise, the states with given values of n and are said to form a subshell. The letters s, p, d, f, g, . . . are used to designate the statesfor which 0, 1, 2, 3, 4, . . . . These notations are summarized in Table 28.1.States that violate the restriction 0 n 1, for a given value of n, can’t exist.A 2d state, for instance, would have n 2 and 2, but can’t exist because thehighest allowed value of is n 1, or 1 in this case. For n 2, 2s and 2p are allowedsubshells, but 2d, 2f, . . . are not. For n 3, the allowed states are 3s, 3p, and 3d.Another modification of the Bohr theory arose when it was discovered that thespectral lines of a gas are split into several closely spaced lines when the gas is placedin a strong magnetic field. (This is called the Zeeman effect, after its discoverer.) Figure28.9 shows a single spectral line being split into three closely spaced lines. This indicatesthat the energy of an electron is slightly modified when the atom is immersed ina magnetic field. In order to explain this observation, a new quantum number, m ,called the orbital magnetic quantum number, was introduced. The theory is in accordwith experimental results when m is restricted to values ranging from to ininteger steps. For a given value of , there are 2 1 possible values of m .Finally, very high resolution spectrometers revealed that spectral lines of gasesare in fact two very closely spaced lines even in the absence of an external magneticfield. This splitting was referred to as fine structure. In 1925 Samuel Goudsmit andGeorge Uhlenbeck introduced the idea of an electron spinning about its own axisto explain the origin of fine structure. The results of their work introduced yet anotherquantum number, m s , called the spin magnetic quantum number.For each electron there are two spin states. A subshell corresponding to a givenfactor of can contain no more than 2(2 1) electrons. This number comesfrom the fact that electrons in a subshell must have unique pairs of the quantumnumbers (m , m s ). There are 2 1 different magnetic quantum numbers m ,and two different spin quantum numbers m s , making 2(2 1) unique pairs(m , m s ). For example, the p subshell ( 1) is filled when it contains 2(2 1 1) 6electrons. This fact can be extended to include all four quantum numbers, as willbe important to us later when we discuss the Pauli exclusion principle.All these quantum numbers (addressed in more detail in upcoming sections)were postulated to account for the observed spectra of elements. Only later werecomprehensive mathematical theories developed that naturally yielded the sameanswers as these empirical models.28.5 DE BROGLIE WAVES ANDTHE HYDROGEN ATOMOne of the postulates made by Bohr in his theory of the hydrogen atom was thatthe angular momentum of the electron is quantized in units of , orm e vr n

28.6 Quantum Mechanics and the Hydrogen Atom 913rABFigure 28.10 (a) Standing-wavepattern for an electron wave in astable orbit of hydrogen. There arethree full wavelengths in this orbit.(b) Standing-wave pattern for avibrating stretched string fixed at itsends. This pattern also has three fullwavelengths.l(a)(b)For more than a decade following Bohr’s publication, no one was able to explain whythe angular momentum of the electron was restricted to these discrete values. Finally,de Broglie gave a direct physical way of interpreting this condition. He assumed thatan electron orbit would be stable (allowed) only if it contained an integral number ofelectron wavelengths. Figure 28.10a demonstrates this point when three completewavelengths are contained in one circumference of the orbit. Similar patterns can bedrawn for orbits containing one wavelength, two wavelengths, four wavelengths, fivewavelengths, and so forth. These waves are analogous to standing waves on a string,discussed in Chapter 14. There, we found that strings have preferred (resonant) frequenciesof vibration. Figure 28.10b shows a standing-wave pattern containing threewavelengths for a string fixed at each end. Now imagine that the vibrating string is removedfrom its supports at A and B and bent into a circular shape that brings thosepoints together. The end result is a pattern such as the one shown in Figure 28.10a.In general, the condition for a de Broglie standing wave in an electron orbit isthat the circumference must contain an integral number of electron wavelengths.We can express this condition as2r n n 1, 2, 3, . . .Because the de Broglie wavelength of an electron is h/m e v, we can write thepreceding equation as 2r nh/m e v, orm e vr nThis is the same as the quantization of angular momentum condition imposed byBohr in his original theory of hydrogen.The electron orbit shown in Figure 28.10a contains three complete wavelengthsand corresponds to the case in which the principal quantum number n 3. Theorbit with one complete wavelength in its circumference corresponds to the firstBohr orbit, n 1; the orbit with two complete wavelengths corresponds to the secondBohr orbit, n 2; and so forth.By applying the wave theory of matter to electrons in atoms, de Broglie was ableto explain the appearance of integers in the Bohr theory as a natural consequenceof standing-wave patterns. This was the first convincing argument that the wave natureof matter was at the heart of the behavior of atomic systems. Although theanalysis provided by de Broglie was a promising first step, gigantic strides weremade subsequently with the development of Schrödinger’s wave equation and itsapplication to atomic systems.28.6 QUANTUM MECHANICSAND THE HYDROGEN ATOMOne of the first great achievements of quantum mechanics was the solution of thewave equation for the hydrogen atom. The details of the solution are far beyondthe level of this course, but we’ll describe its properties and implications foratomic structure.

914 Chapter 28 Atomic PhysicsTABLE 28.2Three Quantum Numbers for the Hydrogen AtomNumber ofQuantumAllowedNumber Name Allowed Values StatesN Principal quantum number 1, 2, 3, . . . Any numberOrbital quantum number 0, 1, 2, . . . , n 1 nm Orbital magnetic quantum , 1, . . . , 2 1number 0, . . . , 1, According to quantum mechanics, the energies of the allowed states are in exactagreement with the values obtained by the Bohr theory (Eq. 28.12) when theallowed energies depend only on the principal quantum number n.In addition to the principal quantum number, two other quantum numbersemerged from the solution of the wave equation: and m . The quantum number is called the orbital quantum number, and m is called the orbital magnetic quantumnumber. As pointed out in Section 28.4, these quantum numbers had alreadyappeared in empirical modifications made to the Bohr theory. The significance ofquantum mechanics is that those numbers and the restrictions placed on their valuesarose directly from mathematics and not from any ad hoc assumptions tomake the theory consistent with experimental observation. Because we will need tomake use of the various quantum numbers in the sections that follow, the allowedranges of their values are repeated:The value of n can range from 1 to in integer steps.The value of can range from 0 to n 1 in integer steps.The value of m can range from to in integer steps.From these rules, it can be seen that for a given value of n, there are n possible valuesof , while for a given value of there are 2 1 possible values of m . For example,if n 1, there is only 1 value of , 0. Because 2 1 2 0 1 1,there is only one value of m , which is m 0. If n 2, the value of may be 0 or 1;if 0, then m 0, but if 1, then m may be 1, 0, or 1. Table 28.2 summarizesthe rules for determining the allowed values of and m for a given value of n.States that violate the rules given in Table 28.2 cannot exist. For instance, onestate that cannot exist is the 2d state, which would have n 2 and 2. This stateis not allowed because the highest allowed value of is n 1, or 1 in this case.Thus, for n 2, 2s and 2p are allowed states, but 2d, 2f, . . . are not. For n 3, theallowed states are 3s, 3p, and 3d.In general, for a given value of n 1 there are n 2 states with distinct pairs of valuesof and m .Quick Quiz 28.3When the principal quantum number is n 5, how many different values of (a) and (b) m are possible? (c) How many states have distinct pairs of values of and m ?EXAMPLE 28.3GoalThe n 2 Level of HydrogenCount states and determine energy based on atomic energy level.Problem (a) Determine the number of states with a unique set of values for and m in the hydrogen atom for n 2.(b) Calculate the energies of these states.Strategy This is a matter of counting, following the quantum rules for n, , and m . “Unique” means that no otherquantum state has the same pair of numbers for and m the energies are all the same because all states have thesame principal quantum number, n 2.

28.7 The Spin Magnetic Quantum Number 915Solution(a) Determine the number of states with a unique set ofvalues for and m in the hydrogen atom for n 2.Determine the different possible values of for n 2: 0 n 1, so, for n 2, 0 1 and 0 or 1Find the different possible values of m for 0: m , so 0 m 0 implies m 0List the distinct pairs of (, m ) for 0: There is only one: (, m ) (0, 0).Find the different possible values of m for 1: m , so 1 m 1 implies m 1, 0, or 1List the distinct pairs of (, m ) for 1: There are three: (, m ) ( 1, 1), (1, 0), and (1, 1).Sum the results for 0 and 1: Number of states 1 3 4(b) Calculate the energies of these states.The common energy of all of the states can be foundwith Equation 28.13:E n 13.6 eV13.6 eVn 2 : E 2 2 2 3.40 eVRemarks While these states normally have the same energy, applying a magnetic field will result in their takingslightly different energies centered around the energy corresponding to n 2. As seen in the next section, there arein fact twice as many states, corresponding to a new quantum number called spin.Exercise 28.3(a) Determine the number of states with a unique pair of values for and m in the n 3 level of hydrogen.(b) Determine the energies of those states.Answers (a) 9 (b) E 3 1.51 eV28.7 THE SPIN MAGNETIC QUANTUM NUMBERAs we’ll see in this section, there actually are eight states corresponding to n 2for hydrogen, not four as given in Example 28.3. This happens because anotherquantum number, m s , the spin magnetic quantum number, has to be introduced toexplain the splitting of each level into two.The need for this new quantum number first came about because of an unusualfeature in the spectra of certain gases, such as sodium vapor. Close examination ofone of the prominent lines of sodium shows that it is, in fact, two very closelyspaced lines. The wavelengths of these lines occur in the yellow region of the spectrum,at 589.0 nm and 589.6 nm. In 1925, when this doublet was first noticed,atomic theory couldn’t explain it. To resolve the dilemma, Samuel Goudsmit andGeorge Uhlenbeck, following a suggestion by the Austrian physicist WolfgangPauli, proposed the introduction of a fourth quantum number to describe atomicenergy levels, called the spin quantum number.In order to describe the spin quantum number, it’s convenient (but technically incorrect)to think of the electron as spinning on its axis as it orbits the nucleus, just asthe Earth spins on its axis as it orbits the Sun. Strangely, there are only two ways inwhich the electron can spin as it orbits the nucleus, as shown in Figure 28.11. If thedirection of spin is as shown in Figure 28.11a, the electron is said to have “spin up.” Ifthe direction of spin is reversed, as in Figure 28.11b, the electron is said to have “spindown.” The energy of the electron is slightly different for the two spin directions, andthis energy difference accounts for the sodium doublet. The quantum numbers associatedwith electron spin are m s for the spin-up state and m s 1 22 for the spin-1down state. As we’ll see in Example 28.5, this new quantum number doubles thenumber of allowed states specified by the quantum numbers n, , and m .Nucleus(a)Nucleus(b)Spin upSpin downFigure 28.11 As an electron movesin its orbit about the nucleus, its spincan be either (a) up or (b) down.

916 Chapter 28 Atomic PhysicsTIP 28.2 The Electron Isn’tReally SpinningThe electron is not physicallyspinning. Electron spin is a purelyquantum effect that gives theelectron an angular momentum as ifit were physically spinning.Any classical description of electron spin is incorrect because quantum mechanicstells us that since the electron can’t be located precisely in space, it cannot beconsidered to be a spinning solid object, as pictured in Figure 28.11. In spite ofthis conceptual difficulty, all experimental evidence supports the fact that an electrondoes have some intrinsic property that can be described by the spin magneticquantum number.The spin quantum number didn’t come from the original formulation of quantummechanics by Schrodinger (and independently, by Heisenberg). The Englishmathematical physicist P. A. M. Dirac developed a relativistic quantum theory inwhich spin appears naturally.EXAMPLE 28.4 The Quantum Numbers for the 2p SubshellGoal List the distinct quantum states of a subshell by their quantum numbers, including spin.ProblemList the unique sets of quantum numbers for electrons in the 2p subshell.Strategy This is again a matter following the quantum rules for n, , and m , and now m s as well. The 2p subshellhas n 2 (that’s the “2” in 2p) and 1 (that’s from the p in 2p).SolutionBecause 1, the magnetic quantum number can havethe values 1, 0, 1, and the spin quantum number is always 1 2 or 1 2. Consequently, there are 3 2 6 possiblesets of quantum numbers with n 2 and 1,listed in the table at right.n m m s2 1 12 1 12 1 02 1 02 1 12 1 1 1 212 1 212 1 212Remark Remember that these quantum states are not just abstractions; they have real physical consequences, suchas which electronic transitions can be made within an atom and, consequently, which wavelengths of radiation can beobserved.Exercise 28.4(a) How many different sets of quantum numbers are there in the 3d subshell? (b) How many sets of quantum numbersare there in a 2d subshell?Answers (a) 10 (b) None. A 2d subshell doesn’t exist because that would imply a quantum state with n 2 and 2, impossible because n 1.28.8 ELECTRON CLOUDSThe solution of the wave equation, discussed in Section 27.7, yields a wave function that depends on the quantum numbers n, , and m . We assume that wehave found such a wave function and see what it may tell us about the hydrogenatom. Let n 1 for the principal quantum number, which corresponds to the lowestenergy state for hydrogen. For n 1, the restrictions placed on the remainingquantum numbers are that 0 and m 0.The quantity 2 has great physical significance. If p is a point and V p a verysmall volume containing that point, then 2 V p is approximately the probability offinding the electron inside the volume V p . Figure 28.12 gives the probability perunit length of finding the electron at various distances from the nucleus in the 1sstate of hydrogen. Some useful and surprising information can be extracted from

28.9 The Exclusion Principle and the Periodic Table 917this curve. First, the curve peaks at a value of r 0.052 9 nm, the Bohr radius forthe first (n 1) electron orbit in hydrogen. This means that there is a maximumprobability of finding the electron in a small interval centered at that distancefrom the nucleus. However, as the curve indicates, there is also a probability offinding the electron in a small interval centered at any other distance from the nucleus.In other words, the electron is not confined to a particular orbital distancefrom the nucleus, as assumed in the Bohr model. The electron may be found atvarious distances from the nucleus, but the probability of finding it at a distancecorresponding to the Bohr radius is a maximum. Quantum mechanics alsopredicts that the wave function for the hydrogen atom in the ground state is sphericallysymmetric; hence the electron can be found in a spherical region surroundingthe nucleus. This is in contrast to the Bohr theory, which confines the positionof the electron to points in a plane. The quantum mechanical result is often interpretedby viewing the electron as a cloud surrounding the nucleus. An attempt atpicturing this cloud-like behavior is shown in Figure 28.13. The densest regions ofthe cloud represent those locations where the electron is most likely to be found.If a similar analysis is carried out for the n 2, 0, state of hydrogen, a peakof the probability curve is found at 4a 0 . Likewise, for the n 3, 0 state, thecurve peaks at 9a 0 . Thus, quantum mechanics predicts a most probable electrondistance to the nucleus that is in agreement with the location predicted by theBohr theory.P1s(r)a 0 = 0.0529 nmFigure 28.12 The probability perunit length of finding the electronversus distance from the nucleusfor the hydrogen atom in the 1s(ground) state. Note that the graphhas its maximum value when r equalsthe first Bohr radius, a 0 .yr28.9 THE EXCLUSION PRINCIPLEAND THE PERIODIC TABLEEarlier, we found that the state of an electron in an atom is specified by four quantumnumbers: n, , m , and m s . For example, an electron in the ground state of hydrogencould have quantum numbers of n 1, 0, m 0, and m s 1 2. As itturns out, the state of an electron in any other atom may also be specified by thissame set of quantum numbers. In fact, these four quantum numbers can be usedto describe all the electronic states of an atom, regardless of the number of electronsin its structure.How many electrons in an atom can have a particular set of quantum numbers?This important question was answered by Pauli in 1925 in a powerful statementknown as the Pauli exclusion principle:No two electrons in an atom can ever have the same set of values for the set of quantumnumbers n, , m , and m s .zFigure 28.13 The sphericalelectron cloud for the hydrogenatom in its 1s state. The Pauli exclusion principlexThe Pauli exclusion principle explains the electronic structure of complex atomsas a succession of filled levels with different quantum numbers increasing in energy,where the outermost electrons are primarily responsible for the chemicalproperties of the element. If this principle weren’t valid, every electron would endup in the lowest energy state of the atom and the chemical behavior of the elementswould be grossly different. Nature as we know it would not exist—and wewould not exist to wonder about it!As a general rule, the order that electrons fill an atom’s subshell is as follows:once one subshell is filled, the next electron goes into the vacant subshell that islowest in energy. If the atom were not in the lowest energy state available to it, itwould radiate energy until it reached that state. A subshell is filled when it contains2(2 1) electrons. This rule is based on the analysis of quantum numbersto be described later. Following the rule, shells and subshells can contain numbersof electrons according to the pattern given in Table 28.3.The exclusion principle can be illustrated by an examination of the electronicarrangement in a few of the lighter atoms.Hydrogen has only one electron, which, in its ground state, can be described by1either of two sets of quantum numbers: 1, 0, 0, 2 or 1, 0, 0, 1 . The electronicconfiguration of this atom is often designated as 1s 1 2. The notation 1s refers to aTIP 28.3 The ExclusionPrinciple is More GeneralThe exclusion principle stated here isa limited form of the more generalexclusion principle, which states thatno two fermions (particles with spin1/2, 3/2, . . .) can be in the samequantum state.

918 Chapter 28 Atomic PhysicsWOLFGANG PAULI (1900–1958)An extremely talented Austrian theoreticalphysicist who made important contributionsin many areas of modern physics,Pauli gained public recognition at the ageof 21 with a masterful review article onrelativity that is still considered one of thefinest and most comprehensive introductionsto the subject. Other major contributionswere the discovery of the exclusionprinciple, the explanation of the connectionbetween particle spin and statistics,and theories of relativistic quantum electrodynamics,the neutrino hypothesis, andthe hypothesis of nuclear spin.CERN/Courtesy of AIP Emilio Segre Visual ArchivesTABLE 28.3Number of Electrons in Filled Subshells and ShellsNumber of Number ofElectrons in Electrons inShell Subshell Filled Subshell Filled ShellK (n 1) s( 0) 2 2s( 0) 2L (n 2)p( 1) 68s( 0) 2M (n 3) p( 1) 6 18}d( 2) 10s( 0) 2}p( 1) 6N (n 4)d( 2) 1032f( 3) 14state for which n 1 and 0, and the superscript indicates that one electron ispresent in this level.Neutral helium has two electrons. In the ground state, the quantum numbers for1these two electrons are 1, 0, 0, and 1, 0, 0, 1 22. No other possible combinations ofquantum numbers exist for this level, and we say that the K shell is filled. The heliumelectronic configuration is designated as 1s 2 .Neutral lithium has three electrons. In the ground state, two of these are in the1s subshell and the third is in the 2s subshell, because the latter is lower in energythan the 2p subshell. Hence, the electronic configuration for lithium is 1s 2 2s 1 .A list of electronic ground-state configurations for a number of atoms is providedin Table 28.4. In 1871 Dmitri Mendeleev (1834–1907), a Russian chemist,arranged the elements known at that time into a table according to their atomicmasses and chemical similarities. The first table Mendeleev proposed containedmany blank spaces, and he boldly stated that the gaps were there only becausethose elements had not yet been discovered. By noting the column in which thesemissing elements should be located, he was able to make rough predictions abouttheir chemical properties. Within 20 years of this announcement, the elementswere indeed discovered.The elements in our current version of the periodic table are still arranged sothat all those in a vertical column have similar chemical properties. For example,consider the elements in the last column: He (helium), Ne (neon), Ar (argon), Kr(krypton), Xe (xenon), and Rn (radon). The outstanding characteristic of theseelements is that they don’t normally take part in chemical reactions, joining withother atoms to form molecules, and are therefore classified as inert. Because ofthis “aloofness,” they are referred to as the noble gases. We can partially understandtheir behavior by looking at the electronic configurations shown in Table 28.4,page 919. The element helium has the electronic configuration 1s 2 . In otherwords, one shell is filled. The electrons in this filled shell are considerably separatedin energy from the next available level, the 2s level.The electronic configuration for neon is 1s 2 2s 2 2p 6 . Again, the outer shell isfilled and there is a large difference in energy between the 2p level and the 3slevel. Argon has the configuration 1s 2 2s 2 2p 6 3s 2 3p 6 . Here, the 3p subshell is filledand there is a wide gap in energy between the 3p subshell and the 3d subshell.Through all the noble gases, the pattern remains the same: a noble gas is formedwhen either a shell or a subshell is filled, and there is a large gap in energy beforethe next possible level is encountered.The elements in the first column of the periodic table are called the alkalimetals and are highly active chemically. Referring to Table 28.4, we can understandwhy these elements interact so strongly with other elements. All of these alkali}

28.9 The Exclusion Principle and the Periodic Table 919TABLE 28.4Electronic Configurations of Some ElementsGround-State Ionization Ground-State IonizationZ Symbol Configuration Energy (eV) Z Symbol Configuration Energy (eV)1 H 1s 1 13.595 19 K [Ar] 4s 1 4.3392 He 1s 2 24.581 20 Ca 4s 2 6.11121 Sc 3d4s 2 6.543 Li [He] 2s 1 5.390 22 Ti 3d 2 4s 2 6.834 Be 2s 2 9.320 23 V 3d 3 4s 2 6.745 B 2s 2 2p 1 8.296 24 Cr 3d 5 4s 1 6.766 C 2s 2 2p 2 11.256 25 Mn 3d 5 4s 2 7.4327 N 2s 2 2p 3 14.545 26 Fe 3d 6 4s 2 7.878 O 2s 2 2p 4 13.614 27 Co 3d 7 4s 2 7.869 F 2s 2 2p 5 17.418 28 Ni 3d 8 4s 2 7.63310 Ne 2s 2 2p 6 21.559 29 Cu 3d 10 4s 1 7.72430 Zn 3d 10 4s 2 9.39111 Na [Ne] 3s 1 5.138 31 Ga 3d 10 4s 2 4p 1 6.0012 Mg 3s 2 7.644 32 Ge 3d 10 4s 2 4p 2 7.8813 Al 3s 2 3p 1 5.984 33 As 3d 10 4s 2 4p 3 9.8114 Si 3s 2 3p 2 8.149 34 Se 3d 10 4s 2 4p 4 9.7515 P 3s 2 3p 3 10.484 35 Br 3d 10 4s 2 4p 5 11.8416 S 3s 2 3p 4 10.357 36 Kr 3d 10 4s 2 4p 6 13.99617 Cl 3s 2 3p 5 13.0118 Ar 3s 2 3p 6 15.755Note: The bracket notation is used as a shorthand method to avoid repetition in indicating inner-shell electrons. Thus, [He] represents 1s 2 , [Ne] represents 1s 2 2s 2 2p 6 ,[Ar] represents 1s 2 2s 2 2p 6 3s 2 3p 6 , and so on.metals have a single outer electron in an s subshell. This electron is shielded fromthe nucleus by all the electrons in the inner shells. Consequently, it’s only looselybound to the atom and can readily be accepted by other atoms that bind it moretightly to form molecules.The elements in the seventh column of the periodic table are called thehalogens and are also highly active chemically. All these elements are lacking oneelectron in a subshell, so they readily accept electrons from other atoms to formmolecules.Quick Quiz 28.4Krypton (atomic number 36) has how many electrons in its next to outer shell(n 3)?(a) 2 (b) 4 (c) 8 (d) 18Applying Physics 28.3The Periodic TableScanning from left to right across one row of theperiodic table, the effective size of the atoms firstdecreases and then increases. What would cause thisbehavior?Explanation Starting on the left side of the periodictable and moving toward the middle, the nuclearcharge is increasing. As a result, there is an increasingCoulomb attraction between the nucleus and theelectrons, and the electrons are pulled into an averageposition that is closer to the nucleus. From the middleof the row to the right side, the increasing number ofelectrons being placed in proximity to each otherresults in a mutual repulsion that increases theaverage distance from the nucleus and causes theatomic size to grow.

920 Chapter 28 Atomic PhysicsIntensity30KβKα40 50 60 70 80 90λ, pmFigure 28.14 The x-ray spectrumof a metal target consists of a broadcontinuous spectrum (bremstrahlung)plus a number of sharp lines that aredue to characteristic x-rays. The datashown were obtained when 35-keVelectrons bombarded a molybdenumtarget. Note that 1 pm 10 12 m 0.001 nm.1/λFigure 28.15 A Moseley plotof versus Z, where is thewavelength of the K x-ray line ofthe element of atomic number Z.√1/Z28.10 CHARACTERISTIC X-RAYSX-rays are emitted when a metal target is bombarded with high-energy electrons.The x-ray spectrum typically consists of a broad continuous band and a series of intensesharp lines that are dependent on the type of metal used for the target, asshown in Figure 28.14. These discrete lines, called characteristic x-rays, were discoveredin 1908, but their origin remained unexplained until the details of atomicstructure were developed.The first step in the production of characteristic x-rays occurs when a bombardingelectron collides with an electron in an inner shell of a target atom with sufficientenergy to remove the electron from the atom. The vacancy created in theshell is filled when an electron in a higher level drops down into the lower energylevel containing the vacancy. The time it takes for this to happen is very short, lessthan 10 9 s. The transition is accompanied by the emission of a photon with energyequaling the difference in energy between the two levels. Typically, the energyof such transitions is greater than 1 000 eV, and the emitted x-ray photons havewavelengths in the range of 0.01 nm to 1 nm.We assume that the incoming electron has dislodged an atomic electron fromthe innermost shell, the K shell. If the vacancy is filled by an electron droppingfrom the next higher shell, the L shell, the photon emitted in the process is referredto as the K line on the curve of Figure 28.14. If the vacancy is filled by anelectron dropping from the M shell, the line produced is called the K line.Other characteristic x-ray lines are formed when electrons drop from upperlevels to vacancies other than those in the K shell. For example, L lines are producedwhen vacancies in the L shell are filled by electrons dropping from highershells. An L line is produced as an electron drops from the M shell to the L shell,and an L line is produced by a transition from the N shell to the L shell.We can estimate the energy of the emitted x-rays as follows: consider two electronsin the K shell of an atom whose atomic number is Z. Each electron partiallyshields the other from the charge of the nucleus, Ze, so each is subject to an effectivenuclear charge Z eff (Z 1)e. We can now use a modified form of Equation28.18 to estimate the energy of either electron in the K shell (with n 1). We havewhere E 0 is the ground-state energy. Substituting Z eff Z 1 givesE K (Z 1) 2 (13.6 eV) [28.20]As Example 28.5 will show, we can estimate the energy of an electron in an L or anM shell in a similar fashion. Taking the energy difference between these two levels,we can then calculate the energy and wavelength of the emitted photon.In 1914, Henry G. J. Moseley plotted the Z values for a number of elementsagainst , where l is the wavelength of the K line for each element. He foundthat such a plot produced a straight line, as in Figure 28.15. This is consistent withour rough calculations of the energy levels based on Equation 28.20. From hisplot, Moseley was able to determine the Z values of other elements, providing aperiodic chart in excellent agreement with the known chemical properties of theelements.√1/E K m e Z 2 effk e 2 e 42 2 Z 2 eff E 0EXAMPLE 28.5 Characteristic X-RaysGoal Calculate the energy and wavelength of characteristic x-rays.Problem Estimate the energy of the characteristic x-ray emitted from a tungsten target when an electron dropsfrom an M shell (n 3 state) to a vacancy in the K shell (n 1 state).Strategy Develop two estimates, one for the electron in the K shell (n 1) and one for the electron in the M shell(n 3). For the K-shell estimate, we can use Equation 28.20. For the M shell, we need a new equation. There is one

28.11 Atomic Transitions 921electron in the K shell (because one is missing) and 8 in the L shell, making 9 electrons shielding the nuclearcharge. This means Z eff 74 9 and E M Z eff 2 E 3 , where E 3 is the energy of the n 3 level in hydrogen. Thedifference E M E K is the energy of the photon.SolutionUse Equation 28.20 to estimate the energy of an electronin the K shell of tungsten, atomic number Z 74:Estimate the energy of an electron in the M shell in thesame way:Calculate the difference in energy between the M and Kshells:Find the wavelength of the emitted light:E K (74 1) 2 (13.6 eV) 72 500 eVE M Z 2 eff E 3 (Z 9) 2 E 0(13.6 eV)32 (74 9)296 380 eVE M E K 6 380 eV ( 72 500 eV) 66 100 eVE hf hc: hcE (6.63 1034 Js)(3.00 10 8 m/s)(6.61 10 4 eV)(1.60 10 19 J/eV) 1.88 10 11 m 0.018 8 nmExercise 28.5Repeat the problem for a 2p electron transiting from the L shell to the K shell. (For technical reasons, the L shellelectron must have 1, so a single 1s electron and two 2s electrons shield the nucleus.)Answer (a) 5.54 10 4 eV (b) 0.022 4 nm28.11 ATOMIC TRANSITIONSWe have seen that an atom will emit radiation only at certain frequencies that correspondto the energy separation between the various allowed states. Consider anatom with many allowed energy states, labeled E 1 , E 2 , E 3 , . . . , as in Figure 28.16.When light is incident on the atom, only those photons whose energy hf matchesthe energy separation E between two levels can be absorbed by the atom. Aschematic diagram representing this stimulated absorption process is shown in ActiveFigure 28.17. At ordinary temperatures, most of the atoms in a sample are inthe ground state. If a vessel containing many atoms of a gas is illuminated with alight beam containing all possible photon frequencies (that is, a continuous spectrum),only those photons of energies E 2 E 1 , E 3 E 1 , E 4 E 1 , and so on, canbe absorbed. As a result of this absorption, some atoms are raised to various allowedhigher energy levels, called excited states.Once an atom is in an excited state, there is a constant probability that it will jumpback to a lower level by emitting a photon, as shown in Active Figure 28.18 (page 922).E 4E 3E 2E 1Figure 28.16 Energy level diagramof an atom with various allowedstates. The lowest energy state, E 1 ,is the ground state. All others areexcited states.hfAtom inground state∆EE 2Atom inexcited stateE 2ACTIVE FIGURE 28.17Diagram representing the process ofstimulated absorption of a photon by an atom.The blue dot represents an electron. Theelectron is transferred from the ground stateto the excited state when the atom absorbs aphoton of energy hf E 2 E 1 .BeforeE 1 E 1AfterLog into PhysicsNow at www.cp7e.com andgo to Active Figure 28.17 to observestimulated absorption.

922 Chapter 28 Atomic PhysicsAtom inexcited stateAtom inground stateAtom inexcited stateAtom inground stateE 2E 2E 2E 2∆Ehf = ∆Ehf = ∆E∆EhfE 1 E 1BeforeAfterACTIVE FIGURE 28.18Diagram representing the process ofspontaneous emission of a photon by anatom that is initially in the excitedstate E 2 . When the electron falls tothe ground state, the atom emits aphoton of energy hf E 2 E 1 .E 1 E 1BeforeAfterACTIVE FIGURE 28.19Diagram representing the process of stimulated emission of a photon by an incoming photon of energyhf. Initially, the atom is in the excited state. The incoming photon stimulates the atom to emit a secondphoton of energy hf E 2 E 1 .hfLog into PhysicsNow atwww.cp7e.com and go to ActiveFigure 28.18 to observe spontaneousemission.Log into PhysicsNow at www.cp7e.com and go to Active Figure 28.19 to observe stimulated emission.This process is known as spontaneous emission. Typically, an atom will remain in anexcited state for only about 10 8 s.A third process that is important in lasers, stimulated emission, was predicted byEinstein in 1917. Suppose an atom is in the excited state E 2 , as in Active Figure28.19, and a photon with energy hf E 2 E 1 is incident on it. The incoming photonincreases the probability that the excited atom will return to the ground stateand thereby emit a second photon having the same energy hf. Note that twoidentical photons result from stimulated emission: the incident photon and theemitted photon. The emitted photon is exactly in phase with the incident photon. Thesephotons can stimulate other atoms to emit photons in a chain of similar processes.The many photons produced in this fashion are the source of the intense, coherent(in-phase) light in a laser.Applying Physics 28.4Streaking MeteoroidsA physics student is watching a meteor shower in theearly morning hours. She notices that the streaks oflight from the meteoroids entering the very highregions of the atmosphere last for as long as 2 or 3seconds before fading. She also notices a lightningstorm off in the distance. The streaks of light from thelightning fade away almost immediately after the flash,certainly in much less than 1 second. Both lightningand meteors cause the air to turn into a plasma becauseof the very high temperatures generated. Thelight is given off when the stripped electrons in theplasma recombine with the ionized atoms. Why wouldthe light last longer for meteors than for lightning?Explanation To answer this question, we examinethe phrase “the streaks of light from the meteoroidsentering the very high regions of the atmosphere.” Inthe very high regions of the atmosphere, the pressure isvery low, so the density is also very low and the atoms ofthe gas are relatively far apart. Low density means thatafter the air is ionized by the passing meteoroid, theprobability of freed electrons finding an ionized atomwith which to recombine is relatively low. As a result, therecombination process occurs over a relatively long time,measured in seconds. Lightning, however, occurs in thelower regions of the atmosphere (the troposphere),where the pressure and density are relatively high. Afterthe ionization by the lightning flash, the electrons andionized atoms are much closer together than in theupper atmosphere. The probability of a recombinationis accordingly much higher, and the time for therecombination to occur is much shorter.28.12 LASERS AND HOLOGRAPHYWe have described how an incident photon can cause atomic transitions either upward(stimulated absorption) or downward (stimulated emission). The twoprocesses are equally probable. When light is incident on a system of atoms, there

28.12 Lasers and Holography 923is usually a net absorption of energy, because when the system is in thermal equilibrium,there are many more atoms in the ground state than in excited states.However, if the situation can be inverted so that there are more atoms in an excitedstate than in the ground state, a net emission of photons can result. Such acondition is called population inversion. This is the fundamental principle involvedin the operation of a laser, an acronym for light amplification by stimulatedemission of r adiation. The amplification corresponds to a buildup of photons inthe system as the result of a chain reaction of events. The following three conditionsmust be satisfied in order to achieve laser action:1. The system must be in a state of population inversion (that is, more atoms inan excited state than in the ground state).2. The excited state of the system must be a metastable state, which means its lifetimemust be long compared with the otherwise usually short lifetimes of excitedstates. When that is the case, stimulated emission will occur before spontaneousemission.3. The emitted photons must be confined within the system long enough to allowthem to stimulate further emission from other excited atoms. This is achievedby the use of reflecting mirrors at the ends of the system. One end is totally reflecting,and the other is slightly transparent to allow the laser beam to escape.One device that exhibits stimulated emission of radiation is the helium–neongas laser. Figure 28.20 is an energy-level diagram for the neon atom in this system.The mixture of helium and neon is confined to a glass tube sealed at the ends bymirrors. A high voltage applied to the tube causes electrons to sweep through it,colliding with the atoms of the gas and raising them into excited states. Neonatoms are excited to state E 3 * through this process and also as a result of collisionswith excited helium atoms. When a neon atom makes a transition to state E 2 , itstimulates emission by neighboring excited atoms. This results in the productionof coherent light at a wavelength of 632.8 nm. Figure 28.21 summarizes the stepsin the production of a laser beam.ENERGYE 3 *E 2Metastable stateλ = 632.8 nmOutputenergyInputenergyE 1Figure 28.20 Energy-level diagramfor the neon atom in a helium–neonlaser. The atom emits 632.8-nm photonsthrough stimulated emission inthe transition E * 3 : E 2 . This is thesource of coherent light in the laser.hfSpontaneous emission—random directionsLaseroutputMirror oneStimulating waveon axisEnergy input(a)Mirror two(b)Courtesy of HRL Laboratories LLC, Malibu, CAFigure 28.21 (a) Steps in theproduction of a laser beam. The tubecontains atoms, which represent theactive medium. An external source ofenergy (optical, electrical, etc.) isneeded to “pump” the atoms toexcited energy states. The parallelend mirrors provide the feedback ofthe stimulating wave. (b) Photographof the first ruby laser, showing theflash lamp surrounding the ruby rod.

924 Chapter 28 Atomic PhysicsM 2M 1LaserL 2BL 1FilmCourtesy of Central Scientific Company(a)(b)Figure 28.22 (a) Experimental arrangement for producing a hologram. (b) Photograph of a hologrammade with a cylindrical film. Note the detail of the Volkswagen image.APPLICATIONLaser TechnologySince the development of the first laser in 1960, laser technology has exhibitedtremendous growth. Lasers that cover wavelengths in the infrared, visible, andultraviolet regions of the spectrum are now available. Applications include thesurgical “welding” of detached retinas, “lasik” surgery, precision surveying andlength measurement, a potential source for inducing nuclear fusion reactions,precision cutting of metals and other materials, and telephone communicationalong optical fibers. These and other applications are possible because of theunique characteristics of laser light. In addition to being highly monochromaticand coherent, laser light is also highly directional and can be sharply focused toproduce regions of extremely intense light energy.Scientist checking the performanceof an experimental laser-cuttingdevice mounted on a robot arm. Thelaser is being used to cut through ametal plate.APPLICATIONHolographyPhilippe Plailly/Photo Researchers, Inc.HolographyOne interesting application of the laser is holography: the production of threedimensionalimages of objects. Figure 28.22a shows how a hologram is made.Light from the laser is split into two parts by a half-silvered mirror at B. One partof the beam reflects off the object to be photographed and strikes an ordinaryphotographic film. The other half of the beam is diverged by lens L 2 , reflects frommirrors M 1 and M 2 , and finally strikes the film. The two beams overlap to form anextremely complicated interference pattern on the film, one that can be producedonly if the phase relationship of the waves is constant throughout the exposure ofthe film. This condition is met through the use of light from a laser, because suchlight is coherent. The hologram records not only the intensity of the light scatteredfrom the object (as in a conventional photograph), but also the phase differencebetween the reference beam and the beam scattered from the object. Becauseof this phase difference, an interference pattern is formed that produces animage with full three-dimensional perspective.A hologram is best viewed by allowing coherent light to pass through the developedfilm while you look back along the direction from which the beam comes.Figure 28.22b is a photograph of a hologram made using a cylindrical film.28.13 ENERGY BANDS IN SOLIDSIn this section we trace the changes that occur in the discrete energy levels ofisolated atoms when the atoms group together and form a solid. We find that insolids, the discrete levels of isolated atoms broaden into allowed energy bands

28.13 Energy Bands in Solids 925Energy2s 2s 2s1sEnergy1sEnergyEquilibriumseparation1sFigure 28.23 (a) Splitting of the1s and 2s states when two atoms arebrought together. (b) Splitting of the1s and 2s states when five atoms arebrought close together. (c) Formationof energy bands when a large numberof sodium atoms are assembled toform a solid.rrr(a)(b)r 0(c)separated by forbidden gaps. The separation and electron population of thehighest bands determines whether a given solid is a conductor, an insulator, or asemiconductor.Consider two identical atoms, initially widely separated, that are brought closerand closer together. If two identical atoms are very far apart, they do not interact,and their electronic energy levels can be considered to be those of isolated atoms.Hence, the energy levels are exactly the same. As the atoms come close together,they essentially become one quantum system, and the Pauli exclusion principle demandsthat the electrons be in different quantum states for this single system. Theexclusion principle manifests itself as a changing or splitting of electron energylevels that were identical in the widely separated atoms, as shown in Figure 28.23a.Figure 28.23b shows that with 5 atoms, each energy level in the isolated atom splitsinto five different, more closely spaced levels.If we extend this argument to the large number of atoms found in solids (onthe order of 10 23 atoms/cm 3 ), we obtain a large number of levels so closely spacedthat they may be regarded as a continuous band of energy levels, as in Figure28.23c. An electron can have any energy within an allowed energy band, but cannothave an energy in the band gap, or the region between allowed bands. Notethat the band gap energy E g is indicated in Figure 28.23c. In practice we are onlyinterested in the band structure of a solid at some equilibrium separation of itsatoms r 0 , and so we remove the distance scale on the x-axis and simply plot the allowedenergy bands of a solid as a series of horizontal bands, as shown in Figure28.24 for sodium.3pConductors and InsulatorsFigure 28.24 shows that the band structure of a particular solid is quite complicatedwith individual atomic levels broadening by varying amounts and some levels(3s and 3p) broadening so much that they overlap. Nevertheless, it is possible togain a qualitative understanding of whether a solid is a conductor, an insulator, ora semiconductor by considering only the structure of the upper or upper two energybands and whether they are occupied by electrons.Deciding whether an energy band is empty (unoccupied by electrons), partiallyfilled, or full is carried out in basically the same way as for the energy-level populationof atoms: we distribute the total number of electrons from the lowest energylevels up in a way consistent with the exclusion principle. While we omit the detailsof this process here, one important case is that shown in Figure 28.25a (page 926),where the highest-energy occupied band is only partially full. The other importantcase, where the highest occupied band is completely full, is shown in Figure 28.25b.Notice that this figure also shows that the highest filled band is called the valenceband and the next higher empty band is called the conduction band. The energyband gap, which varies with the solid, is also indicated as the energy difference E gbetween the top of the valence band and the bottom of the conduction band.3s2p2s1sFigure 28.24 Energy bands ofsodium. Note the energy gaps (whiteregions) between the allowed bands;electrons can’t occupy states that liein these forbidden gaps. Bluerepresents energy bands occupied bythe sodium electrons when the atomis in its ground state. Gold representsenergy bands that are empty. Notethat the 3s and 3p levels broaden somuch that they overlap.

926 Chapter 28 Atomic PhysicsFigure 28.25 (a) Half-filled bandof a metal, an electrical conductor.(b) An electrical insulator at T 0Khas a filled valence band and anempty conduction band. (c) Bandstructure of a semiconductor atordinary temperatures (T 300 K).The energy gap is much smaller thanin an insulator, and many electronsoccupy states in the conduction band.Conduction bandEnergy gapE gE gConduction bandValence bandValence bandMetal(a)InsulatorE g 10 eV(b)SemiconductorE g 1 eV(c)With these ideas and definitions we are now in a position to understandwhat determines, quantum mechanically, whether a solid will be a conductor or aninsulator. When a modest voltage is applied to a good conductor, the electrons accelerateand gain energy. In quantum terms, electron energies increase if there arehigher unoccupied energy levels for electrons to jump to. For example, electrons near thetop of the partially filled band in sodium need to gain very little energy from theapplied voltage to reach one of the nearby, closely spaced, empty states. Thus, it iseasy for a small voltage to kick electrons into higher energy states, and chargeflows easily in sodium, an excellent conductor.Now consider the case of a material in which the highest occupied band is completelyfull of electrons and there is a band gap separating this filled valence bandfrom the vacant conduction band, as in Figure 28.25b. A typical case might be diamond(carbon), in which the band gap is about 10 eV. When a voltage is applied,electrons can’t easily gain energy, because there are no vacant energy states nearbyto which electrons can make transitions. Because the only empty band is the conductionband, an electron must gain an amount of energy at least equal to theband gap in order for it to move through the solid. This large amount of energycan’t be supplied by a modest applied voltage, so no charge flows and diamond isa good insulator. In summary then, a conductor has a highest-energy occupiedband which is partially filled, and in an insulator, has a highest-energy occupiedband which is completely filled with a large energy gap between the valence and conductionbands.SemiconductorsTo this point, we have completely ignored the influence of temperature on theelectronic populations of energy bands. Recalling that the average thermal energyof a particle at temperature T is 3k B T/2, we find that an electron at room temperaturehas an average energy of about 0.04 eV. Because this energy is about 100 timessmaller than the band gap in a typical insulator, very few electrons would haveenough random thermal energy to jump the energy gap in an insulator and contributeto conduction. However things are different for a semiconductor. As we seein Figure 28.25c, a semiconductor is a material with a small band gap of about1 eV whose conductivity results from appreciable thermal excitation of electronsacross the gap into the conduction band at room temperature. The most commonlyused semiconductors are silicon and gallium arsenide, with band gaps of1.14 eV and 1.43 eV, respectively, at 300 K. As you might expect, the resistivity ofsemiconductors usually decreases with increasing temperature, because k B T becomesa larger fraction of the band gap energy.It is interesting that the electrons in the conduction band of a semiconductordon’t carry the entire current when a voltage is applied, as Figure 28.26 shows.(It might be said that conduction electrons do not constitute the “whole” story.)The missing electrons in the valence band, shown as a narrow white band in the

28.13 Energy Bands in Solids 927EnergyConduction electronselectronsholesConduction bandFigure 28.26 Movement ofcharges (holes and electrons) in asemiconductor. The electrons movein the direction opposite the directionof the external electric field, andthe holes move in the direction of thefield.Narrow forbidden gapValence bandApplied E fieldfigure, provide a few empty states called holes for valence band electrons to fill; sosome electrons in the valence band can gain energy and move towards a positiveelectrode and thus also carry the current. Since the valence band electrons that fillholes leave behind other holes, it is equally valid and more common to view theconduction process in the valence band as a flow of positive holes towards the negativeelectrode applied to a semiconductor. Thus, a pure semiconductor, such assilicon, can be viewed in a symmetric way: silicon has equal numbers of mobileelectrons in the conduction band and holes in the valence band. Furthermore,when an external voltage is applied to the semiconductor, electrons move towardthe positive electrode and holes move toward the negative electrode. In the nextsection we will look at the concepts of an electron and a hole in a simpler, moregraphic way as the presence or absence of an outer-shell electron at a particularlocation in a crystal lattice.When small amounts of impurities are added to a semiconductor such assilicon (about one impurity atom per 10 7 silicon atoms), both the band structureof the semiconductor and its resistivity are modified. The process of addingimpurities, called doping, is important in making devices having well-definedregions of different resistivity. For example, when an atom containing fiveouter-shell electrons, such as arsenic, is added to a semiconductor such as silicon,four of the arsenic electrons form shared bonds with atoms of the semiconductorand one is left over. This extra electron is nearly free of its parent atomand has an energy level that lies in the energy gap, just below the conductionband. Such a pentavalent atom in effect donates an electron to the structureand hence is referred to as a donor atom. Because the spacing between the energylevel of the electron of the donor atom and the bottom of the conductionband is very small (typically, about 0.05 eV), only a small amount of thermalenergy is needed to cause this electron to move into the conduction band.(Recall that the average thermal energy of an electron at room temperature is3k B T/2 0.04 eV). Semiconductors doped with donor atoms are called n-typesemiconductors, because the charge carriers are electrons, the charge of whichis negative.If a semiconductor is doped with atoms containing three outer-shell electrons,such as aluminum, the three electrons form shared bonds with neighboring semiconductoratoms, leaving an electron deficiency—a hole—where the fourth bondwould be if an impurity-atom electron was available to form it. The energy level ofthis hole lies in the energy gap, just above the valence band. An electron from thevalence band has enough energy at room temperature to fill that impurity level,leaving behind a hole in the valence band. Because a trivalent atom, in effect, acceptsan electron from the valence band, such impurities are referred to as acceptoratoms. A semiconductor doped with acceptor impurities is known as a p-typesemiconductor, because the majority of charge carriers are positively chargedholes.

928 Chapter 28 Atomic Physics28.14 SEMICONDUCTOR DEVICESThe p–n JunctionNow let us consider what happens when a p -semiconductor is joined to ann -semiconductor to form a p–n junction. The junction consists of the threedistinct regions shown in Figure 28.27a: a p -region, a depletion region, and ann -region.The depletion region, which extends several micrometers to either side of thecenter of the junction, may be visualized as arising when the two halves of thejunction are brought together. Mobile donor electrons from the n side nearest thejunction (the blue area in Fig. 28.27a) diffuse to the p side, leaving behind immobilepositive ions. At the same time, holes from the p side nearest the junction diffuseto the n side and leave behind a region (the red area in Fig. 28.27a) of fixednegative ions. The depletion region is so named because it is depleted of mobilecharge carriers.The depletion region contains an internal electric field (arising from thecharges of the fixed ions) on the order of 10 4 to 10 6 V/cm. This field sweeps mobilecharge out of the depletion region and keeps it truly depleted. This internalelectric field creates an internal potential difference V 0 that prevents further diffusionof holes and electrons across the junction and thereby ensures zero currentin the junction when no external potential difference is applied.Perhaps the most notable feature of the p–n junction is its ability to pass currentin only one direction. Such diode action is easiest to understand in terms ofthe potential-difference graph shown in Figure 28.27c. If an external voltage V isapplied to the junction such that the p side is connected to the positive terminal ofa voltage source as in Figure 28.27a, the internal potential difference V 0 acrossthe junction is decreased, resulting in a current that increases exponentially withincreasing forward voltage, or forward bias. In reverse bias (where the n side of thejunction is connected to the positive terminal of a voltage source), the internalpotential difference V 0 increases with increasing reverse bias. This results in avery small reverse current that quickly reaches a saturation value I 0 . The current–voltage relationship for an ideal diode isI I 0 (e q V/k BT 1)[28.21]Depletion regionFixed ioncores(a)p––––+ +E+ +n––+ ++E(b)0x–Figure 28.27 (a) Physicalarrangement of a p–n junction.(b) Internal electric field versus x forthe p–n junction. (c) Internal electricpotential V versus x for the p–njunction. V 0 represents the potentialdifference across the junction in theabsence of an applied electric field.(c)∆V∆V 0x

28.14 Semiconductor Devices 929I (mA)504030IAp∆V+ –Forward bias(a)nI 0 20 mA2010–1.0 –0.5Reverse bias(b)0.5 1.0Forward biasFigure 28.28 (a) Schematic of a p–n junction under forward bias. (b) The characteristic curve for areal p–n junction.∆V (V)where q is the electron charge, k B is Boltzmann’s constant, and T is the temperaturein kelvins. Figure 28.28 shows an I–V plot characteristic of a real p–n junction,along with a schematic of such a device under forward bias.The most common use of the semiconductor diode is as a rectifier, a device thatchanges 120-V AC voltage supplied by the power company to, say the 12-V DC voltageneeded by your music keyboard. We can understand how a diode rectifies acurrent by considering Figure 28.29a, which shows a diode connected in serieswith a resistor and an AC source. Because appreciable current can pass throughthe diode in just one direction, the alternating current in the resistor is reduced tothe form shown in Figure 28.29b. The diode is said to act as a half-wave rectifier,because there is current in the circuit during only half of each cycle.Figure 28.30a shows a circuit that lowers the AC voltage to 12 V with a stepdowntransformer and then rectifies both halves of the 12-V AC. Such a rectifier iscalled a full-wave rectifier and when combined with a step-down transformer is themost common DC power supply around the home today. A capacitor added in parallelwith the load will yield an even steadier DC voltage.I(a)(b)Figure 28.29 (a) A diode in serieswith a resistor allows current to passin only one direction. (b) Thecurrent versus time for the circuitin (a).RtThe Junction TransistorThe invention of the transistor by John Bardeen (1908–1991), Walter Brattain(1902–1987), and William Shockley (1910–1989) in 1948 totally revolutionizedthe world of electronics. For this work, these three men shared a Nobel prize in1956. By 1960, the transistor had replaced the vacuum tube in many electronic applications.The advent of the transistor created a multitrillion-dollar industry thatproduced such popular devices as pocket radios, handheld calculators, computers,television receivers, and electronic games. In this section we explain how a transistoracts as an amplifier to boost the tiny voltages and currents generated in a microphoneto the ear-splitting levels required to drive a speaker.One simple form of the transistor, called the junction transistor, consists of asemiconducting material in which a very narrow n region is sandwiched betweentwo p regions. This configuration is called a pnp transistor. Another configurationis the npn transistor, which consists of a p region sandwiched between two n regions.Because the operation of the two transistors is essentially the same, we describeonly the pnp transistor. The structure of the pnp transistor, together with itscircuit symbol, is shown in Figure 28.31 (page 930). The outer regions are calledthe emitter and collector, and the narrow central region is called the base.The configuration contains two junctions: the emitter–base interface and thecollector–base interface.IABCD 1Transformer D 2(a)(b)Figure 28.30 (a) A full-wave rectifiercircuit. (b) The current versustime in the resistor R.ItR

930 Chapter 28 Atomic PhysicsEmitter Base Collector–++–∆V eb+ –p––––+ +n+ ++ ++ +––––pEmitterCollectorI bI bbep n pI eI ccR(a)(b)Base+ –∆V ec(c)Figure 28.31 (a) The pnp transistor consists of an n region (base) sandwiched between twop regions (emitter and collector). (b) Circuit symbol for the pnp transistor. (c) A bias voltage V ebapplied to the base as shown produces a small base current I b that is used to control the collectorcurrent I c in a pnp transistor.Suppose a voltage is applied to the transistor so that the emitter is at a higherelectric potential than the collector. (This is accomplished with the battery labeledV ec in Figure 28.31c.) If we think of the transistor as two diodes back to back, wesee that the emitter–base junction is forward biased and the base–collector junctionis reverse biased. The emitter is heavily doped relative to the base, and as a result,nearly all the current consists of holes moving across the emitter–base junction.Most of these holes do not recombine with electrons in the base because it isvery narrow. Instead they are accelerated across the reverse-biased base–collectorjunction, producing the emitter current I e in Figure 28.31c.Although only a small percentage of holes recombine in the base, those that dolimit the emitter current to a small value because positive charge carriers accumulatingin the base prevent holes from flowing in. In order not to limit the emittercurrent, some of the positive charge on the base must be drawn off; this is accomplishedby connecting the base to the battery labeled V eb in Figure 28.31c. Thosepositive charges that are not swept across the base–collector junction leave thebase through this added pathway. This base current I b is very small, but a smallchange in it can significantly change the collector current I c . If the transistor isproperly biased, the collector (output) current is directly proportional to the base(input) current and the transistor acts as a current amplifier. This condition maybe writtenI c I bwhere , the current gain factor, is typically in the range from 10 to 100. Thus, thetransistor may be used to amplify a small signal. The small voltage to be amplifiedis placed in series with the battery V eb . The input signal produces a small variationin the base current, resulting in a large change in the collector current and hencea large change in the voltage across the output resistor.The Integrated CircuitInvented independently by Jack Kilby (b. 1923) at Texas Instruments in late 1958and by Robert Noyce at Fairchild Camera and Instrument in early 1959, the integratedcircuit has been justly called “the most remarkable technology ever to hitmankind.” Kilby’s first device is shown in Figure 28.32a. Integrated circuits have indeedstarted a “second industrial revolution” and are found at the heart of computers,watches, cameras, automobiles, aircraft, robots, space vehicles, and all sortsof communication and switching networks.In simplest terms, an integrated circuit is a collection of interconnected transistors,diodes, resistors, and capacitors fabricated on a single piece of siliconknown as a chip. State-of-the-art chips easily contain several million components in

Summary 931(a)Figure 28.32 (a) Jack Kilby’s first integrated circuit was tested on September 12, 1958.(b) Integrated circuits continue to shrink in size and price while simultaneously growing incapability.Courtesy of Texas Instruments, Inc(b)Courtesy of Intel Corporationa 1-cm 2 area, with the number of components per square inch having doubledevery year since the integrated circuit was invented.Integrated circuits were invented partly to solve the interconnection problemspawned by the transistor. In the era of vacuum tubes, power and size considerationsof individual components set significant limits on the number of componentsthat could be interconnected in a given circuit. With the advent of the tiny,low-power, highly reliable transistor, design limits on the number of componentsdisappeared and were replaced by the problem of wiring together hundreds ofthousands of components. The magnitude of this problem can be appreciatedwhen we consider that second-generation computers (consisting of discrete transistorsrather than integrated circuits) contained several hundred thousand componentsrequiring more than a million hand-soldered joints to be made andtested.In addition to solving the interconnection problem, integrated circuits possessthe advantages of miniaturization and fast response, two attributes critical forhigh-speed computers. The fast response results from the miniaturization andclose packing of components, because the response time of a circuit depends onthe time it takes for electrical signals traveling at about the speed of light to passfrom one component to another. This time is clearly reduced by packing componentsclosely.SUMMARYTake a practice test by logging intoPhysicsNow at www.cp7e.com and clicking on the Pre-Testlink for this chapter.28.3 The Bohr Theory of Hydrogen &28.4 Modification of the Bohr TheoryThe Bohr model of the atom is successful in describingthe spectra of atomic hydrogen and hydrogenlike ions.One of the basic assumptions of the model is that theelectron can exist only in certain orbits such that its angularmomentum mvr is an integral multiple of , where is Planck’s constant divided by 2. Assuming circularorbits and a Coulomb force of attraction between electronand proton, the energies of the quantum states forhydrogen areE n m ek e 2 e 42 2 1 n 2n 1, 2, 3, . . . [28.12]where k e is the Coulomb constant, e is the charge onthe electron, and n is an integer called a quantumnumber.

932 Chapter 28 Atomic PhysicsIf the electron in the hydrogen atom jumps from anorbit having quantum number n i to an orbit havingquantum number n f , it emits a photon of frequency f,given byf m ek e 2 e 44 3 1n f2 1n i2[28.14]Bohr’s correspondence principle states that quantummechanics is in agreement with classical physics whenthe quantum numbers for a system are very large.The Bohr theory can be generalized to hydrogen-likeatoms, such as singly ionized helium or doubly ionizedlithium. This modification consists of replacing e 2 by Ze 2wherever it occurs.28.6 Quantum Mechanics and theHydrogen Atom &28.7 The Spin Magnetic QuantumNumberOne of the many successes of quantum mechanics isthat the quantum numbers n, , and m associated withatomic structure arise directly from the mathematics ofthe theory. The quantum number n is called the principalquantum number, is the orbital quantum number,and m is the orbital magnetic quantum number. Thesequantum numbers can take only certain values: 1 n in integer steps, 0 n 1, and m . Inaddition, a fourth quantum number, called the spinmagnetic quantum number m s , is needed to explain afine doubling of lines in atomic spectra, with m s 21 .28.9 The Exclusion Principle and thePeriodic TableAn understanding of the periodic table of the elementsbecame possible when Pauli formulated the exclusionprinciple, which states that no two electrons in an atomin the same atom can have the same values for the setof quantum numbers n, , m , and m s . A particular setof these quantum numbers is called a quantum state.The exclusion priniciple explains how different energylevels in atoms are populated. Once one subshell isfilled, the next electron goes into the vacant subshellthat is lowest in energy. Atoms with similar configurationsin their outermost shell have similar chemicalproperties and are found in the same column of the periodictable.28.10 Characteristic X-RaysCharacteristic x-rays are produced when a bombardingelectron collides with an electron in an inner shell of anatom with sufficient energy to remove the electron fromthe atom. The vacancy is filled when an electron from ahigher level drops down into the level containing the vacancy,emitting a photon in the x-ray part of the spectrumin the process.28.11 Atomic Transitions &28.12 Lasers and HolographyWhen an atom is irradiated by light of all different wavelengths,it will only absorb only wavelengths equal to thedifference in energy of two of its energy levels. This phenomenon,called stimulated absorption, places anatom’s electrons into excited states. Atoms in an excitedstate have a probability of returning to a lower level ofexcitation by spontaneous emission. The wavelengthsthat can be emitted are the same as the wavelengths thatcan be absorbed. If an atom is in an excited state and aphoton with energy hf E 2 E 1 is incident on it, theprobability of emission of a second photon of this energyis greatly enhanced. The emitted photon is exactlyin phase with the incident photon. This process is calledstimulated emission. The emitted and original photoncan then stimulate more emission, creating an amplifyingeffect.Lasers are monochromatic, coherent light sourcesthat work on the principle of stimulated emission of radiationfrom a system of atoms.CONCEPTUAL QUESTIONS1. In the hydrogen atom, the quantum number n can increasewithout limit. Because of this, does the frequencyof possible spectral lines from hydrogen alsoincrease without limit?2. Does the light emitted by a neon sign constitute acontinuous spectrum or only a few colors? Defendyour answer.3. In an x-ray tube, if the energy with which the electronsstrike the metal target is increased, the wavelengthsof the characteristic x-rays do not change.Why not?4. Must an atom first be ionized before it can emitlight? Discuss.

Problems 9335. Is it possible for a spectrum from an x-ray tube toshow the continuous spectrum of x-rays without thepresence of the characteristic x-rays?6. Suppose that the electron in the hydrogen atomobeyed classical mechanics rather than quantummechanics. Why should such a hypothetical atomemit a continuous spectrum rather than the observedline spectrum?7. When a hologram is produced, the system (includinglight source, object, beam splitter, and so on)must be held motionless within a quarter of thelight’s wavelength. Why?8. If matter has a wave nature, why is it not observablein our daily experience?9. Discuss some consequences of the exclusion principle.10. Can the electron in the ground state of hydrogenabsorb a photon of energy less than 13.6 eV? Can itabsorb a photon of energy greater than 13.6 eV?Explain.11. Why do lithium, potassium, and sodium exhibit similarchemical properties?12. List some ways in which quantum mechanics alteredour view of the atom pictured by the Bohr theory.13. It is easy to understand how two electrons (one withspin up, one with spin down) can fill the 1s shell for ahelium atom. How is it possible that eight more electronscan fit into the 2s, 2p level to complete the1s2s 2 2p 6 shell for a neon atom?14. The ionization energies for Li, Na, K, Rb, and Csare 5.390, 5.138, 4.339, 4.176, and 3.893 eV, respectively.Explain why these values are to be expectedin terms of the atomic structures.15. Why is stimulated emission so important in the operationof a laser?16. The Bohr theory of the hydrogen atom is basedupon several assumptions. Discuss these assumptionsand their significance. Do any of them contradictclassical physics?17. Explain why, in the Bohr model, the total energy ofthe hydrogen atom is negative.18. Consider the quantum numbers n, , m , and m s .(a) Which of these are integers and which are fractional?(b) Which are always positive and which canbe negative? (c) If n 2, what is the largest value of? (d) If 1, what are the possible values of m ?19. Photon A is emitted when an electron in a hydrogenatom drops from the n 3 level to the n 2 level.Photon B is emitted when an electron in a hydrogenatom drops from the n 4 level to the n 2 level.(a) In which case is the wavelength of the emittedphoton greater? (b) In which case is the energy ofthe emitted photon greater?PROBLEMS1, 2, 3 = straightforward, intermediate, challenging = full solution available in Student Solutions Manual/Study Guide= coached problem with hints available at www.cp7e.com = biomedical applicationSection 28.1 Early Models of the AtomSection 28.2 Atomic Spectra1. Use Equation 28.1 to calculate the wavelengthof the first three lines in the Balmer series for hydrogen.2. Show that the wavelengths for the Balmer series satisfythe equation 364.5n2n 2 4 nmwhere n 3, 4, 5, . . .3. The “size” of the atom in Rutherford’s model isabout 1.0 10 10 m. (a) Determine the attractiveelectrostatic force between an electron and a protonseparated by this distance. (b) Determine (in eV)the electrostatic potential energy of the atom.4. The “size” of the nucleus in Rutherford’s model ofthe atom is about 1.0 fm 1.0 10 15 m. (a) Determinethe repulsive electrostatic force betweentwo protons separated by this distance. (b) Determine(in MeV) the electrostatic potential energyof the pair of protons.5.The “size” of the atom in Rutherford’smodel is about 1.0 10 10 m. (a) Determinethe speed of an electron moving about the protonusing the attractive electrostatic force between anelectron and a proton separated by this distance.(b) Does this speed suggest that Einsteinian relativitymust be considered in studying the atom?(c) Compute the de Broglie wavelength of the electronas it moves about the proton. (d) Does thiswavelength suggest that wave effects, such as diffractionand interference, must be considered in studyingthe atom?

934 Chapter 28 Atomic Physics6. In a Rutherford scattering experiment, an -particle(charge 2e) heads directly toward a goldnucleus (charge 79e). The -particle had a kineticenergy of 5.0 MeV when very far ( r : )from the nucleus. Assuming the gold nucleus tobe fixed in space, determine the distance of closestapproach. [Hint: Use conservation of energywith PE k e q 1 q 2 /r.]Section 28.3 The Bohr Theory of Hydrogen7. A hydrogen atom is in its first excited state (n 2).Using the Bohr theory of the atom, calculate (a) theradius of the orbit, (b) the linear momentum of theelectron, (c) the angular momentum of the electron,(d) the kinetic energy, (e) the potential energy,and (f ) the total energy.8. For a hydrogen atom in its ground state, use theBohr model to compute (a) the orbital speed of theelectron, (b) the kinetic energy of the electron, and(c) the electrical potential energy of the atom.9. Show that the speed of the electron in the nth Bohrorbit in hydrogen is given byv n k ee 2n10. A photon is emitted as a hydrogen atom undergoesa transition from the n 6 state to the n 2 state.Calculate (a) the energy, (b) the wavelength, and(c) the frequency of the emitted photon.11. A hydrogen atom emits a photon of wavelength656 nm. From what energy orbit to what lower energyorbit did the electron jump?12. Following are four possible transitions for a hydrogenatomI. n i 2; n f 5 II. n i 5; n f 3III. n i 7; n f 4 IV. n i 4; n f 7(a) Which transition will emit the shortest-wavelengthphoton? (b) For which transition will theatom gain the most energy? (c) For which transition(s)does the atom lose energy?13. What is the energy of a photon that, when absorbedby a hydrogen atom, could cause (a) an electronictransition from the n 3 state to the n 5 stateand (b) an electronic transition from the n 5state to the n 7 state?14. A hydrogen atom initially in its ground state (n 1) absorbs a photon and ends up in the state forwhich n 3. (a) What is the energy of the absorbedphoton? (b) If the atom eventually returnsto the ground state, what photon energies couldthe atom emit?15. Determine both the longest and the shortest wavelengthsin (a) the Lyman series (n f 1) and (b) thePaschen series (n f 3) of hydrogen.16. Show that the speed of the electron in the first(ground-state) Bohr orbit of the hydrogen atom maybe expressed asv (1/137)c.17. A monochromatic beam of light is absorbed by acollection of ground-state hydrogen atoms in such away that six different wavelengths are observedwhen the hydrogen relaxes back to the groundstate. What is the wavelength of the incident beam?18. A particle of charge q and mass m, moving with aconstant speed v, perpendicular to a constant magneticfield, B, follows a circular path. If in this casethe angular momentum about the center of this circleis quantized so that mvr 2n,show that the allowedradii for the particle arer n √2 nqBwhere n 1, 2, 3, . . .19.(a) If an electron makes a transitionfrom the n 4 Bohr orbit to the n 2 orbit,determine the wavelength of the photon created inthe process. (b) Assuming that the atom was initiallyat rest, determine the recoil speed of the hydrogenatom when this photon is emitted.20. Consider a large number of hydrogen atoms, withelectrons all initially in the n 4 state. (a) Howmany different wavelengths would be observed inthe emission spectrum of these atoms? (b) What isthe longest wavelength that could be observed? Towhich series does it belong?21. Analyze the Earth–Sun system by following the Bohrmodel, where the gravitational force between Earth(mass m) and Sun (mass M) replaces the Coulombforce between the electron and proton (so that F GMm/r 2 and PE GMm/r). Show that (a) the totalenergy of the Earth in an orbit of radius r is givenby (a) E GMm/2r, (b) the radius of the nth orbitis given by r n r 0 n 2 , where r 0 2 /GMm 2 2.32 10 138 m, and (c) the energy of the nth orbit is givenby E n E 0 /n 2 , where E 0 G 2 M 2 m 3 /2 2 1.71 10 182 J. (d) Using the Earth–Sun orbit radius ofr 1.49 10 11 m, determine the value of the quan-

Problems 935tum number n. (e) Should you expect to observequantum effects in the Earth–Sun system?22. An electron is in the nth Bohr orbit of the hydrogenatom. (a) Show that the period of the electron isT t o n 3 , and determine the numerical value of t o .(b) On the average, an electron remains in the n 2orbit for about 10 s before it jumps down to then 1 (ground-state) orbit. How many revolutionsdoes the electron make before it jumps to theground state? (c) If one revolution of the electron isdefined as an “electron year” (analogous to anEarth year being one revolution of the Eartharound the Sun), does the electron in the n 2 orbit“live” very long? Explain. (d) How does theabove calculation support the “electron cloud” concept?23. Consider a hydrogen atom. (a) Calculate the frequencyf of the n 2 : n 1 transition, and compareit with the frequency f orb of the electron orbitalmotion in the n 2 state. (b) Make the samecalculation for the n 10 000 : n 9 999 transition.Comment on the results.24. Two hydrogen atoms collide head-on and endup with zero kinetic energy. Each then emits a121.6-nm photon (n 2 to n 1 transition). Atwhat speed were the atoms moving before thecollision?25. Two hydrogen atoms, both initially in the groundstate, undergo a head-on collision. If both atoms areto be excited to the n 2 level in this collision,what is the minimum speed each atom can have beforethe collision?26. (a) Calculate the angular momentum of the Moondue to its orbital motion about the Earth. In yourcalculation, use 3.84 10 8 m as the averageEarth–Moon distance and 2.36 10 6 s as the periodof the Moon in its orbit. (b) If the angularmomentum of the moon obeys Bohr’s quantizationrule ( L n), determine the value of the quantumnumber n. (c) By what fraction would theEarth–Moon radius have to be increased to increasethe quantum number by 1?Section 28.4 Modification of the Bohr TheorySection 28.5 De Broglie Waves andthe Hydrogen Atom27. (a) Find the energy of the electron in the groundstate of doubly ionized lithium, which has anatomic number Z 3. (b) Find the radius of itsground-state orbit.28. (a) Construct an energy level diagram for the He ion, for which Z 2. (b) What is the ionization energyfor He ?29. The orbital radii of a hydrogen-like atom is given bythe equationWhat is the radius of the first Bohr orbit in (a) He ,(b) Li 2 , and (c) Be 3 ?30. (a) Substitute numerical values into Equation 28.19to find a value for the Rydberg constant for singlyionized helium, He . (b) Use the result of part (a)to find the wavelength associated with a transitionfrom the n 2 state to the n 1 state of He . (c)Identify the region of the electromagnetic spectrumassociated with this transition.31.r n2 2Zm e k e e 2Determine the wavelength of anelectron in the third excited orbit of the hydrogenatom, with n 4.32. Using the concept of standing waves, de Brogliewas able to derive Bohr’s stationary orbit postulate.He assumed that a confined electron couldexist only in states where its de Broglie waves formstanding-wave patterns, as in Figure 28.10a. Considera particle confined in a box of length L to beequivalent to a string of length L and fixed at bothends. Apply de Broglie’s concept to show that(a) the linear momentum of this particle is quantizedwith p mv nh/2L and (b) the allowedstates correspond to particle energies of E n n 2 E 0 ,where E 0 h 2 /(8mL 2 ).Section 28.6 Quantum Mechanics andthe Hydrogen AtomSection 28.7 The Spin Magnetic QuantumNumber33. List the possible sets of quantum numbers for electronsin the 3p subshell.34. When the principal quantum number is n 4,how many different values of (a) and (b) m arepossible?35. The -meson has a charge of e, a spin quantumnumber of 1, and a mass 1 507 times that of theelectron. If the electrons in atoms were replaced by -mesons, list the possible sets of quantum numbersfor -mesons in the 3d subshell..

936 Chapter 28 Atomic PhysicsSection 28.9 The Exclusion Principle andthe Periodic Table36. (a) Write out the electronic configuration of theground state for oxygen (Z 8). (b) Write out thevalues for the set of quantum numbers n, , m , andm s for each of the electrons in oxygen.37.Two electrons in the same atom have n 3 and 1. (a) List the quantum numbers for the possiblestates of the atom. (b) How many states wouldbe possible if the exclusion principle did not applyto the atom?38. How many different sets of quantum numbers arepossible for an electron for which (a) n 1,(b) n 2, (c) n 3, (d) n 4, and (e) n 5?Check your results to show that they agree with thegeneral rule that the number of different sets ofquantum numbers is equal to 2n 2 .39. Zirconium (Z 40) has two electrons in an incompleted subshell. (a) What are the values of n and for each electron? (b) What are all possible valuesof m and m s ? (c) What is the electron configurationin the ground state of zirconium?Section 28.10 CharacteristicX-Rays40. The K-shell ionization energy of copper is 8 979 eV.The L-shell ionization energy is 951 eV. Determinethe wavelength of the K emission line of copper.What must the minimum voltage be on an x-raytube with a copper target in order to see the K line?41. The K x-ray is emitted when an electron undergoesa transition from the L shell (n 2) to the K shell(n 1). Use the method illustrated in Example 28.5to calculate the wavelength of the K x-ray from anickel target (Z 28).42. When an electron drops from the M shell (n 3)to a vacancy in the K shell (n 1), the measuredwavelength of the emitted x-ray is found to be0.101 nm. Identify the element.43. The K series of the discrete spectrum of tungstencontains wavelengths of 0.018 5 nm, 0.020 9 nm,and 0.021 5 nm. The K-shell ionization energy is69.5 keV. Determine the ionization energies of theL, M, and N shells. Sketch the transitions that producethe above wavelengths.ADDITIONAL PROBLEMS44. In a hydrogen atom, what is the principle quantumnumber of the electron orbit with a radius closest to1.0 m?45. (a) How much energy is required to cause an electronin hydrogen to move from the n 1 state tothe n 2 state? (b) If the electrons gain this energyby collision between hydrogen atoms in ahigh-temperature gas, find the minimum temperatureof the heated hydrogen gas. The thermal energyof the heated atoms is given by 3k B T/2, wherek B is the Boltzmann constant.46. A pulsed ruby laser emits light at 694.3 nm. For a14.0-ps pulse containing 3.00 J of energy, find (a)the physical length of the pulse as it travels throughspace and (b) the number of photons in it. (c) Ifthe beam has a circular cross section 0.600 cm in diameter,find the number of photons per cubic millimeter.47. The Lyman series for a (new?) one-electron atomis observed in a distant galaxy. The wavelengths ofthe first four lines and the short-wavelength limitof this Lyman series are given by the energy-leveldiagram in Figure P28.47. Based on this information,calculate (a) the energies of the groundstate and first four excited states for this oneelectronatom and (b) the longest-wavelength (alpha)lines and the short-wavelength series limit inthe Balmer series for this atom.ENERGYn = ∞n = 5n = 4n = 3n = 2n = 1= 202.6 nmλ= 170.9 nmλ= 162.1 nmλFigure P28.47= 158.3 nmλ= 152.0 nmλE ∞ = 0E 5E 4E 3E 2E 1

Problems 93748. A dimensionless number that often appears inatomic physics is the fine-structure constant, where k e is the Coulomb constant.(a) Obtain a numerical value for 1/. (b) In termsof , what is the ratio of the Bohr radius a 0 to theCompton wavelength C h/m e c ? (d) In terms of, what is the ratio of the reciprocal of the Rydbergconstant 1/R H to the Bohr radius?49. Mercury’s ionization energy is 10.39 eV. The threelongest wavelengths of the absorption spectrum ofmercury are 253.7 nm, 185.0 nm, and 158.5 nm.(a) Construct an energy-level diagram for mercury.(b) Indicate all emission lines that can occur whenan electron is raised to the third level above theground state. (c) Disregarding recoil of the mercuryatom, determine the minimum speed an electronmust have in order to make an inelastic collisionwith a mercury atom in its ground state.50. Suppose the ionization energy of an atom is 4.100eV. In this same atom, we observe emission linesthat have wavelengths of 310.0 nm, 400.0 nm, and1 378 nm. Use this information to construct the energy-leveldiagram with the least number of levels.Assume the higher energy levels are closer together.51.A laser used in eye surgery emits a3.00-mJ pulse in 1.00 ns, focused to a spot 30.0 min diameter on the retina. (a) Find (in SI units) thepower per unit area at the retina. (This quantity iscalled the irradiance.) (b) What energy is deliveredper pulse to an area of molecular size—say, a circulararea 0.600 nm in diameter.52. An electron has a de Broglie wavelength equal tothe diameter of a hydrogen atom in its ground state.(a) What is the kinetic energy of the electron?(b) How does this energy compare with the groundstateenergy of the hydrogen atom?53. Use Bohr’s model of the hydrogen atom to showthat, when the atom makes a transition from thestate n to the state n 1, the frequency of the emittedlight is given by k e e 2 /cf 2 2 mk e 2 e 4h 3 2n 1(n 1) 2 n 254. Calculate the classical frequency for the light emittedby an atom. To do so, note that the frequency ofrevolution is v/2r, where r is the Bohr radius.Show that as n approaches infinity in the equationof the preceding problem, the expression giventhere varies as 1/n 3 and reduces to the classical frequency.(This is an example of the correspondenceprinciple, which requires that the classical andquantum models agree for large values of n.)55. A pi meson ( ) of charge e and mass 273 timesgreater than that of the electron is captured by ahelium nucleus (Z 2) as shown in FigureP28.55. (a) Draw an energy-level diagram (in unitsof eV) for this “Bohr-type” atom up to the first sixenergy levels. (b) When the -meson makes atransition between two orbits, a photon is emittedthat Compton scatters off a free electron initiallyat rest, producing a scattered photon of wavelength 0.089 929 3 nm at an angle of 42.68°, as shown on the right-hand side of FigureP28.55. Between which two orbits did the -mesonmake a transition?n in f“Pi mesonic” He + atom(Z = 2, m p = 273m e )lFigure P28.55Free electron56. When a muon with charge e is captured by a proton,the resulting bound system forms a “muonicatom,” which is the same as hydrogen, except witha muon (of mass 207 times the mass of an electron)replacing the electron. For this “muonicatom,” determine (a) the Bohr radius and (b) thethree lowest energy levels.57. In this problem, you will estimate the classical lifetimeof the hydrogen atom. An accelerating chargeloses electromagnetic energy at a rate given by 2k e q 2 a 2 /(3c 3 ), where k e is the Coulomb constant,q is the charge of the particle, a is its acceleration,and c is the speed of light in a vacuum. Assume thatthe electron is one Bohr radius (0.052 9 nm) fromthe center of the hydrogen atom. (a) Determine itsacceleration. (b) Show that has units of energyper unit time and determine the rate of energy loss.(c) Calculate the kinetic energy of the electron anddetermine how long it will take for all of this energyto be converted into electromagnetic waves, assumingthat the rate calculated in part (b) remains constantthroughout the electron’s motion.58. An electron in a hydrogen atom jumps from someinitial Bohr orbit n i to some final Bohr orbit n f , as inOul'

938 Chapter 28 Atomic PhysicsFigure P28.58. (a) If the photon emitted in theprocess is capable of ejecting a photoelectron fromtungsten (work function 4.58 eV), determine n f .(b) If a minimum stopping potential of V 0 7.51volts is required to prevent the photoelectron fromhitting the anode, determine the value of n i .ACTIVITIES1. With your partner not looking, use modeling clay tobuild one or more mounds on top of a table. Place apiece of cardboard over your mound(s), and assignyour partner the task of determining the size, shape,and number of mounds without looking. He is to dothis by rolling marbles at the unseen mounds andobserving how they emerge. This experiment modelsthe Rutherford scattering experiment.AnodePhotoelectron–+n fniTungsten2. Your instructor can probably lend you a small plasticdiffraction grating to enable you to examine thespectrum of different light sources. You can usethese gratings to examine a source by holding thegrating very close to your eye and noting the spectrumproduced by glancing out of the corner ofyour eye while looking at a light source. You shouldlook at light sources such as sodium vapor lights andmercury vapor lights used in many parking lots,neon lights used in many signs, black lights, ordinaryincandescent light bulbs, and so forth.V 0Figure P28.58

Aerial view of a nuclear power plantthat generates electrical power.Energy is generated in such plantsfrom the process of nuclear fission, inwhich a heavy nucleus such as 235 Usplits into smaller particles having alarge amount of kinetic energy. Thissurplus energy can be used to heatwater into high pressure steam anddrive a turbine.Nuclear PhysicsIn 1896, the year that marks the birth of nuclear physics, Henri Becquerel (1852–1908)discovered radioactivity in uranium compounds. A great deal of activity followed this discoveryas researchers attempted to understand and characterize the radiation that we now knowto be emitted by radioactive nuclei. Pioneering work by Rutherford showed that the radiationwas of three types, which he called alpha, beta, and gamma rays. These types are classifiedaccording to the nature of their electric charge and their ability to penetrate matter. Laterexperiments showed that alpha rays are helium nuclei, beta rays are electrons, and gammarays are high-energy photons.In 1911 Rutherford and his students Geiger and Marsden performed a number of importantscattering experiments involving alpha particles. These experiments established the ideathat the nucleus of an atom can be regarded as essentially a point mass and point charge andthat most of the atomic mass is contained in the nucleus. Further, such studies demonstrateda wholly new type of force: the nuclear force, which is predominant at distances of less thanabout 10 14 m and drops quickly to zero at greater distances.Other milestones in the development of nuclear physics include• the first observations of nuclear reactions by Rutherford and coworkers in 1919, in whichnaturally occurring particles bombarded nitrogen nuclei to produce oxygen,• the first use of artificially accelerated protons to produce nuclear reactions, by Cockcroftand Walton in 1932,• the discovery of the neutron by Chadwick in 1932,• the discovery of artificial radioactivity by Joliot and Irene Curie in 1933,• the discovery of nuclear fission by Hahn, Strassman, Meitner, and Frisch in 1938, and• the development of the first controlled fission reactor by Fermi and his collaboratorsin 1942.In this chapter we discuss the properties and structure of the atomic nucleus. We start bydescribing the basic properties of nuclei and follow with a discussion of the phenomenon ofradioactivity. Finally, we explore nuclear reactions and the various processes by which nuclei decay.Courtesy of Public Service Electric and Gas CompanyCHAPTER29O U T L I N E29.1 Some Properties of Nuclei29.2 Binding Energy29.3 Radioactivity29.4 The Decay Processes29.5 Natural Radioactivity29.6 Nuclear Reactions29.7 Medical Applicationsof Radiation29.8 Radiation Detectors939

940 Chapter 29 Nuclear PhysicsERNEST RUTHERFORD,New Zealand Physicist(1871 – 1937)Rutherford was awarded the Nobel Prize in1908 for discovering that atoms can bebroken apart by alpha rays and for studyingradioactivity. “On consideration, I realizedthat this scattering backward must be theresult of a single collision, and when I madecalculations I saw that it was impossible toget anything of that order of magnitude unlessyou took a system in which the greaterpart of the mass of the atom was concentratedin a minute nucleus. It was then thatI had the idea of an atom with a minutemassive center carrying a charge.”Definition of the unified mass unit u TIP 29.1 Mass Number is notthe Atomic MassDon’t confuse the mass number A withthe atomic mass. Mass number is aninteger that specifies an isotope andhas no units—it’s simply equal to thenumber of nucleons. Atomic mass isan average of the masses of theisotopes of a given element and hasunits of u.North Wind Picture Archives29.1 SOME PROPERTIES OF NUCLEIAll nuclei are composed of two types of particles: protons and neutrons. The onlyexception is the ordinary hydrogen nucleus, which is a single proton. In describingsome of the properties of nuclei, such as their charge, mass, and radius, wemake use of the following quantities:• the atomic number Z, which equals the number of protons in the nucleus,• the neutron number N, which equals the number of neutrons in the nucleus,• the mass number A, which equals the number of nucleons in the nucleus(nucleon is a generic term used to refer to either a proton or a neutron).AThe symbol we use to represent nuclei is Z X, where X represents the chemical27symbol for the element. For example, 13 Al has the mass number 27 and the atomicnumber 13; therefore, it contains 13 protons and 14 neutrons. When no confusionis likely to arise, we often omit the subscript Z, because the chemical symbol can alwaysbe used to determine Z .The nuclei of all atoms of a particular element must contain the same numberof protons, but they may contain different numbers of neutrons. Nuclei that arerelated in this way are called isotopes. The isotopes of an element have the same Zvalue, but different N and A values. The natural abundances of isotopes can differ11 12 13 14substantially. For example, 6 C, 6 C, 6 C, and 6 C are four isotopes of carbon. Thenatural abundance of the12 13 6 C isotope is about 98.9%, whereas that of the 6 C isotopeis only about 1.1%. Some isotopes don’t occur naturally, but can be producedin the laboratory through nuclear reactions. Even the simplest element, hydrogen,12 3has isotopes: 1H, hydrogen; 1 H, deuterium; and 1 H, tritium.Charge and MassThe proton carries a single positive charge e 1.602 177 33 10 19 C, the electroncarries a single negative charge e, and the neutron is electrically neutral.Because the neutron has no charge, it’s difficult to detect. The proton is about1 836 times as massive as the electron, and the masses of the proton and the neutronare almost equal (Table 29.1).For atomic masses, it is convenient to define the unified mass unit u insuch a way that the mass of one atom of the isotope 12 C is exactly 12 u, where1u 1.660 559 10 27 kg. The proton and neutron each have a mass ofabout 1 u, and the electron has a mass that is only a small fraction of an atomicmass unit.Because the rest energy of a particle is given by E R mc 2 , it is often convenientto express the particle’s mass in terms of its energy equivalent. For one atomicmass unit, we have an energy equivalent ofE R mc 2 (1.660 559 10 27 kg)(2.997 92 10 8 m/s) 2 1.492 431 10 10 J 931.494 MeVIn calculations, nuclear physicists often express mass in terms of the unitMeV/c 2 , where1 u 931.494 MeV/c 2TABLE 29.1Masses of the Proton, Neutron, and Electron in Various UnitsMassParticle kg u MeV/c 2Proton 1.6726 10 27 1.007 276 938.28Neutron 1.6750 10 27 1.008 665 939.57Electron 9.109 10 31 5.486 10 4 0.511

29.1 Some Properties of Nuclei 941The Size of NucleiThe size and structure of nuclei were first investigated in the scattering experimentsof Rutherford, discussed in Section 28.1. Using the principle of conservationof energy, Rutherford found an expression for how close an alpha particlemoving directly toward the nucleus can come to the nucleus before being turnedaround by Coulomb repulsion.In such a head-on collision, the kinetic energy of the incoming alpha particlemust be converted completely to electrical potential energy when the particlestops at the point of closest approach and turns around (Active Fig. 29.1). If weequate the initial kinetic energy of the alpha particle to the maximum electricalpotential energy of the system (alpha particle plus target nucleus), we have12 mv 2 q 1 q 2 k erwhere d is the distance of closest approach. Solving for d, we getd 4k eZe 2mv 2From this expression, Rutherford found that alpha particles approached to within3.2 10 14 m of a nucleus when the foil was made of gold. Thus, the radius of thegold nucleus must be less than this value. For silver atoms, the distance of closestapproach was 2 10 14 m. From these results, Rutherford concluded that thepositive charge in an atom is concentrated in a small sphere, which he called thenucleus, with radius no greater than about 10 14 m. Because such small lengthsare common in nuclear physics, a convenient unit of length is the femtometer (fm),sometimes called the fermi and defined as1 fm 10 15 mSince the time of Rutherford’s scattering experiments, a multitude of otherexperiments have shown that most nuclei are approximately spherical and havean average radius given byr r 0 A 1/3 k e(2e)(Ze)d[29.1]where A is the total number of nucleons and r 0 is a constant equal to 1.2 10 15 m.Because the volume of a sphere is proportional to the cube of its radius, it followsfrom Equation 29.1 that the volume of a nucleus (assumed to be spherical) isdirectly proportional to A, the total number of nucleons. This relationship thensuggests that all nuclei have nearly the same density. Nucleons combine to form anucleus as though they were tightly packed spheres (Fig. 29.2).Ze2e v = 0++ +++ ++ ++ ++ACTIVE FIGURE 29.1An alpha particle on a head-on collisioncourse with a nucleus of chargeZe. Because of the Coulomb repulsionbetween the like charges, the alphaparticle will stop instantaneously at adistance d from the nucleus, calledthe distance of closest approach.Log into PhysicsNow at www.cp7e.comand go to Active Figure 29.1, whereyou can adjust the atomic number ofthe target nucleus and the kineticenergy of the alpha particle. Thenobserve the approach of the alphaparticle toward the nucleus.Figure 29.2 A nucleus can bevisualized as a cluster of tightly packedspheres, each of which is a nucleon.dEXAMPLE 29.1GoalSizing a Neutron StarApply the concepts of nuclear size.Problem One of the end stages of stellar life is a neutron star, where matter collapses and electrons combine with protonsto form neutrons. Some liken neutron stars to a single gigantic nucleus. (a) Approximately how many nucleons arein a neutron star with a mass of 3.00 10 30 kg? (This is the mass number of the star.) (b) Calculate the radius of thestar, treating it as a giant nucleus. (c) Calculate the density of the star, assuming the mass is distributed uniformly.Strategy The effective mass number of the neutron star can be found by dividing the star mass in kg by the mass ofa neutron. Equation 29.1 then gives an estimate of the radius of the star, which together with the mass determinesthe density.Solution(a) Find the approximate number of nucleons in the star.Divide the star’s mass by the mass of a neutron to find A: 1.79 10 57A 3.00 10 30 kg1.675 10 27 kg

942 Chapter 29 Nuclear Physics(b) Calculate the radius of the star, treating it as a giantatomic nucleus.Substitute into Equation 29.1: r r 0 A 1/3 (1.2 10 15 m)(1.79 10 57 ) 1/3 1.46 10 4 m(c) Calculate the density of the star, assuming that itsmass is distributed uniformly.Substitute values into the equation for density andassume the star is a uniform sphere: m V m43r 3 3.00 1030 kg43 (1.46 10 4 m) 3 2.30 10 17 kg/m 3Remarks This density is typical of atomic nuclei as well as of neutron stars. A ball of neutron star matter having aradius of only 1 meter would have a powerful gravity field: it could attract objects a kilometer away at an accelerationof over 50 m/s 2 !Exercise 29.1Estimate the radius of a uranium-235 nucleus.Answer7.41 10 15 mMARIA GOEPPERT-MAYER,German Physicist (1906 – 1972)Goeppert-Mayer was born and educatedin Germany. She is best known for herdevelopment of the shell model of thenucleus, published in 1950. A similarmodel was simultaneously developed byHans Jensen, a German scientist. MariaGoeppert-Mayer and Hans Jensen wereawarded the Nobel Prize in physics in 1963for their extraordinary work inunderstanding the structure of the nucleus.Courtesy of Louise Barker/AIP Niels Bohr LibraryNuclear StabilityGiven that the nucleus consists of a closely packed collection of protons andneutrons, you might be surprised that it can even exist. The very large repulsiveelectrostatic forces between protons should cause the nucleus to fly apart.However, nuclei are stable because of the presence of another, short-range(about 2 fm) force: the nuclear force, an attractive force that acts between allnuclear particles. The protons attract each other via the nuclear force, and at thesame time they repel each other through the Coulomb force. The attractivenuclear force also acts between pairs of neutrons and between neutrons andprotons.The nuclear attractive force is stronger than the Coulomb repulsive forcewithin the nucleus (at short ranges). If this were not the case, stable nucleiwould not exist. Moreover, the strong nuclear force is nearly independent ofcharge. In other words, the nuclear forces associated with proton–proton,proton–neutron, and neutron–neutron interactions are approximately thesame, apart from the additional repulsive Coulomb force for the proton–protoninteraction.There are about 260 stable nuclei; hundreds of others have been observed,but are unstable. A plot of N versus Z for a number of stable nuclei is given inFigure 29.3. Note that light nuclei are most stable if they contain equal numbers ofprotons and neutrons, so that N Z, but heavy nuclei are more stable if N Z.This difference can be partially understood by recognizing that as the number ofprotons increases, the strength of the Coulomb force increases, which tends tobreak the nucleus apart. As a result, more neutrons are needed to keep thenucleus stable, because neutrons are affected only by the attractive nuclearforces. In effect, the additional neutrons “dilute” the nuclear charge density. Eventually,when Z 83, the repulsive forces between protons cannot be compensatedfor by the addition of neutrons. Elements that contain more than 83 protons don’thave stable nuclei, but decay or disintegrate into other particles in variousamounts of time. The masses and some other properties of selected isotopes areprovided in Appendix B.

29.2 Binding Energy 943130120110100Figure 29.3 A plot of the neutronnumber N versus the proton numberZ for the stable nuclei (solid points).The dashed straight line correspondsto the condition N Z. They arecentered on the so-called line ofstability. The shaded area showsradioactive (unstable) nuclei.Neutron number N9080706050N =Z4030201000 10 20 30 40 50 60 70 80 90Proton number Z29.2 BINDING ENERGYThe total mass of a nucleus is always less than the sum of the masses of its nucleons.Also, because mass is another manifestation of energy, the total energy of thebound system (the nucleus) is less than the combined energy of the separatednucleons. This difference in energy is called the binding energy of the nucleus andcan be thought of as the energy that must be added to a nucleus to break it apartinto its separated neutrons and protons.EXAMPLE 29.2GoalThe Binding Energy of the DeuteronCalculate the binding energy of a nucleus.Problem The nucleus of the deuterium atom, called the deuteron, consists of a proton and a neutron. Calculatethe deuteron’s binding energy in MeV, given that its atomic mass—that is, the mass of a deuterium nucleus plus anelectron—is 2.014 102 u.Strategy Calculate the sum of the masses of the individual particles and subtract the mass of the combined particle.The masses of the neutral atoms can be used instead of the nuclei because the electron masses cancel. Use thevalues from Table 29.4 or Table B of the appendix. The mass of an atom given in Appendix B includes the mass of Zelectrons, where Z is the atom’s atomic number.SolutionTo find the binding energy, first sum the masses of thehydrogen atom and neutron and subtract the mass ofthe deuteron:Convert this mass difference to its equivalent in MeV:m (m p m n ) m d (1.007 825 u 1.008 665 u) 2.014 102 u 0.002 388 u931.5 MeVE b (0.002 388 u)1 u2.224 MeVRemarks This result tells us that to separate a deuteron into a proton and a neutron, it’s necessary to add 2.224 MeVof energy to the deuteron to overcome the attractive nuclear force between the proton and the neutron. One way ofsupplying the deuteron with this energy is by bombarding it with energetic particles.

944 Chapter 29 Nuclear PhysicsIf the binding energy of a nucleus were zero, the nucleus would separate into its constituent protons and neutronswithout the addition of any energy; that is, it would spontaneously break apart.Exercise 29.23Calculate the binding energy of 2He.Answer7.718 MeVApplying Physics 29.1Figure 29.4 shows a graph of the amount of energyrequired to remove a nucleon from the nucleus. Thefigure indicates that an approximately constant amountof energy is necessary to remove a nucleon above A 40,whereas we saw in Chapter 28 that widely varyingamounts of energy are required to remove an electronfrom the atom. What accounts for this difference?Explanation In the case of Figure 29.4, the approximatelyconstant value of the nuclear binding energy is aresult of the short-range nature of the nuclear force. Agiven nucleon interacts only with its few nearest neighbors,rather than with all of the nucleons in the nucleus.Thus, no matter how many nucleons are present in thenucleus, pulling any one nucleon out involves separatingIt’s interesting to examine a plot of binding energy per nucleon, E b /A, as afunction of mass number for various stable nuclei (Fig. 29.4). Except for thelighter nuclei, the average binding energy per nucleon is about 8 MeV. Note thatthe curve peaks in the vicinity of A 60, which means that nuclei with mass numbersgreater or less than 60 are not as strongly bound as those near the middle ofthe periodic table. As we’ll see later, this fact allows energy to be released in fissionand fusion reactions. The curve is slowly varying for A 40, which suggests thatthe nuclear force saturates. In other words, a particular nucleon can interact withonly a limited number of other nucleons, which can be viewed as the “nearestneighbors” in the close-packed structure illustrated in Figure 29.2.Binding Nucleons and Electronsit only from its nearest neighbors. The energy to do this,therefore, is approximately independent of how manynucleons are present. For the clearest comparison withthe electron, think of averaging the energies required tostrip all of the electrons out of a particular atom, fromthe outermost valence electron to the innermost K-shellelectron. This average increases steeply with increasingatomic number. The electrical force binding the electronsto the nucleus in an atom is a long-range force. Anelectron in an atom interacts with all the protons in thenucleus. When the nuclear charge increases, there is astronger attraction between the nucleus and the electrons.Therefore, as the nuclear charge increases, moreenergy is necessary to remove an average electron.Region of greatest stability4 He12 C20 Ne62 Ni208 Pb97235 ClGe8723 Na19 F56 Fe98 Mo107 Ag127 I 159 Tb 197Au226 Ra238 UFigure 29.4 Binding energy pernucleon versus the mass number Afor nuclei that are along the line ofstability shown in Figure 29.3. Somerepresentative nuclei appear as bluedots with labels. (Nuclei to the rightof 208 Pb are unstable. The curverepresents the binding energy forthe most stable isotopes.)Binding energy perparticle, MeV6543210014 N116 B Li9 Be2 H20 406080100 120 140Mass number A160180200220240

29.3 Radioactivity 94529.3 RADIOACTIVITYIn 1896, Becquerel accidentally discovered that uranium salt crystals emit an invisibleradiation that can darken a photographic plate even if the plate is covered toexclude light. After several such observations under controlled conditions, he concludedthat the radiation emitted by the crystals was of a new type, one requiringno external stimulation. This spontaneous emission of radiation was soon calledradioactivity. Subsequent experiments by other scientists showed that othersubstances were also radioactive.The most significant investigations of this type were conducted by Marieand Pierre Curie. After several years of careful and laborious chemicalseparation processes on tons of pitchblende, a radioactive ore, the Curiesreported the discovery of two previously unknown elements, both of whichwere radioactive. These were named polonium and radium. Subsequent experiments,including Rutherford’s famous work on alpha-particle scattering,suggested that radioactivity was the result of the decay, or disintegration, of unstablenuclei.Three types of radiation can be emitted by a radioactive substance: alpha ()4particles, in which the emitted particles are 2 He nuclei; beta () particles, in whichand with Becquerel for their studies ofradioactive substances. In 1911 she wasMARIE CURIE, Polish Scientist(1867 – 1934)In 1903 Marie Curie shared the NobelPrize in physics with her husband, Pierre,the emitted particles are either electrons or positrons; and gamma () rays, inwhich the emitted “rays” are high-energy photons. A positron is a particle similarto the electron in all respects, except that it has a charge of e. (The positron issaid to be the antiparticle of the electron.) The symbol e is used to designate anawarded a second Nobel Prize in chemistryfor the discovery of radium and polonium.Marie Curie died of leukemia causedby years of exposure to radioactive substances.“I persist in believing that theelectron, and e designates a positron.ideas that then guided us are the onlyIt’s possible to distinguish these three forms of radiation by using the schemeones which can lead to the true socialdescribed in Figure 29.5. The radiation from a radioactive sample is directed intoprogress. We cannot hope to build a bettera region with a magnetic field, and the beam splits into three components, two world without improving the individual.bending in opposite directions and the third not changing direction. From this Toward this end, each of us must worksimple observation it can be concluded that the radiation of the undeflected beam toward his own highest development,(the gamma ray) carries no charge, the component deflected upward containspositively charged particles (alpha particles), and the component deflected downwardcontains negatively charged particles (e ). If the beam includes a positronaccepting at the same time his share ofresponsibility in the general life ofhumanity.”(e ), it is deflected upward.The three types of radiation have quite different penetrating powers. Alphaparticles barely penetrate a sheet of paper, beta particles can penetrate a fewmillimeters of aluminum, and gamma rays can penetrate several centimeters oflead. arrayThe Decay Constant and Half-LifeObservation has shown that if a radioactive sample contains N radioactive nuclei atsome instant, then the number of nuclei, N, that decay in a small time interval tis proportional to N ; mathematically,NNtorN N t [29.2]where is a constant called the decay constant. The negative sign signifies that Ndecreases with time; that is, N is negative. The value of for any isotope determinesthe rate at which that isotope will decay. The decay rate, or activity R, of asample is defined as the number of decays per second. From Equation 29.2, we seethat the decay rate isR Nt NIsotopes with a large value decay rapidly; those with small decay slowly.[29.3]FPG InternationalLeadRadioactivesourceDetector e– B inFigure 29.5 The radiation from aradioactive source, such as radium,can be separated into three componentsusing a magnetic field to deflectthe charged particles. The detectorarray at the right records theevents. The gamma ray isn’t deflectedby the magnetic field. Decay rate

946 Chapter 29 Nuclear PhysicsA general decay curve for a radioactive sample is shown in Active Figure 29.6. Itcan be shown from Equation 29.2 (using calculus) that the number of nuclei presentvaries with time according to the equationThe hands and numbers of this luminouswatch contain minute amountsof radium salt. The radioactive decayof radium causes the phosphors toglow in the dark.N 012 N 014 N 0N(t)T 1/2N =N 0 e – t2T 1/2ACTIVE FIGURE 29.6Plot of the exponential decay law forradioactive nuclei. The vertical axisrepresents the number of radioactivenuclei present at any time t, and thehorizontal axis is time. The parameterT 1/2 is the half-life of the sample.Log into PhysicsNow at www.cp7e.comand go to Active Figure 29.6, whereyou can observe the decay curves fornuclei with varying half-lives.TIP 29.2 Two Half-Lives Don’tMake a Whole-LifeA half-life is the time it takes for halfof a given number of nuclei to decay.During a second half-life, half theremaining nuclei decay, so in twohalf-lives, three-quarters of theoriginal material has decayed,not all of it.t© Richard Megna/Fundamental PhotographsN N 0 e t[29.4a]where N is the number of radioactive nuclei present at time t, N 0 is the numberpresent at time t 0, and e 2.718 . . . is Euler’s constant. Processes that obeyEquation 29.4a are sometimes said to undergo exponential decay. 1Another parameter that is useful for characterizing radioactive decay is the halflifeT 1/2 . The half-life of a radioactive substance is the time it takes for half of agiven number of radioactive nuclei to decay. Using the concept of half-life, it canbe shown that Equation 29.4a can also be written asN N 0 1 2 n[29.4b]where n is the number of half-lives. The number n can take any non-negative valueand need not be an integer. From the definition, it follows that n is related to timet and the half-life T 1/2 byn Setting N N 0 /2 and t T 1/2 in Equation 29.4a givesN 02 N 0e T 1/2[29.4c]Writing this in the form eT 1/2 2 and taking the natural logarithm of both sides,we getT 1/2 ln 2 0.693[29.5]This is a convenient expression relating the half-life to the decay constant. Notethat after an elapsed time of one half-life, N 0 /2 radioactive nuclei remain (by definition);after two half-lives, half of these will have decayed and N 0 /4 radioactivenuclei will be left; after three half-lives, N 0 /8 will be left; and so on.The unit of activity R is the curie (Ci), defined as1 Ci 3.7 10 10 decays/s[29.6]This unit was selected as the original activity unit because it is the approximateactivity of 1 g of radium. The SI unit of activity is the becquerel (Bq):1Bq 1 decay/s [29.7]Therefore, 1 Ci 3.7 10 10 Bq. The most commonly used units of activity are themillicurie (10 3 Ci) and the microcurie (10 6 Ci).Quick Quiz 29.1tT 1/2What fraction of a radioactive sample has decayed after two half-lives have elapsed?(a) 1/4 (b) 1/2 (c) 3/4 (d) not enough information to sayQuick Quiz 29.2Suppose the decay constant of radioactive substance A is twice the decay constant ofradioactive substance B. If substance B has a half life of 4 hr, what’s the half life ofsubstance A? (a) 8 hr (b) 4 hr (c) 2 hr (d) not enough information to say1 Other examples of exponential decay were discussed in Chapter 18 in connection with RC circuits and in Chapter 20in connection with RL circuits.

29.3 Radioactivity 947INTERACTIVE EXAMPLE 29.3GoalThe Activity of RadiumCalculate the activity of a radioactive substance at different times.Problem The half-life of the radioactive nucleus is 1.6 10 3 yr. If a sample initially contains 3.00 10 16 88 Rasuchnuclei, determine (a) the initial activity in curies, (b) the number of radium nuclei remaining after 4.8 10 3 yr, and(c) the activity at this later time.Strategy For parts (a) and (c), find the decay constant and multiply it by the number of nuclei. Part (b) requiresmultiplying the initial number of nuclei by one-half for every elapsed half-life. (Essentially, this is an application ofEquation 29.4b.)Solution(a) Determine the initial activity in curies.226Convert the half-life to seconds:Substitute this value into Equation 29.5 to get the decayconstant:Calculate the activity of the sample at t 0, usingR 0 N 0 , where R 0 is the decay rate at t 0 and N 0 isthe number of radioactive nuclei present at t 0:Convert to curies to obtain the activity at t 0, usingthe fact that 1 Ci 3.7 10 10 decays/s:T 1/2 (1.6 10 3 yr)(3.156 10 7 s/yr) 5.0 10 10 s 0.693T 1/20.6935.0 10 10 s 1.4 1011 s 1R 0 N 0 (1.4 10 11 s 1 )(3.0 10 16 nuclei) 4.2 10 5 decays/sR 0 (4.2 10 5 decays/s) 1 Ci3.7 10 decays/s101.1 10 5 Ci 11 Ci(b) How many radium nuclei remain after4.8 10 3 years?Calculate the number of half-lives, n:Multiply the initial number of nuclei by the number offactors of one-half:Substitute N 0 3.0 10 16 and n 3.0:(c) Calculate the activity after 4.8 10 3 yr.Multiply the number of remaining nuclei by the decayconstant to find the activity R:n 4.8 10 3 yr1.6 10 3 yr/half-lifeN N 0 1 2 nN (3.0 10 16 nuclei) 12 3.0 3.0 half-lives3.8 10 15 nucleiR N (1.4 10 11 s 1 )(3.8 10 15 nuclei) 5.3 10 4 decays/s 1.4 Ci(1)Remarks The activity is reduced by half every half-life, which is naturally the case because activity is proportional to thenumber of remaining nuclei. The precise number of nuclei at any time is never truly exact, because particles decayaccording to a probability. The larger the sample, however, the more accurate are the predictions from Equation 29.4.Exercise 29.3Find (a) the number of remaining radium nuclei after 3.2 10 3 yr and (b) the activity at this time.Answer (a) 7.5 10 15 nuclei (b) 2.8 CiPractice evaluating the parameters for the radioactive decay of various isotopes of radium by logginginto PhysicsNow at www.cp7e.com and going to Interactive Example 29.3.

948 Chapter 29 Nuclear PhysicsEXAMPLE 29.4GoalRadon GasCalculate the number of nuclei after an arbitrary time and the time required for a given number of nuclei to decay.22286 RnProblem Radon is a radioactive gas that can be trapped in the basements of homes, and its presence in highconcentrations is a known health hazard. Radon has a half-life of 3.83 days. A gas sample contains 4.00 10 8 radonatoms initially. (a) How many atoms will remain after 14.0 days have passed if no more radon leaks in? (b) What is theactivity of the radon sample after 14.0 days? (c) How long before 99% of the sample has decayed?Strategy The activity can be found by substitution into Equation 29.5, as before. Equation 29.4a (or Eq. 29.4b)must be used to find the number of particles remaining after 14.0 days. To obtain the time asked for in part(c), Equation 29.4a must be solved for time.Solution(a) How many atoms will remain after 14.0 days have passed?Determine the decay constant from Equation 29.5: 0.693 0.693T 1/2 3.83 days 0.181 day1Now use Equation 29.4a, taking N 0 4.0 10 8 , and thevalue of just found to obtain the number N remainingafter 14 days:N N 0 e t (4.00 10 8 atoms)e (0.181 day1 )(14.0 days) 3.17 10 7 atoms(b) What is the activity of the radon sample after14.0 days?Express the decay constant in units of s 1 :From Equation 29.3 and this value of , compute theactivity R: (0.181 day 1 ) 1 day8.64 10 4 s 2.09 106 s 1R N (2.09 10 6 s 1 )(3.17 10 7 atoms) 66.3 decays/s 66.3 Bq(c) How much time must pass before 99% of the samplehas decayed?Solve Equation 29.4a for t, using natural logarithms:Substitute values:ln(N ) ln(N 0 e t ) ln(N 0 ) ln(e t )ln(N ) ln(N 0 ) ln(e t ) tt ln(N 0) ln(N ) ln(N 0/N )t ln(N 0/0.01 N 0 )2.09 10 6 s 1 2.20 106 s 25.5 daysRemarks This kind of calculation is useful in determining how long you would have to wait for radioactivity at agiven location to fall to safe levels.Exercise 29.4(a) Find the activity of the radon sample after 12.0 days have elapsed. (b) How long would it take for 85.0% of thesample to decay?Answer (a) 95.3 Bq (b) 9.08 10 5 s 10.5 days29.4 THE DECAY PROCESSESAs stated in the previous section, radioactive nuclei decay spontaneously via alpha,beta, and gamma decay. As we’ll see in this section, these processes are very differentfrom each other.

29.4 The Decay Processes 949Alpha DecayIf a nucleus emits an alpha particle ( 4 2He), it loses two protons and two neutrons.Therefore, the neutron number N of a single nucleus decreases by 2, Z decreasesby 2, and A decreases by 4. The decay can be written symbolically asAZ X : Z2 A4 Y 4 2 He[29.8]where X is called the parent nucleus and Y is known as the daughter nucleus. Asexamples, 238 U and 226 Ra are both alpha emitters and decay according to theschemes23892 U : 23490 Th 4 2 He[29.9]and22688 Ra : 22286 Rn 4 2 He[29.10]The half-life for 238 U decay is 4.47 10 9 years, and the half-life for 226 Ra decay is1.60 10 3 years. In both cases, note that the A of the daughter nucleus is four lessthan that of the parent nucleus, while Z is reduced by two. The differences are accountedfor in the emitted alpha particle (the 4 He nucleus).The decay of 226 Ra is shown in Active Figure 29.7. When one element changesinto another, as happens in alpha decay, the process is called spontaneous decay ortransmutation. As a general rule, (1) the sum of the mass numbers A must be thesame on both sides of the equation, and (2) the sum of the atomic numbers Zmust be the same on both sides of the equation.In order for alpha emission to occur, the mass of the parent must be greaterthan the combined mass of the daughter and the alpha particle. In thedecay process, this excess mass is converted into energy of other forms andappears in the form of kinetic energy in the daughter nucleus and the alphaparticle. Most of the kinetic energy is carried away by the alpha particle becauseit is much less massive than the daughter nucleus. This can be understoodby first noting that a particle’s kinetic energy and momentum p are related asfollows:pRnBefore decayKE Rn226 Ra88222 Rn86After decayKE Ra = 0p Ra = 0KEααp αACTIVE FIGURE 29.7The alpha decay of radium-226. Theradium nucleus is initially at rest. Afterthe decay, the radon nucleus haskinetic energy KE Rn and momentump: Rn , and the alpha particle has kineticenergy KE and momentum .p: Log into PhysicsNow at www.cp7e.comand go to Active Figure 29.7, whereyou can observe the decay ofradium-226.KE p22mBecause momentum is conserved, the two particles emitted in the decay of anucleus at rest must have equal, but oppositely directed, momenta. As a result, thelighter particle, with the smaller mass in the denominator, has more kinetic energythan the more massive particle.Quick Quiz 29.3If a nucleus such as 226 Ra that is initially at rest undergoes alpha decay, which ofthe following statements is true? (a) The alpha particle has more kinetic energythan the daughter nucleus. (b) The daughter nucleus has more kinetic energythan the alpha particle. (c) The daughter nucleus and the alpha particle have thesame kinetic energy.Applying Physics 29.2Energy and Half-lifeIn comparing alpha decay energies from a number ofradioactive nuclides, why is it found that the half-life ofthe decay goes down as the energy of the decay goes up?Explanation It should seem reasonable that thehigher the energy of the alpha particle, the morelikely it is to escape the confines of the nucleus.The higher probability of escape translates to afaster rate of decay, which appears as a shorterhalf-life.

950 Chapter 29 Nuclear PhysicsEXAMPLE 29.5 Decaying RadiumGoal Calculate the energy released during an alpha decay.22688 Ra22286 RnProblem We showed that the nucleus undergoes alpha decay to (Eq. 29.10). Calculate the amount226222of energy liberated in this decay. Take the mass of 88 Ra to be 226.025 402 u, that of 86 Rn to be 222.017 571 u, and4that of 2He to be 4.002 602 u, as found in Appendix B.Strategy This is a matter of subtracting the neutral masses of the daughter particles from the original mass of theradon nucleus.SolutionCompute the sum of the mass of the daughter particle,m d , and the mass of the alpha particle, m :Compute the loss of mass, m, during the decay by subtractingthe previous result from M p , the mass of theoriginal particle:Convert the loss of mass m to its equivalent energyin MeV:m d m 222.017 571 u 4.002 602 u 226.020 173 um M p (m d m ) 226.025 402 u 226.020 173 u 0.005 229 uE (0.005 229 u)(931.494 MeV/u) 4.871 MeVRemark The potential barrier is typically higher than this value of the energy, but quantum tunneling permits theevent to occur, anyway.Exercise 29.5Calculate the energy released when84Be splits into two alpha particles. Beryllium-8 has an atomic mass of 8.005 305 u.Answer0.094 1 MeVEXAMPLE 29.6GoalBeta DecayWhen a radioactive nucleus undergoes beta decay, the daughter nucleus has the samenumber of nucleons as the parent nucleus, but the atomic number is changed by 1:AAZX : Z1 Y e[29.11]AAZX : Z1Y e [29.12]Again, note that the nucleon number and total charge are both conserved in thesedecays. However, as we will see shortly, these processes are not described completelyby these expressions. A typical beta decay event is146 C : 14 7 N e[29.13]The emission of electrons from a nucleus is surprising, because, in all our previousdiscussions, we stated that the nucleus is composed of protons and neutronsonly. This apparent discrepancy can be explained by noting that the emitted electronis created in the nucleus by a process in which a neutron is transformed intoa proton. This process can be represented by the equation10 n : 1 1 p e[29.14]Consider the energy of the system of Equation 29.13 before and after decay.As with alpha decay, energy must be conserved in beta decay. The next example14illustrates how to calculate the amount of energy released in the beta decay of 6 C.The Beta Decay of Carbon-14Calculate the energy released in a beta decay.Problem Find the energy liberated in the beta decay of to , as represented by Equation 29.13. Thatequation refers to nuclei, while Appendix B gives the masses of neutral atoms. Adding six electrons to both sides ofEquation 29.13 yields14 146 C atom : 7 N atom146 C147 N

29.4 The Decay Processes 951Strategy As in preceding problems, finding the released energy involves computing the difference in massbetween the resultant particle(s) and the initial particle(s) and converting to MeV.Solution14 14Obtain the masses of 6 C and 7N from Appendix B and m m C m N 14.003 242 u 14.003 074 u 0.000 168 ucompute the difference between them:Convert the mass difference to MeV:E (0.000 168 u)(931.494 MeV/u) 0.156 MeVRemarks The calculated energy is generally more than the energy observed in this process. The discrepancy led toa crisis in physics, because it appeared that energy wasn’t conserved. As discussed below, this crisis was resolved by thediscovery that another particle was also produced in the reaction.Exercise 29.640Calculate the maximum energy liberated in the beta decay of radioactive potassium to calcium: 19 K : 4020 Ca .Answer1.31 MeVFrom Example 29.6, we see that the energy released in the beta decay of 14 C isapproximately 0.16 MeV. As with alpha decay, we expect the electron to carry awayvirtually all of this energy as kinetic energy because, apparently, it is the lightestparticle produced in the decay. As Figure 29.8 shows, however, only a small numberof electrons have this maximum kinetic energy, represented as KE max on thegraph; most of the electrons emitted have kinetic energies lower than that predictedvalue. If the daughter nucleus and the electron aren’t carrying away this liberatedenergy, then where has the energy gone? As an additional complication,further analysis of beta decay shows that the principles of conservation of bothangular momentum and linear momentum appear to have been violated!In 1930 Pauli proposed that a third particle must be present to carry away the“missing” energy and to conserve momentum. Later, Enrico Fermi developed acomplete theory of beta decay and named this particle the neutrino (“little neutralone”) because it had to be electrically neutral and have little or no mass. Althoughit eluded detection for many years, the neutrino () was finally detected experimentallyin 1956. The neutrino has the following properties:• Zero electric charge• A mass much smaller than that of the electron, but probably not zero. (Recentexperiments suggest that the neutrino definitely has mass, but the value isuncertain—perhaps less than 1 eV/c 2 .)1• A spin of2• Very weak interaction with matter, making it difficult to detectWith the introduction of the neutrino, we can now represent the beta decayprocess of Equation 29.13 in its correct form:146 C : 14 7 N e [29.15]The bar in the symbol indicates an antineutrino. To explain what an antineutrinois, we first consider the following decay:127N : 12 6 C e [29.16] Properties of the neutrinoTIP 29.3 Mass Number of theElectronAnother notation that is sometimes0used for an electron is 1 e . Thisnotation does not imply that theelectron has zero rest energy. Themass of the electron is much smallerthan that of the lightest nucleon, sowe can approximate it as zero whenwe study nuclear decays andreactions.Number of -particlesKinetic energy(a)K maxNumber of -particlesKinetic energy(b)Figure 29.8 (a) Distribution ofbeta particle energies in a typical betadecay. All energies are observed up toa maximum value. (b) In contrast,the energies of alpha particles froman alpha decay are discrete.

952 Chapter 29 Nuclear PhysicsHere, we see that when 12 N decays into 12 C, a particle is produced which is identicalto the electron except that it has a positive charge of e. This particle is calleda positron. Because it is like the electron in all respects except charge, the positronis said to be the antiparticle of the electron. We will discuss antiparticles further inChapter 30; for now, it suffices to say that, in beta decay, an electron and an antineutrinoare emitted or a positron and a neutrino are emitted.Unlike beta decay, which results in a daughter particle with a variety of possiblekinetic energies, alpha decays come in discrete amounts, as seen in Figure 29.8b.This is because the two daughter particles have momenta with equal magnitudeand opposite direction and are each composed of a fixed number of nucleons.ENRICO FERMI, Italian Physicist(1901–1954)Fermi was awarded the Nobel Prize in1938 for producing the transuranicelements by neutron irradiation and forhis discovery of nuclear reactions boughtabout by slow neutrons. He made manyother outstanding contributions to physics,including his theory of beta decay, thefree-electron theory of metals, and thedevelopment of the world’s first fissionreactor in 1942. Fermi was truly a giftedtheoretical and experimental physicist. Hewas also well known for his ability topresent physics in a clear and excitingmanner. “Whatever Nature has in store formankind, unpleasant as it may be, menmust accept, for ignorance is never betterthan knowledge.”APPLICATIONCarbon Dating of theDead Sea ScrollsNational Accelerator LaboratoryGamma DecayVery often a nucleus that undergoes radioactive decay is left in an excited energystate. The nucleus can then undergo a second decay to a lower energy state—perhaps even to the ground state—by emitting one or more high-energy photons.The process is similar to the emission of light by an atom. An atom emits radiationto release some extra energy when an electron “jumps” from a state of high energyto a state of lower energy. Likewise, the nucleus uses essentially the same methodto release any extra energy it may have following a decay or some other nuclearevent. In nuclear de-excitation, the “jumps” that release energy are made by protonsor neutrons in the nucleus as they move from a higher energy level to a lowerlevel. The photons emitted in the process are called gamma rays, which have veryhigh energy relative to the energy of visible light.A nucleus may reach an excited state as the result of a violent collision withanother particle. However, it’s more common for a nucleus to be in an excitedstate as a result of alpha or beta decay. The following sequence of events typifiesthe gamma decay processes:125 B : 12 6C * e [29.17]126 C * 12: 6 C [29.18]Equation 29.17 represents a beta decay in which 12 B decays to 12 C * , where theasterisk indicates that the carbon nucleus is left in an excited state following thedecay. The excited carbon nucleus then decays to the ground state by emitting agamma ray, as indicated by Equation 29.18. Note that gamma emission doesn’tresult in any change in either Z or A.Practical Uses of RadioactivityCarbon Dating The beta decay of 14 C given by Equation 29.15 is commonly usedto date organic samples. Cosmic rays (high-energy particles from outer space) inthe upper atmosphere cause nuclear reactions that create 14 C from 14 N. In fact,the ratio of 14 C to 12 C (by numbers of nuclei) in the carbon dioxide molecules ofour atmosphere has a constant value of about 1.3 10 12 , as determined by measuringcarbon ratios in tree rings. All living organisms have the same ratio of 14 Cto 12 C because they continuously exchange carbon dioxide with their surroundings.When an organism dies, however, it no longer absorbs 14 C from the atmosphere,so the ratio of 14 C to 12 C decreases as the result of the beta decay of 14 C. It’stherefore possible to determine the age of a material by measuring its activity perunit mass as a result of the decay of 14 C. Through carbon dating, samples of wood,charcoal, bone, and shell have been identified as having lived from 1 000 to 25 000years ago. This knowledge has helped researchers reconstruct the history of livingorganism—including human—during that time span.A particularly interesting example is the dating of the Dead Sea Scrolls. Thisgroup of manuscripts was first discovered by a young Bedouin boy in a cave atQumran near the Dead Sea in 1947. Translation showed the manuscripts to bereligious documents, including most of the books of the Old Testament. Becauseof their historical and religious significance, scholars wanted to know their age.Carbon dating applied to fragments of the scrolls and to the material in which

29.4 The Decay Processes 953they were wrapped established that they were about 1950 years old. The scrolls arenow stored at the Israel museum in Jerusalem.Smoke Detectors Smoke detectors are frequently used in homes and industry forfire protection. Most of the common ones are the ionization-type that use radioactivematerials. (See Fig. 29.9.) A smoke detector consists of an ionization chamber,a sensitive current detector, and an alarm. A weak radioactive source ionizes theair in the chamber of the detector, which creates charged particles. A voltage ismaintained between the plates inside the chamber, setting up a small butdetectable current in the external circuit. As long as the current is maintained, thealarm is deactivated. However, if smoke drifts into the chamber, the ions becomeattached to the smoke particles. These heavier particles do not drift as readily asdo the lighter ions, which causes a decrease in the detector current. The externalcircuit senses this decrease in current and sets off the alarm.Radon Detection Radioactivity can also affect our daily lives in harmful ways. Soonafter the discovery of radium by the Curies, it was found that the air in contact withradium compounds becomes radioactive. It was then shown that this radioactivitycame from the radium itself, and the product was therefore called “radium emanation.”Rutherford and Soddy succeeded in condensing this “emanation,” confirmingthat it was a real substance: the inert, gaseous element now called radon (Rn).Later, it was discovered that the air in uranium mines is radioactive because of thepresence of radon gas. The mines must therefore be well ventilated to help protectthe miners. Finally, the fear of radon pollution has moved from uranium minesinto our own homes. (See Example 29.4.) Because certain types of rock, soil, brick,and concrete contain small quantities of radium, some of the resulting radon gasfinds its way into our homes and other buildings. The most serious problems arisefrom leakage of radon from the ground into the structure. One practical remedy isto exhaust the air through a pipe just above the underlying soil or gravel directly tothe outdoors by means of a small fan or blower.APPLICATIONSmoke DetectorsCurrentdetector+ –AlarmRadioactivesourceIonsFigure 29.9 An ionization-typesmoke detector. Smoke enteringthe chamber reduces the detectedcurrent, causing the alarm to sound.APPLICATIONRadon PollutionApplying Physics 29.3In 1991, a German tourist discovered the well-preservedremains of a man trapped in a glacier in the ItalianAlps. (See Fig. 29.10.) Radioactive dating of a sampleof bone from this hunter – gatherer, dubbed the“Iceman,” revealed an age of 5 300 years. Whydid scientists date the sample using the isotope 14 C,rather than 11 C, a beta emitter with a half-life of20.4 min?Explanation 14 C has a long half-life of 5 730 years, sothe fraction of 14 C nuclei remaining after one half-lifeis high enough to accurately measure changes in thesample’s activity. The 11 C isotope, which has a veryshort half-life, is not useful, because its activitydecreases to a vanishingly small value over the ageof the sample, making it impossible to detect.If a sample to be dated is not very old—say, about50 years—then you should select the isotope of someother element with half-life comparable to the age ofthe sample. For example, if the sample containedhydrogen, you could measure the activity of 3 H(tritium), a beta emitter of half-life 12.3 years. As ageneral rule, the expected age of the sampleshould be long enough to measure a change inRadioactive Dating of the IcemanHanny Paul/Gamma Liaisonactivity, but not so long that its activity can’t bedetected.Figure 29.10 (Applying Physics 29.3) The body of an ancient man(dubbed the Iceman) was exposed by a melting glacier in the Alps.

954 Chapter 29 Nuclear PhysicsEXAMPLE 29.7GoalTo use radioactive dating techniques, we need to recast some of the equationsalready introduced. We start by multiplying both sides of Equation 29.4 by :N N 0 e tFrom Equation 29.3, we have N R and N 0 R 0 . Substitute these expressionsinto the above equation and divide through by R 0 :RR 0 e tR is the present activity and R 0 was the activity when the object in question waspart of a living organism. We can solve for time by taking the natural logarithm ofboth sides of the foregoing equation:Should We Report This to Homicide?Apply the technique of carbon-14 dating.ln RR 0 ln(e t ) tt ln RR 0[29.19]Problem A 50.0-g sample of carbon is taken from the pelvis bone of a skeleton and is found to have a carbon-14decay rate of 200.0 decays/min. It is known that carbon from a living organism has a decay rate of 15.0 decays/min gand that 14 C has a half-life of 5 730 yr 3.01 10 9 min. Find the age of the skeleton.Strategy Calculate the original activity and the decay constant, and then substitute those numbers and the currentactivity into Equation 29.19.SolutionCalculate the original activity R 0 from the decay rateand the mass of the sample:Find the decay constant from Equation 29.5:R is given, so now we substitute all values into Equation29.19 to find the age of the skeleton:R 0 15.0 decaysdecays(50.0 g) 7.50 102mingmin 0.693t T 1/2ln RR 0 5.74 10 9 min 0.6933.01 10 9 min 2.30 1010 min 11.322.30 10 10 min 1ln 200.0 decays/min7.50 10 2 decays/min2.30 10 10 min 11.09 10 4 yrRemarkFor much longer periods, other radioactive substances with longer half-lives must be used to develop estimates.Exercise 29.7A sample of carbon of mass 7.60 g taken from an animal jawbone has an activity of 4.00 decays/min. How old is thejawbone?Answer2.77 10 4 yrAPPLICATIONCarbon-14 Dating of theShroud of TurinCarbon-14 and the Shroud of TurinSince the Middle Ages, many people have marveled at a 14-foot-long, yellowingpiece of linen found in Turin, Italy, purported to be the burial shroud of Jesus Christ(Fig. 29.11). The cloth bears a remarkable, full-size likeness of a crucified body, with

29.6 Nuclear Reactions 955wounds on the head that could have been caused by a crown of thorns and anotherwound in the side that could have been the cause of death. Skepticism over theauthenticity of the shroud has existed since its first public showing in 1354; in fact, aFrench bishop declared it to be a fraud at the time. Because of its controversialnature, religious bodies have taken a neutral stance on its authenticity.In 1978 the bishop of Turin allowed the cloth to be subjected to scientific analysis,but notably missing from these tests was carbon-14 dating. The reason for thisomission was that, at the time, carbon-dating techniques required a piece of clothabout the size of a handkerchief. In 1988 the process had been refined to thepoint that pieces as small as one square inch were sufficient, and at that timepermission was granted to allow the dating to proceed. Three labs were selectedfor the testing, and each was given four pieces of material. One of these was apiece of the shroud, and the other three pieces were control pieces similar inappearance to the shroud.The testing procedure consisted of burning the cloth to produce carbon dioxide,which was then converted chemically to graphite. The graphite sample wassubjected to carbon-14 analysis, and in the end all three labs agreed amazingly wellon the age of the shroud. The average of their results gave a date for the cloth ofA.D. 1 320 60 years, with an assurance that the cloth could not be older thanA.D. 1 200. Carbon-14 dating has thus unraveled the most important mystery concerningthe shroud, but others remain. For example, investigators have not yetbeen able to explain how the image was imprinted.Santi Visali/The IMAGE BankFigure 29.11 The Shroud of Turinas it appears in a photographic negativeimage.29.5 NATURAL RADIOACTIVITYRadioactive nuclei are generally classified into two groups: (1) unstable nucleifound in nature, which give rise to what is called natural radioactivity, and (2)nuclei produced in the laboratory through nuclear reactions, which exhibitartificial radioactivity.Three series of naturally occurring radioactive nuclei exist (Table 29.2). Eachstarts with a specific long-lived radioactive isotope with half-life exceeding that ofany of its descendants. The fourth series in Table 29.2 begins with 237 Np, atransuranic element (an element having an atomic number greater than that ofuranium) not found in nature. This element has a half-life of “only” 2.14 10 6 yr.The two uranium series are somewhat more complex than the 232 Th series(Fig. 29.12). Also, there are several other naturally occurring radioactive isotopes,such as 14 C and 40 K, that are not part of either decay series.Natural radioactivity constantly supplies our environment with radioactive elementsthat would otherwise have disappeared long ago. For example, because theSolar System is about 5 10 9 years old, the supply of 226 Ra (with a half-life of only1 600 yr) would have been depleted by radioactive decay long ago were it not forthe decay series that starts with 238 U, with a half-life of 4.47 10 9 yr.U Uranium23892 4.47 10 982 PbN140Rae – e – Ac224 Ra135220 Rn216 Po212Pb 130e – 212Bi208 TlPoe – 125 PbZ80 85 90Figure 29.12 Decay series beginningwith 232 Th.TABLE 29.2The Four Radioactive SeriesSeries Starting Isotope Half-life (years) Stable End ProductActinium2357.04 10 829.6 NUCLEAR REACTIONSIt is possible to change the structure of nuclei by bombarding them with energeticparticles. Such changes are called nuclear reactions. Rutherford was the first toobserve nuclear reactions, using naturally occurring radioactive sources for the20620792 U 82 Pb23220890 Th 82 Pb23793 Np 83 BiThorium 1.41 10 10Neptunium 2.14 10 6 209232Th228 Th

956 Chapter 29 Nuclear Physicsbombarding particles. He found that protons were released when alpha particleswere allowed to collide with nitrogen atoms. The process can be representedsymbolically as42 He 14 7N : X 1 1 H[29.20]This equation says that an alpha particle ( 4 2 He) strikes a nitrogen nucleus and producesan unknown product nucleus (X) and a proton ( 1 1 H) . Balancing atomicnumbers and mass numbers, as we did for radioactive decay, enables us to concludethat the unknown is characterized as 8 X. Because the element with atomic17number 8 is oxygen, we see that the reaction is42 He 14 7 N : 17 8 O 1 1H[29.21]This nuclear reaction starts with two stable isotopes—helium and nitrogen—andproduces two different stable isotopes—hydrogen and oxygen.Since the time of Rutherford, thousands of nuclear reactions have been observed,particularly following the development of charged-particle accelerators inthe 1930s. With today’s advanced technology in particle accelerators and particledetectors, it is possible to achieve particle energies of at least 1 000 GeV 1TeV.These high-energy particles are used to create new particles whose properties arehelping to solve the mysteries of the nucleus (and indeed, of the Universe itself).Quick Quiz 29.4Which of the following are possible reactions?1(a) 0n 23592U : 14054Xe 9438Sr 2( 1 0n)1(b) 0n 23592U : 132 10150Sn 42 Mo 3(1 0 n)(c) 1 0n 23994Pu : 12753I 93 41Nb 3( 1 0 n)EXAMPLE 29.8GoalThe Discovery of the NeutronBalance a nuclear reaction to determine an unknown decay product.Problem A nuclear reaction of significant note occurred in 1932 when Chadwick, in England, bombarded aberyllium target with alpha particles. Analysis of the experiment indicated that the following reaction occurred:What isAZXin this reaction?42He 9 4Be : 12 6C A Z XStrategyBalancing mass numbers and atomic numbers yields the answer.SolutionWrite an equation relating the atomic masses on either side:4 9 12 A : A 1Write an equation relating the atomic numbers: 2 4 6 Z : Z 0Identify the particle:AZX 10 n (a neutron)RemarksThis experiment was the first to provide positive proof of the existence of neutrons.Exercise 29.8Identify the unknown particle in this reaction:42 He 14 7 N : 17 8 O A Z XAnswerAZ X 1 1 H(a neutral hydrogen atom)

29.6 Nuclear Reactions 957EXAMPLE 29.9 Synthetic ElementsGoal Construct equations for a series of radioactive decays.Problem (a) A beam of neutrons is directed at a target of . The reaction products are a gamma ray and anotherisotope. What is the isotope? (b) The isotope 92 U is radioactive and undergoes beta decay. Write the equation239symbolizing this decay and identify the resulting isotope.23892 UStrategyBalance the mass numbers and atomic numbers on both sides of the equations.Solution(a) Identify the isotope produced by the reaction of a238neutron with a target of 92 U, with production of agamma ray.Write an equation for the reaction in terms of theunknown isotope:Write and solve equations for the atomic mass andatomic number:Identify the isotope:10 n 23892 U : AZ X A 1 238 239; Z 0 92 92AZ X 23992 U, iden-(b) Write the equation for the beta decay oftifying the resulting isotope.23992 U23992 Uby beta emis-Write an equation for the decay ofsion in terms of the unknown isotope:23992U :AZ Y e Write and solve equations for the atomic mass andcharge conservation (the electron counts as 1 on theright):A 239; 92 Z 1 : Z 93Identify the isotope: Z Y 239 93 Np (neptunium)Remarks The interesting feature of these reactions is the fact that uranium is the element with the greatest numberof protons (92) which exists in nature in any appreciable amount. The reactions in parts (a) and (b) do occur occasionallyin nature; hence, minute traces of neptunium and plutonium are present. In 1940, however, researchersbombarded uranium with neutrons to produce plutonium and neptunium. These two elements were the first elementsmade in the laboratory. Since then, the list of synthetic elements has been extended to include those up toatomic number 112. Recently, elements 113 and 115 have been observed, but as of this writing, their existence hasnot yet been confirmed.Exercise 29.9The isotope23893 UAnswer23994 Puis also radioactive and decays by beta emission. What is the end product?AQ ValuesWe have just examined some nuclear reactions for which mass numbers andatomic numbers must be balanced in the equations. We will now consider the energyinvolved in these reactions, because energy is another important quantity thatmust be conserved.We illustrate this procedure by analyzing the following nuclear reaction:21 H 14 7 N : 12 6 C 4 2He[29.22]

958 Chapter 29 Nuclear PhysicsThe total mass on the left side of the equation is the sum of the mass of2141H (2.014 102 u) and the mass of 7 N (14.003 074 u), which equals 16.017 176 u.Similarly, the mass on the right side of the equation is the sum of the mass of12 4(12.000 000 u) plus the mass of 2 He6 C(4.002 602 u), for a total of 16.002 602 u.Thus, the total mass before the reaction is greater than the total mass afterthe reaction. The mass difference in the reaction is equal to 16.017 176 u 16.002 602 u 0.014 574 u. This “lost” mass is converted to the kinetic energy ofthe nuclei present after the reaction. In energy units, 0.014 574 u is equivalent to13.576 MeV of kinetic energy carried away by the carbon and helium nuclei.The energy required to balance the equation is called the Q value of the reaction.In Equation 29.22, the Q value is 13.576 MeV. Nuclear reactions in which there is arelease of energy—that is, positive Q values—are said to be exothermic reactions.The energy balance sheet isn’t complete, however: We must also consider thekinetic energy of the incident particle before the collision. As an example, assumethat the deuteron in Equation 29.22 has a kinetic energy of 5 MeV. Adding this toour Q value, we find that the carbon and helium nuclei have a total kinetic energyof 18.576 MeV following the reaction.Now consider the reaction42 He 14 7 N : 17 8O 1 1H[29.23]Before the reaction, the total mass is the sum of the masses of the alpha particle andthe nitrogen nucleus: 4.002 602 u 14.003 074 u 18.005 676 u. After the reaction,the total mass is the sum of the masses of the oxygen nucleus and the proton:16.999 133 u 1.007 825 u 18.006 958 u. In this case, the total mass after the reactionis greater than the total mass before the reaction. The mass deficit is 0.001 282 u,equivalent to an energy deficit of 1.194 MeV. This deficit is expressed by the negativeQ value of the reaction, 1.194 MeV. Reactions with negative Q values are calledendothermic reactions. Such reactions won’t take place unless the incoming particlehas at least enough kinetic energy to overcome the energy deficit.At first it might appear that the reaction in Equation 29.23 can take place if theincoming alpha particle has a kinetic energy of 1.194 MeV. In practice, however,the alpha particle must have more energy than this. If it has an energy of only1.194 MeV, energy is conserved but careful analysis shows that momentum isn’t.This can be understood by recognizing that the incoming alpha particle has somemomentum before the reaction. However, if its kinetic energy is only 1.194 MeV,the products (oxygen and a proton) would be created with zero kinetic energy andthus zero momentum. It can be shown that in order to conserve both energy andmomentum, the incoming particle must have a minimum kinetic energy given byKE min 1 m M Q [29.24]where m is the mass of the incident particle, M is the mass of the target, and theabsolute value of the Q value is used. For the reaction given by Equation 29.23, wefind thatKE min 1 4.002 602 1.194 MeV 1.535 MeV14.003 074This minimum value of the kinetic energy of the incoming particle is called thethreshold energy. The nuclear reaction shown in Equation 29.23 won’t occur if theincoming alpha particle has a kinetic energy of less than 1.535 MeV, but can occurif its kinetic energy is equal to or greater than 1.535 MeV.Quick Quiz 29.5If the Q value of an endothermic reaction is 2.17 MeV, then the minimum kineticenergy needed in the reactant nuclei if the reaction is to occur must be (a) equalto 2.17 MeV, (b) greater than 2.17 MeV, (c) less than 2.17 MeV, or (d) exactly halfof 2.17 MeV.

29.7 Medical Applications of Radiation 95929.7 MEDICAL APPLICATIONS OF RADIATIONRadiation Damage in MatterRadiation absorbed by matter can cause severe damage. The degree and kind ofdamage depend on several factors, including the type and energy of the radiationand the properties of the absorbing material. Radiation damage in biologicalorganisms is due primarily to ionization effects in cells. The normal function of acell may be disrupted when highly reactive ions or radicals are formed as the resultof ionizing radiation. For example, hydrogen and hydroxyl radicals producedfrom water molecules can induce chemical reactions that may break bonds inproteins and other vital molecules. Large acute doses of radiation are especiallydangerous because damage to a great number of molecules in a cell may cause thecell to die. Also, cells that do survive the radiation may become defective, whichcan lead to cancer.In biological systems, it is common to separate radiation damage into two categories:somatic damage and genetic damage. Somatic damage is radiation damageto any cells except the reproductive cells. Such damage can lead to cancer at highradiation levels or seriously alter the characteristics of specific organisms. Geneticdamage affects only reproductive cells. Damage to the genes in reproductive cellscan lead to defective offspring. Clearly, we must be concerned about the effect ofdiagnostic treatments, such as x-rays and other forms of exposure to radiation.Several units are used to quantify radiation exposure and dose. The roentgen(R) is defined as that amount of ionizing radiation which will produce 2.08 10 9ion pairs in 1 cm 3 of air under standard conditions. Equivalently, the roentgen isthat amount of radiation which deposits 8.76 10 3 J of energy into 1 kg of air.For most applications, the roentgen has been replaced by the rad (an acronymfor radiation absorbed dose), defined as follows: One rad is that amount of radiationwhich deposits 10 2 J of energy into 1 kg of absorbing material.Although the rad is a perfectly good physical unit, it’s not the best unit formeasuring the degree of biological damage produced by radiation, because thedegree of damage depends not only on the dose, but also on the type of radiation.For example, a given dose of alpha particles causes about 10 times more biologicaldamage than an equal dose of x-rays. The RBE (relative biological effectiveness)factor is defined as the number of rads of x-radiation or gamma radiation that producesthe same biological damage as 1 rad of the radiation being used. The RBEfactors for different types of radiation are given in Table 29.3. Note that the valuesare only approximate because they vary with particle energy and the form ofdamage.Finally, the rem (roentgen e quivalent in man) is defined as the product of thedose in rads and the RBE factor:Dose in rem dose in rads RBEAccording to this definition, 1 rem of any two kinds of radiation will produce thesame amount of biological damage. From Table 29.3, we see that a dose of 1 rad offast neutrons represents an effective dose of 10 rem and that 1 rad of x-radiation isequivalent to a dose of 1 rem.TABLE 29.3RBE Factors for Several Types of RadiationRadiationRBE FactorX-rays and gamma rays 1.0Beta particles 1.0–1.7Alpha particles 10–20Slow neutrons 4–5Fast neutrons and protons 10Heavy ions 20

960 Chapter 29 Nuclear PhysicsAPPLICATIONOccupational RadiationExposure LimitsAPPLICATIONIrradiation of Foodand Medical EquipmentAPPLICATIONRadioactive Tracersin MedicineAPPLICATIONRadioactive Tracers inAgricultural ResearchLow-level radiation from natural sources, such as cosmic rays and radioactiverocks and soil, delivers a dose of about 0.13 rem/year per person. The upper limitof radiation dose recommended by the U.S. government (apart from backgroundradiation and exposure related to medical procedures) is 0.5 rem/year. Many occupationsinvolve higher levels of radiation exposure, and for individuals in theseoccupations, an upper limit of 5 rem/year has been set for whole-body exposure.Higher upper limits are permissible for certain parts of the body, such as thehands and forearms. An acute whole-body dose of 400 to 500 rem results in a mortalityrate of about 50%. The most dangerous form of exposure is ingestion or inhalationof radioactive isotopes, especially those elements the body retains andconcentrates, such as 90 Sr. In some cases, a dose of 1000 rem can result from ingesting1 mCi of radioactive material.Sterilizing objects by exposing them to radiation has been going on for at least25 years, but in recent years the methods used have become safer to use and moreeconomical. Most bacteria, worms, and insects are easily destroyed by exposure togamma radiation from radioactive cobalt. There is no intake of radioactive nuclei byan organism in such sterilizing processes, as there is in the use of radioactive tracers.The process is highly effective in destroying Trichinella worms in pork, salmonellabacteria in chickens, insect eggs in wheat, and surface bacteria on fruits and vegetablesthat can lead to rapid spoilage. Recently, the procedure has been expanded to includethe sterilization of medical equipment while in its protective covering. Surgical gloves,sponges, sutures, and so forth are irradiated while packaged. Also, bone, cartilage, andskin used for grafting is often irradiated to reduce the chance of infection.TracingRadioactive particles can be used to trace chemicals participating in various reactions.One of the most valuable uses of radioactive tracers is in medicine. For example,131 I is an artificially produced isotope of iodine. (The natural, nonradioactiveisotope is 127 I.) Iodine, a necessary nutrient for our bodies, is obtained largelythrough the intake of seafood and iodized salt. The thyroid gland plays a majorrole in the distribution of iodine throughout the body. In order to evaluate theperformance of the thyroid, the patient drinks a small amount of radioactivesodium iodide. Two hours later, the amount of iodine in the thyroid gland is determinedby measuring the radiation intensity in the neck area.A medical application of the use of radioactive tracers occurring in emergencysituations is that of locating a hemorrhage inside the body. Often the location ofthe site cannot easily be determined, but radioactive chromium can identify thelocation with a high degree of precision. Chromium is taken up by red blood cellsand carried uniformly throughout the body. However, the blood will be dumped ata hemorrhage site, and the radioactivity of that region will increase markedly.The tracer technique is also useful in agricultural research. Suppose the bestmethod of fertilizing a plant is to be determined. A certain material in the fertilizer,such as nitrogen, can be tagged with one of its radioactive isotopes. The fertilizeris then sprayed onto one group of plants, sprinkled on the ground for asecond group, and raked into the soil for a third. A Geiger counter is then used totrack the nitrogen through the three types of plants.Tracing techniques are as wide ranging as human ingenuity can devise. Presentapplications range from checking the absorption of fluorine by teeth to checkingcontamination of food-processing equipment by cleansers to monitoring deteriorationinside an automobile engine. In the last case, a radioactive material is usedin the manufacture of the pistons, and the oil is checked for radioactivity to determinethe amount of wear on the pistons.Computed Axial Tomography (CAT) ScansThe normal x-ray of a human body has two primary disadvantages when used as asource of clinical diagnosis. First, it is difficult to distinguish between various typesof tissue in the body because they all have similar x-ray absorption properties.

29.7 Medical Applications of Radiation 961Second, a conventional x-ray absorption picture is indicative of the averageamount of absorption along a particular direction in the body, leading to somewhatobscured pictures. To overcome these problems, an instrument called a CATscanner was developed in England in 1973; the device is capable of producing picturesof much greater clarity and detail than previously possible.The operation of a CAT scanner can be understood by considering the followinghypothetical experiment: suppose a box consists of four compartments,labeled A, B, C, and D, as in Figure 29.13a. Each compartment has a differentamount of absorbing material from any other compartment. What set of experimentalprocedures will enable us to determine the relative amounts of material ineach compartment? The following steps outline one method that will provide thisinformation: first, a beam of x-rays is passed through compartments A and C, as inFigure 29.13b. The intensity of the exiting radiation is reduced by absorption bysome number that we assign as 8. (The number 8 could mean, for example, thatthe intensity of the exiting beam is reduced by eight-tenths of 1% from its initialvalue.) Because we don’t know which of the compartments, A or C, was responsiblefor this reduction in intensity, half the loss is assigned to each compartment, asin Figure 29.13c. Next, a beam of x-rays is passed through compartments B and D,as in Figure 29.13b. The reduction in intensity for this beam is 10, and again we assignhalf the loss to each compartment. We now redirect the x-ray source so that itsends one beam through compartments A and B and another through compartmentsC and D, as in Figure 29.13d. Once more, we measure the absorption. Supposethe absorption through compartments A and B in this experiment is measuredto be 7 units. On the basis of our first experiment, we would have guessed itwould be 9 units: 4 by compartment A and 5 by compartment B. Thus, we have reducedthe guessed absorption for each compartment by 1 unit, so that the sum is 7rather than 9, giving the numbers shown in Figure 29.13e. Likewise, when thebeam is passed through compartments C and D, as in Figure 29.13d, we may findthe total absorption to be 11 as compared to our first experiment of 9. In this case,we add 1 unit of absorption to each compartment to give a sum of 11, as in Figure29.13e. This somewhat crude procedure could be improved by measuring the absorptionalong other paths. However, these simple measurements are sufficient toenable us to conclude that compartment D contains the most absorbing materialand A the least. A visual representation of these results can be obtained by assigning,to each compartment, a shade of gray corresponding to the particular numberAPPLICATIONCAT ScansExitbeam810ABABCD(a)CD(b)Incidentbeam4 5A B4 5C D(c)IncidentbeamACBD(d)711Exitbeam3 4A B5 6C D(e)Figure 29.13 An experimental procedure fordetermining the relative amounts of x-ray absorptionby four different compartments in a box.

962 Chapter 29 Nuclear Physicsassociated with the absorption. In this example, compartment D would be verydark and compartment A would be very light.The steps outlined previously are representative of how a CAT scanner producesimages of the human body. A thin slice of the body is subdivided into perhaps10 000 compartments, rather than 4 as in our simple example. The functionof the CAT scanner is to determine the relative absorption in each of these 10 000compartments and to display a picture of its calculations in various shades of gray.Note that “CAT” stands for computed axial tomography. The term axial is usedbecause the slice of the body to be analyzed corresponds to a plane perpendicularto the head-to-toe axis. Tomos is the Greek word for slice and graph is the Greekword for picture. In a typical diagnosis, the patient is placed in the position shownin Figure 29.14 and a narrow beam of x-rays is sent through the plane of interest.The emerging x-rays are detected and measured by photomultiplier tubes behindthe patient. The x-ray tube is then rotated a few degrees, and the intensity isrecorded again. An extensive amount of information is obtained by rotating thebeam through 180° at intervals of about 1° per measurement, resulting in a set ofnumbers assigned to each of the 10 000 “compartments” in the slice. These numbersare then converted by the computer to a photograph in various shades of grayfor the segment of the body that is under observation.A brain scan of a patient can now be made in about 2 s, and a full-body scanrequires about 6 s. The final result is a picture containing much greater quantitativeinformation and clarity than a conventional x-ray photograph. Because CATscanners use x-rays, which are an ionizing form of radiation, the technique presentsa modest health risk to the patient being diagnosed.APPLICATIONMagnetic ResonanceImaging (MRI)Magnetic Resonance Imaging (MRI)At the heart of magnetic resonance imaging (MRI) is the fact that when a nucleushaving a magnetic moment is placed in an external magnetic field, its momentprecesses about the magnetic field with a frequency that is proportional to thefield. For example, a proton, with a spin of 1/2, can occupy one of two energystates when placed in an external magnetic field. The lower energy state correspondsto the case in which the spin is aligned with the field, whereas the higherenergy state corresponds to the case in which the spin is opposite the field. Transitionsbetween these two states can be observed with a technique known as nuclearmagnetic resonance. A DC magnetic field is applied to align the magnetic moments,and a second, weak oscillating magnetic field is applied perpendicular tothe DC field. When the frequency of the oscillating field is adjusted to match theprecessional frequency of the magnetic moments, the nuclei will “flip” betweenX-raydetectorsX-rayPatient(a)Figure 29.14 (a) CAT scanner detector assembly. (b) Photograph of a patient undergoing a CATscan in a hospital.Jay Freis/The Image Bank(b)

29.8 Radiation Detectors 963Figure 29.15 Computer-enhancedMRI images of (a) a normal humanbrain and (b) a human brain with aglioma tumor.a, SBHA/Getty Images; b, Scott Camazine/ScienceSource/Photo Researchers, Inc.(a)(b)the two spin states. These transitions result in a net absorption of energy by thespin system, which can be detected electronically.In MRI, image reconstruction is obtained using spatially varying magnetic fieldsand a procedure for encoding each point in the sample being imaged. Some MRIimages taken on a human head are shown in Figure 29.15. In practice, a computer-controlledpulse-sequencing technique is used to produce signals that arecaptured by a suitable processing device. The signals are then subjected to appropriatemathematical manipulations to provide data for the final image. The mainadvantage of MRI over other imaging techniques in medical diagnostics is that itcauses minimal damage to cellular structures. Photons associated with the rf signalsused in MRI have energies of only about 10 7 eV. Because molecular bondstrengths are much larger (on the order of 1 eV), the rf photons cause little cellulardamage. In comparison, x-rays or -rays have energies ranging from 10 4 to 10 6 eVand can cause considerable cellular damage.29.8 RADIATION DETECTORSAlthough most medical applications of radiation require instruments to makequantitative measurements of radioactive intensity, we have not yet explained howsuch instruments operate. Various devices have been developed to detect the energeticparticles emitted when a radioactive nucleus decays. The Geiger counter(Fig. 29.16) is perhaps the most common device used to detect radioactivity. It canbe considered the prototype of all counters that use the ionization of a medium asthe basic detection process. A Geiger counter consists of a thin wire electrodealigned along the central axis of a cylindrical metallic tube filled with a gas at lowpressure. The wire is maintained at a high positive voltage of about 1 000 V relativeto the tube. When an energetic charged particle or gamma-ray photon entersthe tube through a thin window at one end, some of the gas atoms are ionized.The electrons removed from these atoms are attracted toward the wire electrode,Wire electrodeGasThinwindowTo amplifierandcounter+(a)–Metallictube© Hank Morgan/Photo Researchers(b)Figure 29.16 (a) Diagram of aGeiger counter. The voltage betweenthe wire electrode and the metallictube is usually about 1 000 V. (b) AGeiger counter.

964 Chapter 29 Nuclear Physicsand in the process they ionize other atoms in their path. This sequential ionizationresults in an avalanche of electrons that produces a current pulse. After the pulsehas been amplified, it can either be used to trigger an electronic counter or deliveredto a loudspeaker that clicks each time a particle is detected. Although aGeiger counter reliably detects the presence and quantity of radiation, it cannotbe used to measure the energy of the detected radiation.A semiconductor diode detector is essentially a reverse biased p–n junction. Asan energetic particle passes through the junction, it produces electron–hole pairsthat are separated by the internal electric field. This movement of electrons andholes creates a brief pulse of current that is measured with an electronic counter.In a typical device, the duration of the pulse is 10 8 s.A scintillation counter usually uses a solid or liquid material having atoms thatare easily excited by radiation. The excited atoms then emit photons of visiblelight when they return to their ground state. Common materials used as scintillatorsare transparent crystals of sodium iodide and certain plastics. If the scintillatormaterial is attached to one end of a device called a photomultiplier (PM) tube, asshown in Figure 29.17, the photons emitted by the scintillator can be converted toan electrical signal. The PM tube consists of numerous electrodes, called dynodes,whose electric potentials increase in succession along the length of the tube.Between the top of the tube and the scintillator material is a plate called aphotocathode. When photons leaving the scintillator hit this plate, electrons areemitted because of the photoelectric effect. As one of these emitted electronsstrikes the first dynode, the electron has sufficient kinetic energy to eject severalother electrons from the surface of the dynode. When these electrons areaccelerated to the second dynode, many more electrons are ejected, and amultiplication process occurs. The end result is 1 million or more electronsstriking the last dynode. Hence, one particle striking the scintillator produces asizable electrical pulse at the PM output, and this pulse is sent to an electroniccounter.Both the scintillator and the semiconductor diode detector are much more sensitivethan a Geiger counter, mainly because of the higher mass density of thedetecting medium. Both can also be used to measure particle energy from theheight of the pulses produced.Track detectors are various devices used to view the tracks or paths of chargedparticles directly. High-energy particles produced in particle accelerators may haveenergies ranging from 10 9 to 10 12 eV. The energy of such particles can’t bemeasured with the small detectors already mentioned. Instead, their energy andScintillationcrystalPhotocathode0 VIncomingparticle+200 V+400 V+600 V+800 V+1 000 V+1 200 V+1 400 V+1 600 VVacuumFigure 29.17 Diagram of a scintillation counterconnected to a photomultiplier tube.Outputto counter

Summary 965Photo Researchers, Inc./Science Photo Library(a)Dan McCoy/Rainbow(b)Figure 29.18 (a) Artificially coloredphotograph showing tracks ofparticles that have passed through abubble chamber. (b) This researchscientist is studying a photograph ofparticle tracks made in a bubblechamber at Fermilab.momentum are found from the curvature of their paths in a magnetic field ofknown magnitude and direction.A photographic emulsion is the simplest example of a track detector. A chargedparticle ionizes the atoms in an emulsion layer. The path of the particle correspondsto a family of points at which chemical changes have occurred in the emulsion.When the emulsion is developed, the particle’s track becomes visible.A cloud chamber contains a gas that has been supercooled to just below itsusual condensation point. An energetic charged particle passing through ionizesthe gas along its path. The ions serve as centers for condensation of the supercooledgas. The track can be seen with the naked eye and can be photographed. Amagnetic field can be applied to determine the charges of the radioactive particles,as well as their momentum and energy.A device called a bubble chamber, invented in 1952 by D. Glaser, uses a liquid(usually liquid hydrogen) maintained near its boiling point. Ions produced by incomingcharged particles leave bubblelike tracks, which can be photographed(Fig. 29.18). Because the density of the liquid in a bubble chamber is much higherthan the density of the gas in a cloud chamber, the bubble chamber has a muchhigher sensitivity.A wire chamber consists of thousands of closely spaced parallel wires that collectthe electrons created by a passing ionizing particle. A second grid, with wires perpendicularto the first, allows the x,y position of the particle in the plane of the twosets of wires to be determined. Finally several such x,y grids arranged parallel toeach other along the z-axis can be used to determine the particle’s track in threedimensions. Wire chambers form a part of most detectors used at high-energyaccelerator labs and provide electronic readouts to a computer for rapid reconstructionand display of tracks.SUMMARYTake a practice test by logging intoPhysicsNow at www.cp7e.com and clicking on the Pre-Testlink for this chapter.29.1 Some Properties of Nuclei &29.2 Binding EnergyNuclei are represented symbolically as Z X , where X representsthe chemical symbol for the element. The quantity AAis the mass number, which equals the total number ofnucleons (neutrons plus protons) in the nucleus. Thequantity Z is the atomic number, which equals the numberof protons in the nucleus. Nuclei that contain the samenumber of protons but different numbers of neutrons arecalled isotopes. In other words, isotopes have the same Zvalue but different A values.Most nuclei are approximately spherical, with anaverage radius given byr r 0 A 1/3 [29.1]

966 Chapter 29 Nuclear Physicswhere A is the mass number and r 0 is a constant equal Note that in this decay, as in all radioactive decayto 1.2 10 15 m.processes, the sum of the Z values on the left equals theThe total mass of a nucleus is always less than the sum sum of the Z values on the right; the same is true for theof the masses of its individual nucleons. This mass differencem, multiplied by c 2 , gives the binding energy of A typical beta decay isA values.the nucleus.146 C : 14 7 N e [29.15]When a nucleus undergoes beta decay, an antineutrino29.3 Radioactivityis emitted along with an electron, or a neutrino alongThe spontaneous emission of radiation by certain nuclei with a positron. A neutrino has zero electric charge andis called radioactivity. There are three processes by a small mass (which may be zero) and interacts weaklywhich a radioactive substance can decay: alpha () decay, with matter.4in which the emitted particles are 2 He nuclei; beta () Nuclei are often in an excited state following radioactivedecay, and they release their extra energy by emit-decay, in which the emitted particles are electrons orpositrons; and gamma () decay, in which the emitted ting a high-energy photon called a gamma ray (). Aparticles are high-energy photons.typical gamma ray emission isThe decay rate, or activity, R, of a sample is given by126 C * : 12 6 C [29.18]R Nt N[29.3]where N is the number of radioactive nuclei at some instantand is a constant for a given substance called thedecay constant.Nuclei in a radioactive substance decay in such a waythat the number of nuclei present varies with time accordingto the expressionwhere the asterisk indicates that the carbon nucleus wasin an excited state before gamma emission.29.6 Nuclear ReactionsNuclear reactions can occur when a bombarding particlestrikes another nucleus. A typical nuclear reaction isN N 0 e t[29.4a]42 He 14 7 N : 17 8O 1 1 H[29.21]where N is the number of radioactive nuclei present attime t, N 0 is the number at time t 0, and e 2.718 . . . isthe base of the natural logarithms.The half-life T 1/2 of a radioactive substance is thetime required for half of a given number of radioactivenuclei to decay. The half-life is related to the decay constantbyT 1/2 0.693[29.5]In this reaction, an alpha particle strikes a nitrogen nucleus,producing an oxygen nucleus and a proton. As inradioactive decay, atomic numbers and mass numbersbalance on the two sides of the arrow.Nuclear reactions in which energy is released are saidto be exothermic reactions and are characterized bypositive Q values. Reactions with negative Q values,called endothermic reactions, cannot occur unless theincoming particle has at least enough kinetic energy toovercome the energy deficit. In order to conserve bothenergy and momentum, the incoming particle musthave a minimum kinetic energy, called the thresholdenergy, given by29.4 The Decay ProcessesIf a nucleus decays by alpha emission, it loses two protonsand two neutrons. A typical alpha decay is23892 U : 23490 Th 4 2 He[29.9]KE min 1 m M Q[29.24]where m is the mass of the incident particle and M is themass of the target atom.

Problems 967CONCEPTUAL QUESTIONS1. Isotopes of a given element have different physicalproperties, such as mass, but the same chemicalproperties. Why is this?2. If a heavy nucleus that is initially at rest undergoesalpha decay, which has more kinetic energy after thedecay, the alpha particle or the daughter nucleus?3. A student claims that a heavy form of hydrogen decaysby alpha emission. How do you respond?4. Explain the main differences between alpha, beta,and gamma rays.5. In beta decay, the energy of the electron or positronemitted from the nucleus lies somewhere in a relativelylarge range of possibilities. In alpha decay,however, the alpha particle energy can only havediscrete values. Why is there this difference?6. If film is kept in a box, alpha particles from a radioactivesource outside the box cannot expose thefilm, but beta particles can. Explain.7. In positron decay, a proton in the nucleus becomesa neutron, and the positive charge is carried away bythe positron. But a neutron has a larger rest energythan a proton. How is this possible?8. An alpha particle has twice the charge of a beta particle.Why does the former deflect less than the latterwhen passing between electrically charged plates, assumingthey both have the same speed?9. Can carbon-14 dating be used to measure the age ofa stone?10. Pick any beta-decay process and show that the neutrinomust have zero charge.11. Why do heavier elements require more neutrons inorder to maintain stability?12. Suppose it could be shown that the intensity of cosmicrays was much greater 10 000 years ago. Howwould this affect the ages we assign to ancient samplesof once-living matter?13. Compare and contrast a photon and a neutrino.14. Why is carbon dating unable to provide accurate estimatesof very old materials?15. Two samples of the same radioactive nuclide areprepared. Sample A has twice the intial activity ofsample B. How does the half-life of A compare withthe half-life of B ? After each has passed through fivehalf-lives, what is the ratio of their activities?16. (a) Describe what happens to the number of protonsand neutrons in a nucleus when the nucleusundergoes alpha decay. (b) Repeat for beta decay.PROBLEMS1, 2, 3 = straightforward, intermediate, challenging full solution available in Student Solutions Manual/Study Guide coached problem with hints available at www.cp7e.com biomedical applicationTable 29.4 will be useful for many of these problems. A more TABLE 29.4complete list of atomic masses is given in Appendix B.Some Atomic Masses0231. Compare the nuclear radii of the following nuclides:,( 1 n ) 1.008 665 12Mg 22.994 1272 60 197 2391 H 27 Co , 79 Au , 94 Pu127.1 H 1.007 825 13Al 26.981 53823012. What is the order of magnitude of the number of2.014 102 15P 29.978 310440protons in your body? Of the number of neutrons?2He 4.002 602 20Ca 39.962 591742Of the number of electrons?3Li 7.016 003 20Ca 41.958 6229434Be 9.012 174 20 Ca 42.958 7703. Using the result of Example 29.1, find the radius of 10 565 B 10.012 936 26Fe 55.934 940a sphere of nuclear matter that would have a mass 12 64equal to that of the Earth. The Earth has a mass of6 C 12.000 000 30Zn 63.929 1445.98 10 24 kg and average radius of 6.37 10 6 13 64m.6 C 13.003 355 29Cu 63.929 59914 937 N 14.003 074 41Nb 92.906 3774. Consider the hydrogen atom to be a sphere of radiusequal to the Bohr radius, 0.53 10 10 7 N 15.000 108 79Au 196.966 54315 197m, and 15 2028 O 15.003 065 80Hg 201.970 617calculate the approximate value of the ratio of the 17 216nuclear density to the atomic density.8 O 16.999 131 84 Po 216.001 79018 2208 O 17.999 160 Rn 220.011 4015. An alpha particle (Z 2, mass 6.64 10 27 86kg) approachesto within 1.00 10 14 9018 2349 F 18.000 937 Th 234.043 583m of a carbon nucleus(Z 6). What are (a) the maximum Coulomb2023810Ne 19.992 435 92U 238.050 784Element Atomic Mass (u) Element Atomic Mass (u)Section 29.1 Some Properties of Nuclei0( 1 e ) 0.000 5492311Na 22.989 770

968 Chapter 29 Nuclear Physicsforce on the alpha particle, (b) the acceleration of thealpha particle at this time, and (c) the potential energyof the alpha particle at the same time?6. Singly ionized carbon atoms are acceleratedthrough 1 000 V and passed into a mass spectrometerto determine the isotopes present. (See Chapter19.) The magnetic field strength in the spectrometeris 0.200 T. (a) Determine the orbital radii for the12 C and the 13 C isotopes as they pass through thefield. (b) Show that the ratio of the radii may bewritten in the formand verify that your radii in part (a) satisfy thisformula.7. (a) Find the speed an alpha particle requires tocome within 3.2 10 14 m of a gold nucleus.(b) Find the energy of the alpha particle in MeV.48. Find the radius of a nucleus of (a) 2 He and238(b) 92 U .Section 29.2 BindingEnergy9. Calculate the average binding energy per nucleon93197of and .10. Calculate the binding energy per nucleon for(a) 2 H, (b) 4 He, (c) 56 Fe, and (d) 238 U.11. A pair of nuclei for which Z 1 N 2 and Z 2 N 1 arecalled mirror isobars. (The atomic and neutron numbersare interchangeable.) Binding-energy measurementson such pairs can be used to obtain evidenceof the charge independence of nuclear forces.Charge independence means that the proton–proton, proton–neutron, and neutron–neutronforces are approximately equal. Calculate the differencein binding energy for the two mirror nucleiand .158 O41 Nb157 N79 Aur 1r 2 √m 1m 212. The peak of the stability curve occurs at 56 Fe. This iswhy iron is prominent in the spectrum of the Sunand stars. Show that56 Fe has a higher binding energyper nucleon than its neighbors 55 Mn and .Compare your results with Figure 29.4.13. Two nuclei having the same mass number areknown as isobars. Calculate the difference in binding2323energy per nucleon for the isobars and 12 Mg .How do you account for this difference?11 Na 59 Co14. Calculate the binding energy of the last neutron43in the 20 Ca nucleus. [Hint: You should compare the43mass of 20 Ca with the mass of 20 Ca plus the mass of a42neutron. The mass of 41.958 622 u.]20 Ca 42Section 29.3 Radioactivity15.The half-life of an isotope of phosphorusis 14 days. If a sample contains 3.0 10 16such nuclei, determine its activity. Express youranswer in curies.9916. A drug tagged with 43 Tc (half-life 6.05 h) is preparedfor a patient. If the original activity of the samplewas 1.1 10 4 Bq, what is its activity after it has sat onthe shelf for 2.0 h?17. The half-life of 131 I is 8.04 days. (a) Calculate the decayconstant for this isotope. (b) Find the number of131 I nuclei necessary to produce a sample with an activityof 0.50 Ci.18. After 2.00 days, the activity of a sample of an unknowntype of radioactive material has decreased to84.2% of the initial activity. (a) What is the half-lifeof this material? (b) Can you identify it by using thetable of isotopes in Appendix B?19. Suppose that you start with 1.00 10 3 g of a pureradioactive substance and 2.0 h later determine thatonly 0.25 10 3 g of the substance remains. Whatis the half-life of this substance?20. Radon gas has a half-life of 3.83 days. If 3.00 g ofradon gas is present at time t 0, what mass ofradon will remain after 1.50 days have passed?21. Many smoke detectors use small quantities of theisotope241 Am in their operation. The half-life of241 Am is 432 yr. How long will it take for the activityof this material to decrease to 1.00 10 3 of theoriginal activity?22. After a plant or animal dies, its 14 C content decreaseswith a half-life of 5 730 yr. If an archaeologistfinds an ancient firepit containing partiallyconsumed firewood, and the 14 C content of thewood is only 12.5% that of an equal carbon samplefrom a present-day tree, what is the age of theancient site?23. A freshly prepared sample of a certain radioactiveisotope has an activity of 10.0 mCi. After 4.00 h, theactivity is 8.00 mCi. (a) Find the decay constant andhalf-life of the isotope. (b) How many atoms of theisotope were contained in the freshly preparedsample? (c) What is the sample’s activity 30 h after itis prepared?

Problems 96924. A building has become accidentally contaminated 31. Determine which of the following suggested decayswith radioactivity. The longest-lived material in can occur spontaneously:the building is strontium-90. (The atomic mass of9038 Sr 40is 89.907 7.) If the building initially contained (a) : e 40144 4 14020 Ca 19 K ; (b) 60 Nd : 2 He 58 Ce5.0 kg of this substance, and the safe level is less6632. (mass 65.929 1 u) undergoes beta decaythan 10.0 counts/min, how long will the building28 Ni 66to (mass 65.9289 u). (a) Write the completebe unsafe?29 Cudecay formula for this process. (b) Find the maximumkinetic energy of the emerging electrons.Section 29.4 The Decay Processes25. Complete the following radioactive decay formulas:21283 Bi : ? 4 2 He95 Kr : ? e 3626. Complete the following radioactive decay formulas:23423090 Th : 88 Ra ?? : 14 7 N e 27. Complete the following radioactive decay formulas:9444 Ru : 4 2 He ?140Nd : ? 58 Ce14460? : 4 2 He 140 58 Ce125 B : ? e ? : e 4019 K28. Figure P29.28 shows the steps by which 92 U decays207to . Enter the correct isotope symbol in eachsquare.235 U 9282 Pb 235e – e – e – e – e – e –33. An 3 H nucleus beta decays into 3 He by creatingan electron and an antineutrino according to thereaction31 H3: 2 He e Use Appendix B to determine the total energy releasedin this reaction.34. A piece of charcoal used for cooking is found at theremains of an ancient campsite. A 1.00-kg sample ofcarbon from the wood has an activity of 2.00 10 3decays per minute. Find the age of the charcoal.[Hint: Living material has an activity of 15.0 decays/minuteper gram of carbon present.]35. A wooden artifact is found in an ancient tomb. Its carbon-14( 6 C) activity is measured to be 60.0% of that14in a fresh sample of wood from the same region. Assumingthe same amount of 14 C was initially presentin the wood from which the artifact was made, determinethe age of the artifact.36. A living specimen in equilibrium with the atmospherecontains one atom of 14 C (half-life 5 730 yr)for every 7.70 10 11 stable carbon atoms. An archeologicalsample of wood (cellulose, C 12 H 22 O 11 )contains 21.0 mg of carbon. When the sample isplaced inside a shielded beta counter with 88.0%counting efficiency, 837 counts are accumulated inone week. Assuming that the cosmic-ray flux and theEarth’s atmosphere have not changed appreciablysince the sample was formed, find the age of thesample.Figure P29.28e – e –20782 Pb29.The mass of 56 Fe is 55.934 9 u andthe mass of 56 Co is 55.939 9 u. Which isotope decaysinto the other and by what process?23830. Find the energy released in the alpha decay of 92 U .234The following mass value will be useful: 90Th has amass of 234.043 583 u.Section 29.6 Nuclear Reactions37. The first known reaction in which the product nucleuswas radioactive (achieved in 1934) was one27in which 13 Al was bombarded with alpha particles.Produced in the reaction were a neutron and aproduct nucleus. (a) What was the product nucleus?(b) Find the Q value of the reaction.38. Complete the following nuclear reactions:? 14 7 N : 1 1 H 17 8 O73 Li 1 1 H : 4 2 He ?

970 Chapter 29 Nuclear Physics39. Identify the unknown particles X and X in the followingnuclear reactions:424 1X 2He : 12Mg 0n235 190192U 0n : 38Sr X 2n 0122H 1 : 1H X X40. The first nuclear reaction utilizing particle acceleratorswas performed by Cockcroft and Walton. Acceleratedprotons were used to bombard lithium nuclei,producing the following reaction:1 7H Li :4He 41 32 2HeSince the masses of the particles involved in the reactionwere well known, these results were used toobtain an early proof of the Einstein mass–energyrelation. Calculate the Q value of the reaction.1041. (a) Suppose 5 B is struck by an alpha particle, releasinga proton and a product nucleus in the reaction.What is the product nucleus? (b) An alpha particleand a product nucleus are produced when 6 C13is struck by a proton. What is the product nucleus?742. (a) Determine the product of the reaction 3Li 42He : ? n. (b) What is the Q value of the reaction?19743. Natural gold has only one isotope: 79 Au . If goldis bombarded with slow neutrons, e particlesare emitted. (a) Write the appropriate reactionequation. (b) Calculate the maximum energy of198the emitted beta particles. The mass of 80 Hg is197.966 75 u.44. Find the threshold energy that an incident neutronmust have to produce the reaction1 42 30n 2He : 1H 1H45.When 18 O is struck by a proton, 18 Fand another particle are produced. (a) What is theother particle? (b) The reaction has a Q value of2.453 MeV, and the atomic mass of 18 O is17.999 160 u. What is the atomic mass of 18 F?Section 29.7 Medical Applications of Radiation46. In terms of biological damage, how many rad ofheavy ions is equivalent to 100 rad of x-rays?47. A person whose mass is 75.0 kg is exposed to awhole-body dose of 25.0 rads. How many joules ofenergy are deposited in the person’s body?48. A 200-rad dose of radiation is administered to a patientin an effort to combat a cancerous growth. Assumingall of the energy deposited is absorbed by thegrowth, (a) calculate the amount of energy deliveredper unit mass. (b) Assuming the growth has a mass of0.25 kg and a specific heat equal to that of water,calculate its temperature rise.49. A “clever” technician decides to heat some water forhis coffee with an x-ray machine. If the machineproduces 10 rad/s, how long will it take to raise thetemperature of a cup of water by 50°C. Ignore heatlosses during this time.50. An x-ray technician works 5 days per week, 50 weeksper year. Assume that the technician takes an averageof eight x-rays per day and receives a dose of5.0 rem/yr as a result. (a) Estimate the dose in remper x-ray taken. (b) How does this result comparewith the amount of low-level background radiationthe technician is exposed to?51.A patient swallows a radiopharmaceuticaltagged with phosphorus-32 ( 15P), a emit-32ter with a half-life of 14.3 days. The average kineticenergy of the emitted electrons is 700 keV. If the initialactivity of the sample is 1.31 MBq, determine(a) the number of electrons emitted in a 10-dayperiod, (b) the total energy deposited in the bodyduring the 10 days, and (c) the absorbed dose ifthe electrons are completely absorbed in 100 g oftissue.52. A particular radioactive source produces 100 mrad of2-MeV gamma rays per hour at a distance of 1.0 m.(a) How long could a person stand at this distancebefore accumulating an intolerable dose of 1 rem?(b) Assuming the gamma radiation is emitteduniformly in all directions, at what distance woulda person receive a dose of 10 mrad/h from thissource?ADDITIONAL PROBLEMS53. A 200.0-mCi sample of a radioactive isotope is purchasedby a medical supply house. If the sample hasa half-life of 14.0 days, how long will it keep beforeits activity is reduced to 20.0 mCi?54. A sample of organic material is found to contain 18 gof carbon. The investigators believe the material to be20 000 years old, based on samples of pottery found atthe site. If so, what is the expected activity of theorganic material? Take data from Example 29.7.55. Deuterons that have been accelerated are used tobombard other deuterium nuclei, resulting in thereaction2 23 11H 1H : 2He 0nDoes this reaction require a threshold energy? If so,what is its value?56. Free neutrons have a characteristic half-life of 12min. What fraction of a group of free neutrons at athermal energy of 0.040 eV will decay before travelinga distance of 10.0 km?

Problems 97157. A by-product of some fission reactors is the isotope23994Pu, an alpha emitter having a half-life of24 120 yr. The reaction involved isPu : U 239Consider a sample of 1.00 kg of pure 94Pu at t 0.Calculate (a) the number of 94Pu nuclei present att 0 and (b) the initial activity in the sample.(c) How long does the sample have to be stored if a“safe” activity level is 0.100 Bq?58. (a) Find the radius of the C nucleus. (b) Find theforce of repulsion between a proton at the surfaceof a C nucleus and the remaining five protons.(c) How much work (in MeV) has to be done to overcomethis electrostatic repulsion in order to put thelast proton into the nucleus? (d) Repeat (a), (b),238and (c) for U.59.239941261262399223592In a piece of rock from the Moon, the 87 Rb contentis assayed to be 1.82 10 10 atoms per gram of materialand the 87 Sr content is found to be 1.07 10 9atoms per gram. (The relevant decay is 87 Rb :87 Sr e . The half-life of the decay is 4.8 10 10 yr.)(a) Determine the age of the rock. (b) Could the materialin the rock actually be much older? What assumptionis implicit in using the radioactive-datingmethod?60. Many radioisotopes have important industrial, medical,and research applications. One such radioisotopeis 60 Co, which has a half-life of 5.2 yr and decaysby the emission of a beta particle (energy0.31 MeV) and two gamma photons (energies1.17 MeV and 1.33 MeV). A scientist wishes to preparea 60 Co sealed source that will have an activity ofat least 10 Ci after 30 months of use. What is theminimum initial mass of 60 Co required?61. A medical laboratory stock solution is prepared withan initial activity due to 24 Na of 2.5 mCi/ml, and10.0 ml of the stock solution is diluted at t 0 0 to aworking solution whose total volume is 250 ml. After48 h, a 5.0-ml sample of the working solution ismonitored with a counter. What is the measured activity?(Note that 1 ml 1 milliliter.)62. A theory of nuclear astrophysics is that all theheavy elements such as uranium are formed in supernovaexplosions of massive stars, which immediatelyrelease the elements into space. If we assumethat at the time of an explosion there wereequal amounts of 235 U and 238 U, how long agowere the elements that formed our Earth released,given that the present 235 U/ 238 U ratio is 0.007?(The half-lives of 235 U and 238 U are 0.70 10 9 yrand 4.47 10 9 yr, respectively.)63. A fission reactor is hit by a nuclear weapon, causing5.0 10 6 Ci of 90 Sr (T 1/2 28.7 yr) to evaporateinto the air. The 90 Sr falls out over an area of 10 4km 2 . How long will it take the activity of the 90 Sr toreach the agriculturally “safe” level of 2.0 Ci/m 2 ?64. After the sudden release of radioactivity from theChernobyl nuclear reactor accident in 1986, theradioactivity of milk in Poland rose to 2 000 Bq/Ldue to iodine-131, with a half-life of 8.04 days. Radioactiveiodine is particularly hazardous, because the thyroidgland concentrates iodine. The Chernobyl accidentcaused a measurable increase in thyroidcancers among children in Belarus. (a) For comparison,find the activity of milk due to potassium. Assumethat 1 liter of milk contains 2.00 g of potassium,of which 0.011 7% is the isotope 40 K, whichhas a half-life of 1.28 10 9 yr. (b) After what lengthof time would the activity due to iodine fall belowthat due to potassium?65. During the manufacture of a steel engine component,radioactive iron ( 59 Fe) is included in the totalmass of 0.20 kg. The component is placed in a testengine when the activity due to the isotope is20.0 Ci. After a 1 000-h test period, oil is removedfrom the engine and is found to contain enough59 Fe to produce 800 disintegrations/min per liter ofoil. The total volume of oil in the engine is 6.5 L.Calculate the total mass worn from the engine componentper hour of operation. (The half-life of 59 Feis 45.1 days.)66. After determining that the Sun has existed for hundredsof millions of years, but before the discoveryof nuclear physics, scientists could not explain whythe Sun has continued to burn for such a long time.For example, if it were a coal fire, the Sun wouldhave burned up in about 3 000 yr. Assume that theSun, whose mass is 1.99 10 30 kg, originally consistedentirely of hydrogen and that its total poweroutput is 3.76 10 26 W. (a) If the energy-generatingmechanism of the Sun is the transforming of hydrogeninto helium via the net reaction11424H 2e : He 2 calculate the energy (in joules) given off by thisreaction. (b) Determine how many hydrogen atomsconstitute the Sun. Take the mass of one hydrogenatom to be 1.67 10 27 kg. (c) Assuming that thetotal power output remains constant, after whattime will all the hydrogen be converted into helium,making the Sun die? The actual projected lifetimeof the Sun is about 10 billion years, because only thehydrogen in a relatively small core is available as afuel. (Only in the Sun’s core are temperatures anddensities high enough for the fusion reaction to beself-sustaining).

972 Chapter 29 Nuclear PhysicsACTIVITIES1. This experiment will take a little longer to do thanmost that we have suggested, but the time spent isworthwhile to help you understand the concept ofhalf-life. Obtain a box of sugar cubes and with apencil make a mark on one side of each of about200 cubes. Each of these cubes will represent thenucleus of a radioactive substance. Thus, at t 0,you have 200 undecayed nuclei. Now, put the 200marked cubes in a box and roll them out on a table,just as you would roll dice. Next, count and removeany cubes that have landed marked-side up. Thesecubes represent nuclei that emitted radiation duringthe roll. They are no longer radioactive and thusdo not participate in the rest of the action. Recordthe number of undecayed cubes remaining as thenumber of undecayed nuclei at t 1 roll.Continue rolling, counting, and removing untilyou have completed 12 to 15 rolls. By then, youshould have only a few cubes remaining. Plot agraph of undecayed cubes versus the roll numberand from this determine the “half-roll” of the cubes.2. Use a nail to punch a hole in the bottom of a largetin can. Hold the can beneath a faucet and adjustthe water flow from the faucet to a fine constantstream. Although water flows from the hole at thebottom, you will note that the level of the water inthe can rises. As it does so, however, the flow of waterleaving the can increases due to increased waterpressure caused by the greater depth of water. Unlessthe flow of water is too great, an equilibriumpoint will be reached at which the amount of waterflowing out of the can each second exactly equalsthe amount flowing in each second. When this happens,the level of water in the can is constant. Asnoted in the text, carbon-14 is continually beingproduced in the atmosphere and is also continuallydisappearing as it decays into nitrogen. What is theanalogy between water entering the can, remainingin the can, and flowing out of the can and the behaviorof carbon-14 in the atmosphere?

This photo shows scientist MelissaDouglas and part of the Z machine,an inertial-electrostatic confinementfusion apparatus at Sandia NationalLaboratories. In the device, giantcapacitors discharge through a gridof tungsten wires finer than humanhairs, vaporizing them. The tungstenions implode inward at a million milesan hour. Braking strongly in the gripof a “Z-pinch,” they emit powerfulx-rays that compress a deuteriumpellet, causing collisions between thedeuterium atoms that lead to fusionevents.Nuclear Energy andElementary ParticlesIn this concluding chapter we discuss the two means by which energy can be derived fromnuclear reactions: fission, in which a nucleus of large mass number splits into two smallernuclei, and fusion, in which two light nuclei fuse to form a heavier nucleus. In either case,there is a release of large amounts of energy, which can be used destructively through bombsor constructively through the production of electric power. We end our study of physics byexamining the known subatomic particles and the fundamental interactions that govern theirbehavior. We also discuss the current theory of elementary particles, which states that all matterin nature is constructed from only two families of particles: quarks and leptons. Finally, wedescribe how such models help us understand the evolution of the Universe.30.1 NUCLEAR FISSIONNuclear fission occurs when a heavy nucleus, such as 235 U, splits, or fissions, intotwo smaller nuclei. In such a reaction, the total mass of the products is less thanthe original mass of the heavy nucleus.Nuclear fission was first observed in 1939 by Otto Hahn and Fritz Strassman, followingsome basic studies by Fermi. After bombarding uranium (Z 92) with neutrons,Hahn and Strassman discovered two medium-mass elements, barium andlanthanum, among the reaction products. Shortly thereafter, Lise Meitner andOtto Frisch explained what had happened: the uranium nucleus had split into twonearly equal fragments after absorbing a neutron. This was of considerable interestto physicists attempting to understand the nucleus, but it was to have even morefar-reaching consequences. Measurements showed that about 200 MeV of energy isreleased in each fission event, and this fact was to affect the course of humanhistory.Sandia National Laboratories. Photo by Randy MontoyaCHAPTER30O U T L I N E30.1 Nuclear Fission30.2 Nuclear Reactors30.3 Nuclear Fusion30.4 Elementary Particles30.5 The Fundamental Forcesof Nature30.6 Positrons and OtherAntiparticles30.7 Mesons and the Beginningof Particle Physics30.8 Classification of Particles30.9 Conservation Laws30.10 Strange Particles andStrangeness30.11 The Eightfold Way30.12 Quarks30.13 Colored Quarks30.14 Electroweak Theoryand the Standard Model30.15 The Cosmic Connection30.16 Problems and Perspectives973

974 Chapter 30 Nuclear Energy and Elementary ParticlesSequence of events ina nuclear fission process The fission of 235 U by slow (low-energy) neutrons can be represented by thesequence of events10 n 23592 U : 23692 U* : X Y neutrons [30.1]where 236 U * is an intermediate state that lasts only for about 10 12 s beforesplitting into nuclei X and Y, called fission fragments. There are many combinationsof X and Y that satisfy the requirements of conservation of energy andcharge. In the fission of uranium, about 90 different daughter nuclei can beformed. The process also results in the production of several (typically two orthree) neutrons per fission event. On the average, 2.47 neutrons are released perevent.A typical reaction of this type is10 n 23592 U : 141 9256 Ba 36 Kr 31 0 n[30.2]The fission fragments, barium and krypton, and the released neutrons have agreat deal of kinetic energy following the fission event.The breakup of the uranium nucleus can be compared to what happens to adrop of water when excess energy is added to it. All of the atoms in the drop haveenergy, but not enough to break up the drop. However, if enough energy is addedto set the drop vibrating, it will undergo elongation and compression until theamplitude of vibration becomes large enough to cause the drop to break apart. Inthe uranium nucleus, a similar process occurs (Fig. 30.1). The sequence of eventsis as follows:1. The 235 U nucleus captures a thermal (slow-moving) neutron.2. The capture results in the formation of 236 U * , and the excess energy of thisnucleus causes it to undergo violent oscillations.3. The 236 U * nucleus becomes highly elongated, and the force of repulsionbetween protons in the two halves of the dumbbell-shaped nucleus tends toincrease the distortion.4. The nucleus splits into two fragments, emitting several neutrons in the process.The energy released in a typical fission process Q can be estimated. From Figure29.4, we see that the binding energy per nucleon is about 7.2 MeV for heavy nuclei(those having a mass number of approximately 240) and about 8.2 MeV for nucleiof intermediate mass. This means that the nucleons in the fission fragments aremore tightly bound, and therefore have less mass, than the nucleons in the originalheavy nucleus. The decrease in mass per nucleon appears as released energywhen fission occurs. The amount of energy released is (8.2 7.2) MeV per nucleon.Assuming a total of 240 nucleons, we find that the energy released per fissionevent isQ 240 nucleons/(8.2 MeV/nucleon 7.2 MeV/nucleon) 240 MeVThis is a large amount of energy relative to the amount released in chemicalprocesses. For example, the energy released in the combustion of one molecule ofthe octane used in gasoline engines is about one hundred-millionth the energyreleased in a single fission event!Figure 30.1 A nuclear fissionevent as described by the liquid-dropmodel of the nucleus. (a) A slowneutron approaches a 235 U nucleus.(b) The neutron is absorbed by the235 U nucleus, changing it to 236 U*,which is a 236 U nucleus in an excitedstate. (c) The nucleus deforms andoscillates like a liquid drop. (d) Thenucleus undergoes fission, resultingin two lighter nuclei X and Y, alongwith several neutrons.236 U*235UX(a) (b) (c) (d)Y

30.1 Nuclear Fission 975Applying Physics 30.1Unstable ProductsIf a heavy nucleus were to fission into just two productnuclei, they would be very unstable. Why is this?Explanation According to Figure 29.3, the ratio of thenumber of neutrons to the number of protons increaseswith Z. As a result, when a heavy nucleus splits in afission reaction to two lighter nuclei, the lighter nucleitend to have too many neutrons. This leads to instability,as the nucleus returns to the curve in Figure 29.3 bydecay processes that reduce the number of neutrons.EXAMPLE 30.1 The Fission of UraniumGoal Balance a nuclear equation to determine details of the fission fragments.Problem When 235 U is struck by a neutron, there are various possible fission fragments. Determine the number ofneutrons produced when the fission fragments are 140 Xe and 94 Sr (isotopes of xenon and strontium).Strategy This is something like balancing chemical equations: the atomic numbers and mass numbers shouldbalance on either side of the equation.SolutionWrite the equation describing the process, with anunknown number x of neutrons:The atomic numbers balance already, as they should.Write an equation relating the mass numbers:10 n 23592 U : 14054Xe 9438 Sr x(1 0 n)1 235 140 94 x : x 2Remark In this case, the number of protons balanced automatically. If that were not the case, there might be otherpossible daughter particles, such as protons or helium nuclei (also called alpha particles).Exercise 30.1Find the number of neutrons released if the two major fragments are 132 Sn and 101 Mo.AnswerThree neutronsQuick Quiz 30.1In the first atomic bomb, the energy released was equivalent to about 30 kilotonsof TNT, where a ton of TNT releases an energy of about 4.0 10 9 J. Estimate theamount of mass converted into energy in this event. (a) 1 g (b) 1 mg (c) 1 g(d) 1 kg (e) 20 kilotonsEXAMPLE 30.2GoalA Fission-Powered WorldRelate raw material to energy output.Problem (a) Calculate the total energy released if 1.00 kg of 235 U undergoes fission, taking the disintegrationenergy per event to be Q 208 MeV (a more accurate value than the estimate given previously). (b) How manykilograms of 235 U would be needed to satisfy the world’s annual energy consumption (about 4 10 20 J)?Strategy In part (a), use the concept of a mole and Avogadro’s number to obtain the total number of nuclei.Multiplying by the energy per reaction then gives the total energy released. Part (b) requires some light algebra.Solution(a) Calculate the total energy released from 1.00 kg of 235 U.Find the total number of nuclei in 1.00 kg of uranium:N 6.02 10 23 nuclei/mol235 g/mol (1.00 103 g) 2.56 10 24 nuclei

976 Chapter 30 Nuclear Energy and Elementary ParticlesMultiply N by the energy yield per nucleus, obtainingthe total disintegration energy:E NQ (2.56 10 24 nuclei) 208 5.32 10 26 MeVMeVnucleus(b) How many kilograms would provide for the world’sannual energy needs?Set the energy per kilogram, E kg , times the number ofkilograms, N kg , equal to the total annual energy consumption.Solve for N kg :E kg N kg E totN kg E totE kg 5 10 6 kg4 10 20 J(5.32 10 32 eV/kg)(1.60 10 19 J/eV)Remark The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems.There are enough known uranium deposits to provide the world’s current energy requirements for a few hundredyears.Exercise 30.2How long can one kilogram of U-235 keep a 100-watt lightbulb burning if all its released energy is converted to electricalenergy?Answer30 000 yr30.2 NUCLEAR REACTORSWe have seen that neutrons are emitted when 235 U undergoes fission. These neutronscan in turn trigger other nuclei to undergo fission, with the possibility of achain reaction (Active Fig. 30.2). Calculations show that if the chain reaction isn’tcontrolled, it will proceed too rapidly and possibly result in the sudden release ofan enormous amount of energy (an explosion), even from only 1 g of 235 U. If theenergy in 1 kg of 235 U were released, it would equal that released by the detonation23592 U9214156 Ba 36 KrNeutron138I5395138 9538 Sr 54 Xe 39 Y 23592 U9841 Nb13551 SbACTIVE FIGURE 30.2A nuclear chain reaction initiated by the capture of a neutron.Log into PhysicsNow at www.cp7e.com and go to Active Figure 30.2 to observe a chain reaction.

30.2 Nuclear Reactors 977Courtesy of Chicago Historical SocietyPainting of the world’s first nuclearreactor. Because of wartime secrecy,there are no photographs of the completedreactor, which was composedof layers of graphite interspersed withuranium. A self-sustained chain reactionwas first achieved on December 2,1942. Word of the success was telephonedimmediately to Washingtonwith this message: “The Italian navigatorhas landed in the New Worldand found the natives very friendly.”The historic event took place in animprovised laboratory in the racquetcourt under the west stands of theUniversity of Chicago’s Stagg Field.The Italian navigator was Fermi.of about 20 000 tons of TNT! An uncontrolled fission reaction, of course, is theprinciple behind the first nuclear bomb.A nuclear reactor is a system designed to maintain what is called a selfsustainedchain reaction. This important process was first achieved in 1942 by agroup led by Fermi at the University of Chicago, with natural uranium as the fuel.Most reactors in operation today also use uranium as fuel. Natural uranium containsonly about 0.7% of the 235 U isotope, with the remaining 99.3% being the238 U isotope. This is important to the operation of a reactor because 238 U almostnever undergoes fission. Instead, it tends to absorb neutrons, producing neptuniumand plutonium. For this reason, reactor fuels must be artificially enriched sothat they contain several percent of the 235 U isotope.Earlier we mentioned that an average of about 2.5 neutrons are emitted in eachfission event of 235 U. In order to achieve a self-sustained chain reaction, one ofthese neutrons must be captured by another 235 U nucleus and cause it to undergofission. A useful parameter for describing the level of reactor operation is thereproduction constant K, defined as the average number of neutrons from eachfission event that will cause another event. As we have seen, K can have a maximumvalue of 2.5 in the fission of uranium. In practice, however, K is less than thisbecause of several factors, which we soon discuss.A self-sustained chain reaction is achieved when K 1. Under this condition,the reactor is said to be critical. When K is less than one, the reactor is subcriticaland the reaction dies out. When K is greater than one the reactor is said to be supercritical,and a runaway reaction occurs. In a nuclear reactor used to furnishpower to a utility company, it is necessary to maintain a K value close to one.The basic design of a nuclear reactor is shown in Figure 30.3. The fuel elementsconsist of enriched uranium. The functions of the remaining parts of the reactorand some aspects of its design are described next.ControlrodsFuel elementsRadiationshieldModeratormaterialFigure 30.3 Cross section of areactor core showing the controlrods, fuel elements containingenriched fuel, and moderatingmaterial, all surrounded by aradiation shield.Neutron LeakageIn any reactor, a fraction of the neutrons produced in fission will leak out of thecore before inducing other fission events. If the fraction leaking out is too large,the reactor will not operate. The percentage lost is large if the reactor is very smallbecause leakage is a function of the ratio of surface area to volume. Therefore, acritical requirement of reactor design is choosing the correct surface-area-tovolumeratio so that a sustained reaction can be achieved.Regulating Neutron EnergiesThe neutrons released in fission events are highly energetic, with kinetic energiesof about 2 MeV. It is found that slow neutrons are far more likely than fastneutrons to produce fission events in 235 U. Further, 238 U doesn’t absorb slow

978 Chapter 30 Nuclear Energy and Elementary Particlesneutrons. In order for the chain reaction to continue, therefore, the neutronsmust be slowed down. This is accomplished by surrounding the fuel with a substancecalled a moderator.In order to understand how neutrons are slowed down, consider a collisionbetween a light object and a massive one. In such an event, the light objectrebounds from the collision with most of its original kinetic energy. However, ifthe collision is between objects having masses that are nearly the same, the incomingprojectile transfers a large percentage of its kinetic energy to the target. In thefirst nuclear reactor ever constructed, Fermi placed bricks of graphite (carbon)between the fuel elements. Carbon nuclei are about 12 times more massive thanneutrons, but after about 100 collisions with carbon nuclei, a neutron is slowedsufficiently to increase its likelihood of fission with 235 U. In this design the carbonis the moderator; most modern reactors use heavy water (D 2 O) as the moderator.Neutron CaptureIn the process of being slowed down, neutrons may be captured by nuclei that donot undergo fission. The most common event of this type is neutron capture by 238 U.The probability of neutron capture by 238 U is very high when the neutrons have highkinetic energies and very low when they have low kinetic energies. The slowing downof the neutrons by the moderator serves the dual purpose of making them availablefor reaction with 235 U and decreasing their chances of being captured by 238 U.APPLICATIONNuclear Reactor DesignControl of Power LevelIt is possible for a reactor to reach the critical stage (K 1) after all neutron lossesdescribed previously are minimized. However, a method of control is needed to adjustK to a value near one. If K were to rise above this value, the heat produced inthe runaway reaction would melt the reactor. To control the power level, controlrods are inserted into the reactor core. (See Fig. 30.3.) These rods are made of materialssuch as cadmium that are highly efficient in absorbing neutrons. By adjustingthe number and position of the control rods in the reactor core, the K value can bevaried and any power level within the design range of the reactor can be achieved.A diagram of a pressurized-water reactor is shown in Figure 30.4. This type ofreactor is commonly used in electric power plants in the United States. FissionSecondary loopSteam turbineand electric generatorControl rod+–Uraniumfuel rodNuclearreactorSteamCondenser (steam from turbineis condensed by cold water)HeatexchangerMolten sodiumor liquid waterunder highpressure (carriesenergy to steamgenerator)PumpFigure 30.4PrimaryloopCold waterMain components of a pressurized-water nuclear reactor.Warm water

30.2 Nuclear Reactors 979events in the reactor core supply heat to the water contained in the primary(closed) system, which is maintained at high pressure to keep it from boiling. Thiswater also serves as the moderator. The hot water is pumped through a heatexchanger, and the heat is transferred to the water contained in the secondary system.There the hot water is converted to steam, which drives a turbine–generatorto create electric power. Note that the water in the secondary system is isolatedfrom the water in the primary system in order to prevent contamination of the secondarywater and steam by radioactive nuclei from the reactor core.Reactor Safety 1The safety aspects of nuclear power reactors are often sensationalized by themedia and misunderstood by the public. The 1979 near disaster of Three MileIsland in Pennsylvania and the accident at the Chernobyl reactor in the Ukrainerightfully focused attention on reactor safety. Yet the safety record in the UnitedStates is enviable. The records show no fatalities attributed to commercial nuclearpower generation in the history of the United States nuclear industry.Commercial reactors achieve safety through careful design and rigid operatingprocedures. Radiation exposure and the potential health risks associated with suchexposure are controlled by three layers of containment. The fuel and radioactivefission products are contained inside the reactor vessel. Should this vessel rupture,the reactor building acts as a second containment structure to prevent radioactivematerial from contaminating the environment. Finally, the reactor facilities mustbe in a remote location to protect the general public from exposure should radiationescape the reactor building.According to the Oak Ridge National Laboratory Review, “The health risk of livingwithin 8 km (5 miles) of a nuclear reactor for 50 years is no greater than the riskof smoking 1.4 cigarettes, drinking 0.5 liters of wine, traveling 240 km by car, flying9 600 km by jet, or having one chest x-ray in a hospital. Each of these activities isestimated to increase a person’s chances of dying in any given year by one in amillion.”Another potential danger in nuclear reactor operations is the possibility thatthe water flow could be interrupted. Even if the nuclear fission chain reactionwere stopped immediately, residual heat could build up in the reactor to the pointof melting the fuel elements. The molten reactor core would melt its way to thebottom of the reactor vessel and conceivably into the ground below—the socalledChina syndrome. Although it might appear that this deep undergroundburial site would be an ideal safe haven for a radioactive blob, there would be dangerof a steam explosion should the molten mass encounter water. This nonnuclearexplosion could spread radioactive material to the areas surrounding thepower plant. To prevent such an unlikely chain of events, nuclear reactors aredesigned with emergency core cooling systems, requiring no power, that automaticallyflood the reactor with water in the event of a loss of coolant. The emergencycooling water moderates heat build-up in the core, which in turn prevents themelting of the reactor vessel.A continuing concern in nuclear fission reactors is the safe disposal of radioactivematerial when the reactor core is replaced. This waste material contains longlived,highly radioactive isotopes and must be stored for long periods of time insuch a way that there is no chance of environmental contamination. At present,sealing radioactive wastes in waterproof containers and burying them in deep saltmines seems to be the most promising solution.Transportation of reactor fuel and reactor wastes poses additional safety risks.However, neither the waste nor the fuel of nuclear power reactors can be used toconstruct a nuclear bomb.Accidents during transportation of nuclear fuel could expose the public toharmful levels of radiation. The Department of Energy requires stringent crash1 The authors are grateful to Professor Gene Skluzacek of the University of Nebraska at Omaha for rewriting this section.

980 Chapter 30 Nuclear Energy and Elementary Particlestests on all containers used to transport nuclear materials. Container manufacturersmust demonstrate that their containers will not rupture, even in high-speedcollisions.The safety issues associated with nuclear power reactors are complex and oftenemotional. All sources of energy have associated risks. Coal, for example, exposesworkers to health hazards (including radioactive radon) and produces atmosphericpollution (including greenhouse gases). In each case, the risks must beweighed against the benefits and the availability of the energy source.This photograph of the Sun, taken onDecember 19, 1973, during the thirdand final manned Skylab mission,shows one of the most spectacularsolar flares ever recorded, spanningmore than 588 000 km (365 000 mi)across the solar surface. Several activeregions can be seen on the easternside of the disk. The photograph wastaken in the light of ionized heliumby the extreme ultraviolet spectroheliographinstrument of the U.S.Naval Research Laboratory.NASA30.3 NUCLEAR FUSIONFigure 29.4 shows that the binding energy of light nuclei (those having a massnumber lower than 20) is much smaller than the binding energy of heavier nuclei.This suggests a process that is the reverse of fission. When two light nuclei combineto form a heavier nucleus, the process is called nuclear fusion. Because the massof the final nucleus is less than the masses of the original nuclei, there is a loss ofmass, accompanied by a release of energy. Although fusion power plants have notyet been developed, a worldwide effort is under way to harness the energy fromfusion reactions in the laboratory.Fusion in the SunAll stars generate their energy through fusion processes. About 90% of stars,including the Sun, fuse hydrogen, whereas some older stars fuse helium or otherheavier elements. Stars are born in regions of space containing vast clouds of dustand gas. Recent mathematical models of these clouds indicate that star formationis triggered by shock waves passing through a cloud. These shock waves are similarto sonic booms and are produced by events such as the explosion of a nearby star,called a supernova. The shock wave compresses certain regions of the cloud, causingthem to collapse under their own gravity. As the gas falls inward toward thecenter, the atoms gain speed, which causes the temperature of the gas to rise. Twoconditions must be met before fusion reactions in the star can sustain its energyneeds: (1) The temperature must be high enough (about 10 7 K for hydrogen) toallow the kinetic energy of the positively charged hydrogen nuclei to overcometheir mutual Coulomb repulsion as they collide, and (2) the density of nuclei mustbe high enough to ensure a high rate of collision.It’s interesting that temperatures inside stars like the Sun are not sufficient toallow colliding protons to overcome Coulomb repulsion. In a certain percentageof collisions, the nuclei pass through the barrier anyway, an example of quantumtunneling. So a quantum effect is key in making sunshine.When fusion reactions occur at the core of a star, the energy that is liberatedeventually becomes sufficient to prevent further collapse of the star under its owngravity. The star then continues to live out the remainder of its life under a balancebetween the inward force of gravity pulling it toward collapse and the outwardforce due to thermal effects and radiation pressure.The proton–proton cycle is a series of three nuclear reactions that are believedto be the stages in the liberation of energy in the Sun and other stars rich inhydrogen. An overall view of the proton–proton cycle is that four protonscombine to form an alpha particle and two positrons, with the release of 25 MeVof energy in the process.The specific steps in the proton–proton cycle are11 H 1 1 H : 2 1 D e and11 H 2 1 D : 3 2 He [30.3]where D stands for deuterium, the isotope of hydrogen having one proton and one2neutron in the nucleus. (It can also be written as .) The second reaction is1 H

30.3 Nuclear Fusion 981followed by either hydrogen–helium fusion or helium–helium fusion:or11H 3 2He :42He e 32He 3 2He : 4 2He 2( 1 1H)The energy liberated is carried primarily by gamma rays, positrons, and neutrinos,as can be seen from the reactions. The gamma rays are soon absorbed by thedense gas, thus raising its temperature. The positrons combine with electrons toproduce gamma rays, which in turn are also absorbed by the gas within a few centimeters.The neutrinos, however, almost never interact with matter; hence, theyescape from the star, carrying about 2% of the energy generated with them. Theseenergy-liberating fusion reactions are called thermonuclear fusion reactions. Thehydrogen (fusion) bomb, first exploded in 1952, is an example of an uncontrolledthermonuclear fusion reaction.Fusion ReactorsThe enormous amount of energy released in fusion reactions suggests the possibilityof harnessing this energy for useful purposes on Earth. A great deal of effort isunder way to develop a sustained and controllable thermonuclear reactor—afusion power reactor. Controlled fusion is often called the ultimate energy sourcebecause of the availability of its fuel source: water. For example, if deuterium, theisotope of hydrogen consisting of a proton and a neutron, were used as the fuel,0.06 g of it could be extracted from 1 gal of water at a cost of about four cents.Such rates would make the fuel costs of even an inefficient reactor almost insignificant.An additional advantage of fusion reactors is that comparatively few radioactiveby-products are formed. As noted in Equation 30.3, the end product of thefusion of hydrogen nuclei is safe, nonradioactive helium. Unfortunately, a thermonuclearreactor that can deliver a net power output over a reasonable timeinterval is not yet a reality, and many problems must be solved before a successfuldevice is constructed.We have seen that the Sun’s energy is based, in part, on a set of reactions in whichordinary hydrogen is converted to helium. Unfortunately, the proton–protoninteraction is not suitable for use in a fusion reactor because the event requiresvery high pressures and densities. The process works in the Sun only because ofthe extremely high density of protons in the Sun’s interior. In fact, even at thedensities and temperatures that exist at the center of the Sun, the average protontakes 14 billion years to react!The fusion reactions that appear most promising in the construction of a fusionpower reactor involve deuterium (D) and tritium (T), which are isotopes of hydrogen.These reactions areAPPLICATIONFusion Reactorsand21D 2 1D : 3 2He 1 0nQ 3.27 MeV21 D 2 1 D : 3 1 T 1 1H Q 4.03 MeV21D 3 1T : 4 2He 1 0n Q 17.59 MeV[30.4]where the Q values refer to the amount of energy released per reaction. As notedearlier, deuterium is available in almost unlimited quantities from our lakes andoceans and is very inexpensive to extract. Tritium, however, is radioactive (T 1/2 12.3 yr) and undergoes beta decay to 3 He. For this reason, tritium doesn’t occurnaturally to any great extent and must be artificially produced.One of the major problems in obtaining energy from nuclear fusion is the factthat the Coulomb repulsion force between two charged nuclei must be overcomebefore they can fuse. The fundamental challenge is to give the two nuclei enoughkinetic energy to overcome this repulsive force. This can be accomplished by

982 Chapter 30 Nuclear Energy and Elementary ParticlesLawson’s criterion heating the fuel to extremely high temperatures (about 10 8 K, far greater than theinterior temperature of the Sun). As you might expect, such high temperaturesare not easy to obtain in a laboratory or a power plant. At these high temperatures,the atoms are ionized and the system consists of a collection of electrons andnuclei, commonly referred to as a plasma.In addition to the high temperature requirements, there are two other critical factorsthat determine whether or not a thermonuclear reactor will function: the plasmaion density n and the plasma confinement time — the time the interacting ions aremaintained at a temperature equal to or greater than that required for the reaction toproceed. The density and confinement time must both be large enough to ensurethat more fusion energy will be released than is required to heat the plasma.Lawson’s criterion states that a net power output in a fusion reactor is possibleunder the following conditions:n 10 14 s/cm 3 Deuterium–tritium interaction [30.5]n 10 16 s/cm 3 Deuterium–deuterium interactionThe problem of plasma confinement time has yet to be solved. How can a plasma beconfined at a temperature of 10 8 K for times on the order of 1 s? The basic plasmaconfinementtechnique under investigation is discussed following Example 30.3.EXAMPLE 30.3 Astrofuel on the MoonGoal Calculate the energy released in a fusion reaction.ProblemStrategyFind the energy released in the reaction of helium-3 with deuterium:32 He 2 1 D : 4 2 He 1 1 HThe energy released is the difference between the mass energy of the reactants and the products.SolutionAdd the masses on the left-hand side, and subtract themasses on the right, obtaining m in atomic mass units:Convert the mass difference to an equivalent amount ofenergy in MeV:m m He-3 m D m He-4 m H 3.016 029 u 2.014 102 u 4.002 602 u 1.007 825 u 0.019 704 uE (0.019 704 u) 931.5 MeV1u 18.35 MeVRemarks This is a large amount of energy per reaction. Helium-3 is rare on Earth, but plentiful on the Moon,where it has become trapped in the fine dust of the lunar regolith. Helium-3 has the advantage of producing moreprotons than neutrons (some neutrons are still produced by side reactions, such as D–D), but has the disadvantageof a higher ignition temperature. If fusion power plants using helium-3 became a reality, studies indicate that itwould be economically advantageous to mine helium-3 robotically and return it to Earth. The energy return per dollarwould be far greater than for mining coal or drilling for oil!Exercise 30.3Find the energy yield in the fusion of two helium-3 nuclei:32He 3 2He : 4 2He 2( 1 1H)Answer12.9 MeVMagnetic Field ConfinementMost fusion experiments use magnetic field confinement to contain a plasma. Onedevice, called a tokamak, has a doughnut-shaped geometry (a toroid), as shownin Figure 30.5a. This device, first developed in the former Soviet Union, uses a

30.3 Nuclear Fusion 983VacuumCurrentBPlasma(a)Courtesy of Princeton Plasma Physics Laboratory(b)Figure 30.5 (a) Diagram of a tokamak used in the magnetic confinement scheme. The plasma istrapped within the spiraling magnetic field lines as shown. (b) Interior view of the Tokamak Fusion TestReactor (TFTR) vacuum vessel located at the Princeton Plasma Physics Laboratory, PrincetonUniversity, Princeton, New Jersey. (c) The National Spherical Torus Experiment (NSTX) that beganoperation in March 1999.Courtesy of Princeton University(c)combination of two magnetic fields to confine the plasma inside the doughnut. Astrong magnetic field is produced by the current in the windings, and a weakermagnetic field is produced by the current in the toroid. The resulting magneticfield lines are helical, as shown in the figure. In this configuration, the field linesspiral around the plasma and prevent it from touching the walls of the vacuumchamber.In order for the plasma to reach ignition temperature, some form of auxiliaryheating is necessary. A successful and efficient auxiliary heating technique that hasbeen used recently is the injection of a beam of energetic neutral particles into theplasma.When it was in operation, the Tokamak Fusion Test Reactor (TFTR) at Princetonreported central ion temperatures of 510 million degrees Celsius, more than30 times hotter than the center of the Sun. TFTR n values for the D–T reactionwere well above 10 13 s/cm 3 and close to the value required by Lawson’s criterion.In 1991, reaction rates of 6 10 17 D–T fusions per second were reached in theJET tokamak at Abington, England.One of the new generations of fusion experiments is the National SphericalTorus Experiment (NSTX) shown in Figure 30.5c. Rather than generating thedonut-shaped plasma of a tokamak, the NSTX produces a spherical plasma thathas a hole through its center. The major advantage of the spherical configurationis its ability to confine the plasma at a higher pressure in a given magnetic field.This approach could lead to the development of smaller and more economicalfusion reactors.There are a number of other methods of creating fusion events. In inertial laserconfinement fusion, the fuel is put into the form of a small pellet and then collapsedby ultrahigh-power lasers. Fusion can also take place in a device the size ofa TV set, and in fact was invented by Philo Farnsworth, one of the fathers of elec-

984 Chapter 30 Nuclear Energy and Elementary Particlestronic television. In this method, called inertial electrostatic confinement, positivelycharged particles are rapidly attracted towards a negatively charged grid.Some of the positive particles then collide and fuse.An international collaborative effort involving four major fusion programs iscurrently under way to build a fusion reactor called the International ThermonuclearExperimental Reactor (ITER). This facility will address the remaining technologicaland scientific issues concerning the feasibility of fusion power. Thedesign is completed, and site and construction negotiations are under way. If theplanned device works as expected, the Lawson number for ITER will be about sixtimes greater than the current record holder, the JT-60U tokamak in Japan.30.4 ELEMENTARY PARTICLESThe word “atom” is from the Greek word atomos, meaning “indivisible.” At onetime, atoms were thought to be the indivisible constituents of matter; that is, theywere regarded as elementary particles. Discoveries in the early part of the 20thcentury revealed that the atom is not elementary, but has protons, neutrons, andelectrons as its constituents. Until 1932, physicists viewed these three constituentparticles as elementary because, with the exception of the free neutron, they arehighly stable. The theory soon fell apart, however, and beginning in 1937, manynew particles were discovered in experiments involving high-energy collisions betweenknown particles. These new particles are characteristically unstable and havevery short half-lives, ranging between 10 23 s and 10 6 s. So far more than 300 ofthem have been cataloged.Until the 1960s, physicists were bewildered by the large number and variety ofsubatomic particles being discovered. They wondered whether the particles werelike animals in a zoo or whether a pattern could emerge that would provide abetter understanding of the elaborate structure in the subnuclear world. In thelast 30 years, physicists have made tremendous advances in our knowledge of thestructure of matter by recognizing that all particles (with the exception of electrons,photons, and a few others) are made of smaller particles called quarks.Protons and neutrons, for example, are not truly elementary but are systems oftightly bound quarks. The quark model has reduced the bewildering array of particlesto a manageable number and has predicted new quark combinations thatwere subsequently found in many experiments.30.5 THE FUNDAMENTAL FORCES OF NATUREThe key to understanding the properties of elementary particles is to be able todescribe the forces between them. All particles in nature are subject to four fundamentalforces: strong, electromagnetic, weak, and gravitational.The strong force is responsible for the tight binding of quarks to form neutronsand protons and for the nuclear force, a sort of residual strong force, bindingneutrons and protons into nuclei. This force represents the “glue” that holdsthe nucleons together and is the strongest of all the fundamental forces. It is avery short-range force and is negligible for separations greater than about10 15 m (the approximate size of the nucleus). The electromagnetic force, whichis about 10 2 times the strength of the strong force, is responsible for the bindingof atoms and molecules. It is a long-range force that decreases in strength as theinverse square of the separation between interacting particles. The weak force is ashort-range nuclear force that tends to produce instability in certain nuclei. It isresponsible for beta decay, and its strength is only about 10 6 times that of thestrong force. (As we discuss later, scientists now believe that the weak and electromagneticforces are two manifestations of a single force called the electroweakforce). Finally, the gravitational force is a long-range force with a strength onlyabout 10 43 times that of the strong force. Although this familiar interaction isthe force that holds the planets, stars, and galaxies together, its effect on elementary

30.6 Positrons and Other Antiparticles 985TABLE 30.1Particle InteractionsInteraction Relative Mediating(Force) Strength a Range of Force Field ParticleStrong 1 Short (1 fm) GluonElectromagnetic 10 2 Long (1/r 2 ) PhotonWeak 10 6 Short (10 3 fm) W and Z bosonsGravitational 10 43 Long (1/r 2 ) Gravitona For two quarks separated by 3 10 17 m.particles is negligible. The gravitational force is by far the weakest of all the fundamentalforces.Modern physics often describes the forces between particles in terms of theactions of field particles or quanta. In the case of the familiar electromagneticinteraction, the field particles are photons. In the language of modern physics, theelectromagnetic force is mediated (carried) by photons, which are the quanta of theelectromagnetic field. Likewise, the strong force is mediated by field particlescalled gluons, the weak force is mediated by particles called the W and Z bosons,and the gravitational force is thought to be mediated by quanta of the gravitationalfield called gravitons. All of these field quanta have been detected except forthe graviton, which may never be found directly because of the weakness of thegravitational field. These interactions, their ranges, and their relative strengths aresummarized in Table 30.1.30.6 POSITRONS AND OTHER ANTIPARTICLESIn the 1920s, the theoretical physicist Paul Adrien Maurice Dirac (1902–1984)developed a version of quantum mechanics that incorporated special relativity.Dirac’s theory successfully explained the origin of the electron’s spin and its magneticmoment. But it had one major problem: its relativistic wave equationrequired solutions corresponding to negative energy states even for free electrons,and if negative energy states existed, we would expect a normal free electron in astate of positive energy to make a rapid transition to one of these lower states,emitting a photon in the process. Normal electrons would not exist and we wouldbe left with a universe of photons and electrons locked up in negative energystates.Dirac circumvented this difficulty by postulating that all negative energy statesare normally filled. The electrons that occupy the negative energy states are said tobe in the “Dirac sea” and are not directly observable when all negative energystates are filled. However, if one of these negative energy states is vacant, leaving ahole in the sea of filled states, the hole can react to external forces and thereforecan be observed as the electron’s positive antiparticle. The general and profoundimplication of Dirac’s theory is that for every particle, there is an antiparticle withthe same mass as the particle, but the opposite charge. For example, the electron’santiparticle, the positron, has a mass of 0.511 MeV/c 2 and a positive charge of1.6 10 19 C. As noted in Chapter 29, we usually designate an antiparticle with abar over the symbol for the particle. For example, p denotes the antiproton andthe antineutrino. In this book, the notation e is preferred for the positron.The positron was discovered by Carl Anderson in 1932, and in 1936 he wasawarded the Nobel prize for his achievement. Anderson discovered the positronwhile examining tracks created by electron-like particles of positive charge in acloud chamber. (These early experiments used cosmic rays—mostly energeticprotons passing through interstellar space—to initiate high-energy reactionson the order of several GeV.) In order to discriminate between positive andnegative charges, the cloud chamber was placed in a magnetic field, causingmoving charges to follow curved paths. Anderson noted that some of theCourtesy AIP Emilio Segre Visual ArchivesPAUL ADRIEN MAURICE DIRAC(1902 – 1984)Dirac was instrumental in the understandingof antimatter and in the unification ofquantum mechanics and relativity. Hemade numerous contributions to thedevelopment of quantum physics andcosmology, and won the Nobel Prize forphysics in 1933.

986 Chapter 30 Nuclear Energy and Elementary ParticlesTIP 30.1 AntiparticlesAn antiparticle is not identified solelyon the basis of opposite charge: evenneutral particles have antiparticles.APPLICATIONPositron EmissionTomographyHIDEKI YUKAWA, JapanesePhysicist (1907 – 1981)Yukawa was awarded the Nobel Prize in1949 for predicting the existence ofmesons. This photograph of Yukawa atwork was taken in 1950 in his office atColumbia University.TIP 30.2 The Nuclear Forceand the Strong ForceThe nuclear force discussed inChapter 29 was originally called thestrong force. Once the quark theory wasestablished, however, the phrase strongforce was reserved for the force betweenquarks. We will follow this convention:the strong force is between quarks andthe nuclear force is between nucleons.UPI/Corbis-Bettmanelectronlike tracks deflected in a direction corresponding to a positively chargedparticle.Since Anderson’s initial discovery, the positron has been observed in a numberof experiments. Perhaps the most common process for producing positrons is pairproduction, introduced in Chapter 26. In this process, a gamma ray with sufficientlyhigh energy collides with a nucleus, creating an electron–positron pair.Because the total rest energy of the pair is 2m e c 2 1.02 MeV, the gamma ray musthave at least this much energy to create such a pair.Practically every known elementary particle has a distinct antiparticle. Amongthe exceptions are the photon and the neutral pion ( 0 ), which are their own antiparticles.Following the construction of high-energy accelerators in the 1950s,many of these antiparticles were discovered. They included the antiproton p, discoveredby Emilio Segrè and Owen Chamberlain in 1955, and the antineutron n,discovered shortly thereafter.The process of electron–positron annihilation is used in the medical diagnostictechnique of positron emission tomography (PET). The patient is injected with aglucose solution containing a radioactive substance that decays by positron emission.Examples of such substances are oxygen-15, nitrogen-13, carbon-11, andfluorine-18. The radioactive material is carried to the brain. When a decay occurs,the emitted positron annihilates with an electron in the brain tissue, resulting intwo gamma ray photons. With the assistance of a computer, an image can be createdof the sites in the brain at which the glucose accumulates.The images from a PET scan can point to a wide variety of disorders in thebrain, including Alzheimer’s disease. In addition, because glucose metabolizesmore rapidly in active areas of the brain, the PET scan can indicate which areas ofthe brain are involved in various processes such as language, music, and vision.30.7 MESONS AND THE BEGINNINGOF PARTICLE PHYSICSPhysicists in the mid-1930s had a fairly simple view of the structure of matter. Thebuilding blocks were the proton, the electron, and the neutron. Three other particleswere known or postulated at the time: the photon, the neutrino, and thepositron. These six particles were considered the fundamental constituents of matter.Although the accepted picture of the world was marvelously simple, no onewas able to provide an answer to the following important question: Because themany protons in proximity in any nucleus should strongly repel each other due totheir like charges, what is the nature of the force that holds the nucleus together?Scientists recognized that this mysterious nuclear force must be much strongerthan anything encountered up to that time.The first theory to explain the nature of the nuclear force was proposed in 1935by the Japanese physicist Hideki Yukawa (1907–1981), an effort that later earnedhim the Nobel prize. In order to understand Yukawa’s theory, it is useful to firstnote that two atoms can form a covalent chemical bond by the exchange ofelectrons. Similarly, in the modern view of electromagnetic interactions, chargedparticles interact by exchanging a photon. Yukawa used this same idea to explainthe nuclear force by proposing a new particle that is exchanged by nucleons in thenucleus to produce the strong force. Further, he demonstrated that the range ofthe force is inversely proportional to the mass of this particle, and predicted thatthe mass would be about 200 times the mass of the electron. Because the new particlewould have a mass between that of the electron and the proton, it was called ameson (from the Greek meso, meaning “middle”).In an effort to substantiate Yukawa’s predictions, physicists began looking forthe meson by studying cosmic rays that enter the Earth’s atmosphere. In 1937,Carl Anderson and his collaborators discovered a particle with mass 106 MeV/c 2 ,about 207 times the mass of the electron. However, subsequent experimentsshowed that the particle interacted very weakly with matter and hence could not bethe carrier of the nuclear force. This puzzling situation inspired several theoreticians

30.7 Mesons and the Beginning of Particle Physics 987to propose that there are two mesons with slightly different masses, an idea thatwas confirmed in 1947 with the discovery of the pi meson (), or simply pion,by Cecil Frank Powell (1903–1969) and Guiseppe P. S. Occhialini (1907–1993).The lighter meson discovered earlier by Anderson, now called a muon (),has only weak and electromagnetic interactions and plays no role in the stronginteraction.The pion comes in three varieties, corresponding to three charge states: , , and 0 . The and particles have masses of 139.6 MeV/c 2 , and the 0 hasa mass of 135.0 MeV/c 2 . Pions and muons are highly unstable particles. For example,the , which has a lifetime of about 2.6 10 8 s, decays into a muon and anantineutrino. The muon, with a lifetime of 2.2 s, then decays into an electron, aneutrino, and an antineutrino. The sequence of decays is [30.6]The interaction between two particles can be understood in general with a simpleillustration called a Feynman diagram, developed by Richard P. Feynman(1918–1988). Figure 30.6 is a Feynman diagram for the electromagnetic interactionbetween two electrons. In this simple case, a photon is the field particle thatmediates the electromagnetic force between the electrons. The photon transfersenergy and momentum from one electron to the other in the interaction. Such aphoton, called a virtual photon, can never be detected directly because it is absorbedby the second electron very shortly after being emitted by the first electron.The existence of a virtual photon might be expected to violate the law of conservationof energy, but it does not because of the time–energy uncertainty principle.Recall that the uncertainty principle says that the energy is uncertain or not conservedby an amount E for a time t such that E t .Now consider the pion exchange between a proton and a neutron via thenuclear force (Fig. 30.7). The energy needed to create a pion of mass m is givenby E m c 2 . Again, the existence of the pion is allowed in spite of conservationof energy if this energy is surrendered in a short enough time t, the time it takesthe pion to transfer from one nucleon to the other. From the uncertainty principle,E t , we get[30.7]Because the pion can’t travel faster than the speed of light, the maximum distanced it can travel in a time t is c t. Using Equation 30.7 and d c t, we find thismaximum distance to bed [30.8]m cThe measured range of the nuclear force is about 1.5 10 15 m. Using this valuefor d in Equation 30.8, the rest energy of the pion is calculated to bem c 2 cd:: e t E m c 2 (1.05 1034 Js)(3.00 10 8 m/s)1.5 10 15 m 2.1 10 11 J 130 MeV This corresponds to a mass of 130 MeV/c 2 (about 250 times the mass of the electron),which is in good agreement with the observed mass of the pion.The concept we have just described is quite revolutionary. In effect, it says that aproton can change into a proton plus a pion, as long as it returns to its originalstate in a very short time. High-energy physicists often say that a nucleon undergoes“fluctuations” as it emits and absorbs pions. As we have seen, these fluctuationsare a consequence of a combination of quantum mechanics (through theuncertainty principle) and special relativity (through Einstein’s energy–mass relationE mc 2 ).Timee – Virtual e –photone – e –Figure 30.6 Feynman diagramrepresenting a photon mediating theelectromagnetic force between twoelectrons.© Shelly Gazin/CORBISRICHARD FEYNMAN, AmericanPhysicist (1918 – 1988)Feynman, together with Julian S. Schwingerand Shinichiro Tomonaga, won the 1965Nobel Prize for physics for fundamentalwork in the principles of quantumelectrodynamics. His many important contributionsto physics include work on the firstatomic bomb in the Manhattan project, theinvention of simple diagrams to representparticle interactions graphically, the theoryof the weak interaction of subatomic particles,a reformulation of quantum mechanics,and the theory of superfluid helium. Later heserved on the commission investigating theChallenger tragedy, demonstrating the problemwith the O-rings by dipping a scalemodelO-ring in his glass of ice water andthen shattering it with a hammer. He alsocontributed to physics education through themagnificent three-volume text The FeynmanLectures on Physics.Timepp0Pion (p )Figure 30.7 Feynman diagramrepresenting a proton interactingwith a neutron via the strong force.In this case, the pion mediates thenuclear force.nn

988 Chapter 30 Nuclear Energy and Elementary ParticlesTABLE 30.2Some Particles and Their PropertiesThis section has dealt with the early theory of Yukawa of particles that mediatethe nuclear force, pions, and the mediators of the electromagnetic force, photons.Although his model led to the modern view, it has been superseded by the morebasic quark–gluon theory, as explained in Sections 30.12 and 30.13.30.8 CLASSIFICATION OF PARTICLESHadronsAll particles other than photons can be classified into two broad categories,hadrons and leptons, according to their interactions. Particles that interact throughthe strong force are called hadrons. There are two classes of hadrons, known asmesons and baryons, distinguished by their masses and spins. All mesons are knownto decay finally into electrons, positrons, neutrinos, and photons. The pion is thelightest of known mesons, with a mass of about 140 MeV/c 2 and a spin of 0. Anotheris the K meson, with a mass of about 500 MeV/c 2 and spin 0 also.Baryons have masses equal to or greater than the proton mass (the name baryonmeans “heavy” in Greek), and their spin is always a non-integer value (1/2 or 3/2).Protons and neutrons are baryons, as are many other particles. With the exceptionof the proton, all baryons decay in such a way that the end products include a proton.For example, the baryon called the hyperon first decays to a 0 in about10 10 s. The 0 then decays to a proton and a p in about 3 10 10 s.Today it is believed that hadrons are composed of quarks. (Later, we will havemore to say about the quark model.) Some of the important properties of hadronsare listed in Table 30.2.L PrincipalAnti- Mass DecayCategory Particle Name Symbol particle (MeV/c 2 ) B L eS Lifetime(s) Modes aLeptons Electron e e 0.511 0 1 0 0 0 StableElectron–neutrino e e 7eV/c 2 0 1 0 0 0 StableMuon 105.7 0 0 1 0 0 2.20 10 6 e e Muon–neutrino 0.3 0 0 1 0 0 StableTau 1 784 0 0 0 1 0 4 10 13Tau –neutrino 30 0 0 0 1 0 StableHadronsMesons Pion 139.6 0 0 0 0 0 2.60 10 8Self 135.0 0 0 0 0 0 0.83 10 16 2Kaon K K 493.7 0 0 0 0 1 1.24 10 8K 0 S K 0 S 497.7 0 0 0 0 1 0.89 10 10497.7 0 0 0 0 1 0K 0 L K 0 L5.2 10 8pe eEta Self 548.8 0 0 0 0 0 10 18 2, 3 Self 958 0 0 0 0 0 2.2 10 21Baryons Proton p p 938.3 1 0 0 0 0 StableNeutron n n 939.6 1 0 0 0 0 920Lambda 0 0 1 115.6 1 0 0 0 1 2.6 10 10 p , n0Sigma 1 189.4 1 0 0 0 1 0.80 10 10 p 0 , n 0 0 1 192.5 1 0 0 0 1 6 10 20 0 1 197.3 1 0 0 0 1 1.5 10 10 nXi 0 0 1 315 1 0 0 0 2 2.9 10 10 0 0 1 321 1 0 0 0 2 1.64 10 10 0 Omega 1 672 1 0 0 0 3 0.82 10 10 0 0 , 0 K a Notations in this column, such as p , n 0 mean two possible decay modes. In this case, the two possible decays are 0 : p and 0 : n 0 .L , 20 , ,0 3 e e, 0 e ev

30.9 Conservation Laws 989LeptonsLeptons (from the Greek leptos, meaning “small” or “light”) are a group of particlesthat participate in the weak interaction. All leptons have a spin of 1/2.Included in this group are electrons, muons, and neutrinos, which are less massivethan the lightest hadron. Although hadrons have size and structure, leptonsappear to be truly elementary, with no structure down to the limit of resolution ofexperiment (about 10 19 m).Unlike hadrons, the number of known leptons is small. Currently, scientistsbelieve there are only six leptons (each having an antiparticle): the electron, themuon, the tau, and a neutrino associated with each: e e The tau lepton, discovered in 1975, has a mass about twice that of the proton.Although neutrinos have masses of about zero, there is strong indirect evidencethat the electron neutrino has a nonzero mass of about 3 eV/c 2 , or 1/180 000 ofthe electron mass. A firm knowledge of the neutrino’s mass could have great significancein cosmological models and in our understanding of the future of theUniverse.30.9 CONSERVATION LAWSA number of conservation laws are important in the study of elementary particles.Although the two described here have no theoretical foundation, they are supportedby abundant empirical evidence. Baryon NumberThe law of conservation of baryon number tells us that whenever a baryon is createdin a reaction or decay, an antibaryon is also created. This information can bequantified by assigning a baryon number: B 1 for all baryons, B 1 for allantibaryons, and B 0 for all other particles. Thus, the law of conservation ofbaryon number states that whenever a nuclear reaction or decay occurs, the sumof the baryon numbers before the process equals the sum of the baryon numbersafter the process.Note that if the baryon number is absolutely conserved, the proton must beabsolutely stable: if it were not for the law of conservation of baryon number, theproton could decay into a positron and a neutral pion. However, such a decay hasnever been observed. At present, we can only say that the proton has a half-life ofat least 10 31 years. (The estimated age of the Universe is about 10 10 years.) In onerecent version of a so-called grand unified theory (GUT), physicists have predictedthat the proton is actually unstable. According to this theory, the baryon number(sometimes called the baryonic charge) is not absolutely conserved, whereas electriccharge is always conserved. Conservation of baryon numberEXAMPLE 30.4 Checking Baryon NumbersGoal Use conservation of baryon number to determine whether a given reaction can occur.ProblemDetermine whether the following reaction can occur based on the law of conservation of baryon number.p n: p p n pStrategy Count the baryons on both sides of the reaction, recalling that that B 1 for baryons and B 1 forantibaryons.SolutionCount the baryons on the left: The neutron and proton are both baryons; hence, 1 1 2.

990 Chapter 30 Nuclear Energy and Elementary ParticlesCount the baryons on the right:There are three baryons and one antibaryon, so1 1 1 (1) 2.Remarkenergy.Baryon number is conserved in this reaction, so it can occur, provided the incoming proton has sufficientExercise 30.4Can the following reaction occur, based on the law of conservation of baryon number?p n : p p pAnswerNo. (Show this by computing the baryon number on both sides and finding that they’re not equal.)Conservation of lepton number Neutron decay Lepton NumberThere are three conservation laws involving lepton numbers, one for each varietyof lepton. The law of conservation of electron-lepton number states that the sumof the electron-lepton numbers before a reaction or decay must equal the sum ofthe electron-lepton numbers after the reaction or decay. The electron and theelectron neutrino are assigned a positive electron-lepton number L e 1, theantileptons e and e are assigned the electron-lepton number L e 1, and allother particles have L e 0. For example, consider neutron decay:Before the decay, the electron-lepton number is L e 0; after the decay, it is0 1 (1) 0, so the electron-lepton number is conserved. It’s important to recognizethat the baryon number must also be conserved. This can easily be seen by notingthat before the decay B 1, whereas after the decay B 1 0 0 1.Similarly, when a decay involves muons, the muon-lepton number L is conserved.The and the are assigned L 1, the antimuons and are assignedL 1, and all other particles have L 0. Finally, the tau-lepton number L isconserved, and similar assignments can be made for the lepton and its neutrino.n : p e e EXAMPLE 30.5 Checking Lepton NumbersGoal Use conservation of lepton number to determine whether a given process is possible.ProblemStrategyDetermine which of the following decay schemes can occur on the basis of conservation of lepton number.: e : e Count the leptons on either side and see if the numbers are equal. e(1)(2)SolutionBecause decay 1 involves both a muon and an electron, L and L e must both be conserved. Before the decay, L 1and L e 0. After the decay, L 0 0 1 1 and L e 1 1 0 0. Both lepton numbers are conserved,and on this basis, the decay mode is possible.Before decay 2 occurs, L 0 and L e 0. After the decay, L 1 1 0 0, but L e 1. This decay isn’tpossible because the electron-lepton number is not conserved.Exercise 30.5Determine whether the decay : e ecan occur.AnswerNo. (Show this by computing muon-lepton numbers on both sides and showing they’re not equal.)

30.10 Strange Particles and Strangeness 991Quick Quiz 30.2Which of the following reactions cannot occur?(a) p p : p p p (b) n : p e e(c) : e (d) e : Quick Quiz 30.3Which of the following reactions cannot occur?(a) p p : 2 (b)(c) (d) p : K p : n 0 0 n : K Quick Quiz 30.4Suppose a claim is made that the decay of a neutron is given by n : p e .Which of the following conservation laws are necessarily violated by this proposeddecay scheme? (a) energy (b) linear momentum (c) electric charge (d) leptonnumber (e) baryon number30.10 STRANGE PARTICLES AND STRANGENESSMany particles discovered in the 1950s were produced by the nuclear interactionof pions with protons and neutrons in the atmosphere. A group of these particles,namely the K, , and particles, was found to exhibit unusual properties in theirproduction and decay and hence were called strange particles.One unusual property of strange particles is that they are always producedin pairs. For example, when a pion collides with a proton, two neutral strange particlesare produced with high probability (Fig. 30.8) following the reaction p : K 0 0 p : K 0 nOn the other hand, the reactionhas never occurred, eventhough it violates no known conservation laws and the energy of the pion is sufficientto initiate the reaction.The second peculiar feature of strange particles is that although they are producedby the strong interaction at a high rate, they do not decay into particles thatinteract via the strong force at a very high rate. Instead, they decay very slowly,which is characteristic of the weak interaction. Their half-lives are in the rangefrom 10 10 s to 10 8 s; most other particles that interact via the strong force havelifetimes on the order of 10 23 s.To explain these unusual properties of strange particles, a law called conservationof strangeness was introduced, together with a new quantum number S calledstrangeness. The strangeness numbers for some particles are given in Table 30.2.The production of strange particles in pairs is explained by assigning S 1to one of the particles and S 1 to the other. All nonstrange particles areassigned strangeness S 0. The law of conservation of strangeness states thatwhenever a nuclear reaction or decay occurs, the sum of the strangeness numbersbefore the process must equal the sum of the strangeness numbers after theprocess.The slow decay of strange particles can be explained by assuming that thestrong and electromagnetic interactions obey the law of conservation of strangeness,whereas the weak interaction does not. Because the decay reaction involvesthe loss of one strange particle, it violates strangeness conservation and hence proceedsslowly via the weak interaction.Courtesy Lawrence Berkeley Laboratory, University of CaliforniaFigure 30.8 This drawing representstracks of many events obtainedby analyzing a bubble-chamber photograph.The strange particles 0and K 0 are formed (at the bottom)as the interacts with a protonaccording to the interaction. (Note thatthe neutral particles leave no tracks,as is indicated by the dashed lines.)The 0 and K 0 then decay accordingto the interactions 0 : p andK 0 : . p : 0 K 0 Conservation of strangenessnumber

992 Chapter 30 Nuclear Energy and Elementary ParticlesApplying Physics 30.2A student claims to have observed a decay of an electroninto two neutrinos traveling in opposite directions.What conservation laws would be violated bythis decay?Explanation Several conservation laws are violated.Conservation of electric charge is violated because thenegative charge of the electron has disappeared.Conservation of electron lepton number is alsoviolated, because there is one lepton before the decayand two afterward. If both neutrinos were electronneutrinos,electron lepton number conservationBreaking Conservation Lawswould be violated in the final state. However, if one of theproduct neutrinos were other than an electron-neutrino,then another lepton conservation law would be violated,because there were no other leptons in the initial state.Other conservation laws are obeyed by this decay.Energy can be conserved—the rest energy of the electronappears as the kinetic energy (and possibly somesmall rest energy) of the neutrinos. The oppositedirections of the velocities of the two neutrinos allowfor the conservation of momentum. Conservation ofbaryon number and conservation of other leptonnumbers are also upheld in this decay.EXAMPLE 30.6 Is Strangeness Conserved?Goal Apply conservation of strangeness to determine whether a process can occur.ProblemStrategyDetermine whether the following reactions can occur on the basis of conservation of strangeness: 0 n : K p : Count strangeness on each side of a given process. If strangeness is conserved, the reaction is possible.(1)(2)SolutionIn the first process, the neutral pion and neutron both have strangeness of zero, so S initial 0 0 0. Because thestrangeness of the K is S 1, and the strangeness of the is S 1, the total strangeness of the final state isS final 1 1 0. Strangeness is conserved and the reaction is allowed.In the second process, the initial state has strangeness S initial 0 0 0, but the final state has strangenessS final 0 (1) 1. Strangeness is not conserved and the reaction isn’t allowed.Exercise 30.6Does the reactionp : K 0 0 obey the law of conservation of strangeness? Show why or why not.AnswerYes. (Show this by computing the strangeness on both sides.)30.11 THE EIGHTFOLD WAYQuantities such as spin, baryon number, lepton number, and strangeness are labelswe associate with particles. Many classification schemes that group particlesinto families based on such labels have been proposed. First, consider the firsteight baryons listed in Table 30.2, all having a spin of 1/2. The family consists ofthe proton, the neutron, and six other particles. If we plot their strangeness versustheir charge using a sloping coordinate system, as in Figure 30.9a, a fascinatingpattern emerges: six of the baryons form a hexagon, and the remaining two are atthe hexagon’s center. (Particles with spin quantum number 1/2 or 3/2 are calledfermions.)Now consider the family of mesons listed in Table 30.2 with spins of zero. (Particleswith spin quantum number 0 or 1 are called bosons.) If we count both particlesand antiparticles, there are nine such mesons. Figure 30.9b is a plot ofstrangeness versus charge for this family. Again, a fascinating hexagonal patternemerges. In this case, the particles on the perimeter of the hexagon lie oppositetheir antiparticles, and the remaining three (which form their own antiparticles)

30.12 Quarks 993n pΣ _ Σ 0 Σ + S = _ 1Ξ _ Ξ 0S = _ 2Q = +1Q = _ 1 Q = 0(a)Λ 0 S = 0K 0 K +K _ 0K 0 S = +1S = 0S = _ 1π _ η π π +η'Q = _ 1 Q = 0(b)Q = +1Figure 30.9 (a) The hexagonaleightfold-way pattern for the eightspin 1 2 baryons. This strangeness versuscharge plot uses a horizontal axisfor the strangeness values S, but asloping axis for the charge number Q.(b) The eightfold-way pattern for thenine spin-zero mesons.are at its center. These and related symmetric patterns, called the eightfold way,were proposed independently in 1961 by Murray Gell-Mann and Yuval Ne’eman.The groups of baryons and mesons can be displayed in many other symmetricpatterns within the framework of the eightfold way. For example, the family ofspin-3/2 baryons contains ten particles arranged in a pattern like the tenpins in abowling alley. After the pattern was proposed, one of the particles was missing—ithad yet to be discovered. Gell-Mann predicted that the missing particle, which hecalled the omega minus ( ), should have a spin of 3/2, a charge of 1, a strangenessof 3, and a mass of about 1 680 MeV/c 2 . Shortly thereafter, in 1964, scientistsat the Brookhaven National Laboratory found the missing particle throughcareful analyses of bubble chamber photographs and confirmed all its predictedproperties.The patterns of the eightfold way in the field of particle physics have much incommon with the periodic table. Whenever a vacancy (a missing particle or element)occurs in the organized patterns, experimentalists have a guide for their investigations.30.12 QUARKSAs we have noted, leptons appear to be truly elementary particles because theyhave no measurable size or internal structure, are limited in number, and do notseem to break down into smaller units. Hadrons, on the other hand, are complexparticles with size and structure. Further, we know that hadrons decay into otherhadrons and are many in number. Table 30.2 lists only those hadrons that are stableagainst hadronic decay; hundreds of others have been discovered. These factsstrongly suggest that hadrons cannot be truly elementary but have some substructure.The Original Quark ModelIn 1963 Gell-Mann and George Zweig independently proposed that hadrons havean elementary substructure. According to their model, all hadrons are compositesystems of two or three fundamental constitutents called quarks, which rhymeswith “forks” (though some rhyme it with “sharks”). Gell-Mann borrowed the wordquark from the passage “Three quarks for Muster Mark” in James Joyce’s bookFinnegans Wake. In the original model there were three types of quarks designatedby the symbols u, d, and s. These were given the arbitrary names up, down, andsideways (or, now more commonly, strange).An unusual property of quarks is that they have fractional electronic charges, asshown—along with other properties—in Table 30.3 (page 994). Associated witheach quark is an antiquark of opposite charge, baryon number, and strangeness.The compositions of all hadrons known when Gell-Mann and Zweig presentedtheir models could be completely specified by three simple rules:Photo courtesy of Michael R. DresslerMURRAY GELL-MANN,American Physicist (1929–)Gell-Mann was awarded the Nobel Prizein 1969 for his theoretical studies dealingwith subatomic particles.

994 Chapter 30 Nuclear Energy and Elementary ParticlesTABLE 30.4Quark Compositionof Several HadronsParticle K K K 0pn 0 0 0 QuarkCompositionMesonsBaryonsduudsuussduududdudsuusudsddsussdsssssTABLE 30.3Properties of Quarks and AntiquarksQuarksBaryonName Symbol Spin Charge Number Strangeness Charm Bottomness Topness11Up u 2 2 0 0 0 01Down d 1 3 e 3120 0 0 01Strange s 1 3 e 3121 0 0 01Charmed c 2 3 e 31230 1 0 011Bottom b2 1 3 e30 0 1 011Top t 2 3 e3 e0 0 0 12AntiquarksBaryonName Symbol Spin Charge Number Strangeness Charm Bottomness TopnessAnti-up1u 2 2 3 e 1 3 0 0 0 0Anti-down1d 2 1 3 e 1 3 0 0 0 0Anti-strange1s 2 1 3 e 1 3 1 0 0 0Anti-charmed1c 2 2 3 e 1 3 0 1 0 01Anti-bottom b 1 1 20 0 1 01Anti-top t 2 3 e 33 e0 0 0 123 1 31. Mesons consist of one quark and one antiquark, giving them a baryon numberof 0, as required.2. Baryons consist of three quarks.3. Antibaryons consist of three antiquarks.Table 30.4 lists the quark compositions of several mesons and baryons. Note thatjust two of the quarks, u and d, are contained in all hadrons encountered in ordinarymatter (protons and neutrons). The third quark, s, is needed only to constructstrange particles with a strangeness of either 1 or 1. Active Figure 30.10is a pictorial representation of the quark compositions of several particles.Applying Physics 30.3We have seen a law of conservation of lepton numberand a law of conservation of baryon number. Why isn’tthere a law of conservation of meson number?Explanation We can argue this from the point ofview of creating particle–antiparticle pairs from availableenergy. If energy is converted to the rest energyof a lepton–antilepton pair, then there is no netchange in lepton number, because the lepton has alepton number of 1 and the antilepton 1. Energycould also be transformed into the rest energy of aConservation of Meson Number?baryon–antibaryon pair. The baryon has baryon number1, the antibaryon 1, and there is no netchange in baryon number.But now suppose energy is transformed into therest energy of a quark–antiquark pair. By definition inquark theory, a quark–antiquark pair is a meson.There was no meson before, and now there’s a meson,so already there is violation of conservation of mesonnumber. With more energy, we can create moremesons, with no restriction from a conservation lawother than that of energy.Charm and Other Recent DevelopmentsAlthough the original quark model was highly successful in classifying particlesinto families, there were some discrepancies between predictions of the model andcertain experimental decay rates. Consequently, a fourth quark was proposed byseveral physicists in 1967. The fourth quark, designated by c, was given a propertycalled charm. A charmed quark would have the charge 2e/3, but its charmwould distinguish it from the other three quarks. The new quark would have a

30.12 Quarks 995charm C 1, its antiquark would have a charm C 1, and all other quarkswould have C 0, as indicated in Table 30.3. Charm, like strangeness, would beconserved in strong and electromagnetic interactions but not in weak interactions.In 1974 a new heavy meson called the J/ particle (or simply, ) was discoveredindependently by a group led by Burton Richter at the Stanford Linear Accelerator(SLAC) and another group led by Samuel Ting at the Brookhaven NationalLaboratory. Richter and Ting were awarded the Nobel Prize in 1976 for this work.The J/ particle didn’t fit into the three-quark model, but had the properties of acombination of a charmed quark and its antiquark ( cc). It was much heavier thanthe other known mesons (3 100 MeV/c 2 ) and its lifetime was much longer thanthose of other particles that decay via the strong force. In 1975, researchers atStanford University reported strong evidence for the existence of the tau ()lepton, with a mass of 1 784 MeV/c 2 . Such discoveries led to more elaborate quarkmodels and the proposal of two new quarks, named top (t) and bottom (b). Todistinguish these quarks from the old ones, quantum numbers called topness andbottomness were assigned to these new particles and are included in Table 30.3. In1977 researchers at the Fermi National Laboratory, under the direction of LeonLederman, reported the discovery of a very massive new meson with compositionbb. In March of 1995, researchers at Fermilab announced the discovery of thetop quark (supposedly the last of the quarks to be found) having mass 173 GeV/c 2 .You are probably wondering whether such discoveries will ever end. How many“building blocks” of matter really exist? The numbers of different quarks and leptonshave implications for the primordial abundance of certain elements, so atpresent it appears there may be no further fundamental particles. Some propertiesof quarks and leptons are given in Table 30.5.Despite extensive experimental efforts, no isolated quark has ever been observed.Physicists now believe that quarks are permanently confined inside ordinaryparticles because of an exceptionally strong force that prevents them from escaping.This force, called the color force (which will be discussed in Section 30.13), increaseswith separation distance (similar to the force of a spring). The greatstrength of the force between quarks has been described by one author as follows: 2Mesonsπ+uuK _dsBaryonspACTIVE FIGURE 30.10Quark compositions of two mesonsand two baryons. Note that themesons on the left contain twoquarks, and the baryons on the rightcontain three quarks.Log into PhysicsNow at www.cp7e.comand go to Active Figure 30.10 toobserve the quark compositions forthe mesons and baryons.uunddudQuarks are slaves of their own color charge, . . . bound like prisoners of a chaingang. . . . Any locksmith can break the chain between two prisoners, but no locksmith isexpert enough to break the gluon chains between quarks. Quarks remain slaves forever.TABLE 30.5The Fundamental Particles and Some of Their PropertiesParticle Rest Energy ChargeQuarksudcstb360 MeV360 MeV1500 MeV540 MeV173 GeV5 GeV 2 3 e 1 3 e 2 3 e 1 3 e 2 3 e 1 3 eLeptonse 511 keV e 107 MeV e 1784 MeV e e 30 eV 0 0.5 MeV 0 250 MeV 02 Harald Fritzsch, Quarks: The Stuff of Matter (London: Allen Lane, 1983).

996 Chapter 30 Nuclear Energy and Elementary ParticlesComputers at Fermilab create apictorial representation such asthis of the paths of particles aftera collision.Courtesy of Fermi National Accelerator LaboratoryTIP 30.3 Color is NotReally ColorWhen we use the word color todescribe a quark, it has nothing to dowith visual sensation from light. It issimply a convenient name for aproperty analgous to electric charge.q(a)(b)qMesonBaryonFigure 30.11 (a) A green quark isattracted to an anti-green quark toform a meson with quark structure( qq). (b) Three different-coloredquarks attract each other to form abaryon.30.13 COLORED QUARKSShortly after the theory of quarks was proposed, scientists recognized that certainparticles had quark compositions that were in violation of the Pauli exclusion principle.Because all quarks have spins of 1/2, they are expected to follow the exclusionprinciple. One example of a particle that violates the exclusion principle isthe (sss) baryon, which contains three s quarks having parallel spins, giving it atotal spin of 3/2. Other examples of baryons that have identical quarks with parallelspins are the (uuu) and the (ddd). To resolve this problem, Moo-Young Han and Yoichiro Nambu suggested in 1965 that quarks possess a new propertycalled color or color charge. This “charge” property is similar in many respectsto electric charge, except that it occurs in three varieties, labeled red, green, andblue! (The antiquarks are labeled anti-red, anti-green, and anti-blue.) To satisfy theexclusion principle, all three quarks in a baryon must have different colors. Just asa combination of actual colors of light can produce the neutral color white, a combinationof three quarks with different colors is also “white,” or colorless. A mesonconsists of a quark of one color and an antiquark of the corresponding anticolor.The result is that baryons and mesons are always colorless (or white).Although the concept of color in the quark model was originally conceived tosatisfy the exclusion principle, it also provided a better theory for explaining certainexperimental results. For example, the modified theory correctly predicts thelifetime of the 0 meson. The theory of how quarks interact with each other bymeans of color charge is called quantum chromodynamics, or QCD, to parallelquantum electrodynamics (the theory of interactions among electric charges). InQCD, the quark is said to carry a color charge, in analogy to electric charge. Thestrong force between quarks is often called the color force. The force is carried bymassless particles called gluons (which are analogous to photons for the electromagneticforce). According to QCD, there are eight gluons, all with color charge.When a quark emits or absorbs a gluon, its color changes. For example, a bluequark that emits a gluon may become a red quark, and a red quark that absorbsthis gluon becomes a blue quark. The color force between quarks is analogous tothe electric force between charges: Like colors repel and opposite colors attract.Therefore, two red quarks repel each other, but a red quark will be attracted to ananti-red quark. The attraction between quarks of opposite color to form a meson(qq) is indicated in Figure 30.11a.Different-colored quarks also attract each other, but with less intensity thanopposite colors of quark and antiquark. For example, a cluster of red, blue, andgreen quarks all attract each other to form baryons, as indicated in Figure 30.11b.Every baryon contains three quarks of three different colors.Although the color force between two color-neutral hadrons (such as a protonand a neutron) is negligible at large separations, the strong color force betweentheir constituent quarks does not exactly cancel at small separations of about 1 fm.This residual strong force is in fact the nuclear force that binds protons and

30.14 Electroweak Theory and the Standard Model 997Timenp(a) Yukawa’s pion modeld u dTimepπ –nnd u upuuu–annihilationdu–uu u–pairproductionupdud d unFigure 30.12 (a) A nuclear interactionbetween a proton and a neutronexplained in terms of Yukawa’spion exchange model. Because thepion carries charge, the proton andneutron switch identities. (b) Thesame interaction explained in termsof quarks and gluons. Note that theexchanged ud quark pair makes up a meson.(b) Quark modelneutrons to form nuclei. It is similar to the residual electromagnetic force thatbinds neutral atoms into molecules. According to QCD, a more basic explanationof nuclear force can be given in terms of quarks and gluons, as shown in Figure30.12, which shows contrasting Feynman diagrams of the same process. Eachquark within the neutron and proton is continually emitting and absorbing virtualgluons and creating and annihilating virtual (qq) pairs. When the neutron andproton approach within 1 fm of each other, these virtual gluons and quarks can beexchanged between the two nucleons, and such exchanges produce the nuclearforce. Figure 30.12b depicts one likely possibility or contribution to the processshown in Figure 30.12a: a down quark emits a virtual gluon (represented by a wavyline in Fig. 30.12b), which creates a uu pair. Both the recoiling d quark and the uare transmitted to the proton where the u annihilates a proton u quark (with thecreation of a gluon) and the d is captured.30.14 ELECTROWEAK THEORY ANDTHE STANDARD MODELRecall that the weak interaction is an extremely short range force having an interactiondistance of approximately 10 18 m (Table 30.1). Such a short-range interactionimplies that the quantized particles which carry the weak field (the spin oneW , W , and Z 0 bosons) are extremely massive, as is indeed the case. These amazingbosons can be thought of as structureless, pointlike particles as massive as kryptonatoms! The weak interaction is responsible for the decay of the c, s, b, and tquarks into lighter, more stable u and d quarks, as well as the decay of the massive and leptons into (lighter) electrons. The weak interaction is very important becauseit governs the stability of the basic particles of matter.A mysterious feature of the weak interaction is its lack of symmetry, especiallywhen compared to the high degree of symmetry shown by the strong, electromagnetic,and gravitational interactions. For example, the weak interaction, unlike thestrong interaction, is not symmetric under mirror reflection or charge exchange.(Mirror reflection means that all the quantities in a given particle reaction are exchangedas in a mirror reflection—left for right, an inward motion toward themirror for an outward motion, etc. Charge exchange means that all the electriccharges in a particle reaction are converted to their opposites—all positives tonegatives and vice versa.) When we say that the weak interaction is not symmetric,we mean that the reaction with all quantities changed occurs less frequentlythan the direct reaction. For example, the decay of the K 0 , which is governedby the weak interaction, is not symmetric under charge exchange because thereaction K 0 : e e occurs much more frequently than the reactionK 0 : e e.In 1979, Sheldon Glashow, Abdus Salam, and Steven Weinberg won a Nobelprize for developing a theory called the electroweak theory that unified theelectromagnetic and weak interactions. This theory postulates that the weak andelectromagnetic interactions have the same strength at very high particle energies,

998 Chapter 30 Nuclear Energy and Elementary ParticlesFigure 30.13 The Standard Modelof particle physics.MATTER AND ENERGYFORCESCONSTITUENTSStrongElectromagneticWeakGravityGluonPhotonW and Z bosonsGravitonQuarksu c td s bLeptonse m t emtA view from inside the LargeElectron–Positron (LEP) collider tunnel,which is 27 km in circumference.Courtesy of CERNand are different manifestations of a single unifying electroweak interaction. Thephoton and the three massive bosons (W and Z 0 ) play a key role in the electroweaktheory. The theory makes many concrete predictions, but perhaps themost spectacular is the prediction of the masses of the W and Z particles at about82 GeV/c 2 and 93 GeV/c 2 , respectively. A 1984 Nobel Prize was awarded to CarloRubbia and Simon van der Meer for their work leading to the discovery of theseparticles at just those energies at the CERN Laboratory in Geneva, Switzerland.The combination of the electroweak theory and QCD for the strong interactionform what is referred to in high energy physics as the Standard Model. Although thedetails of the Standard Model are complex, its essential ingredients can be summarizedwith the help of Figure 30.13. The strong force, mediated by gluons, holdsquarks together to form composite particles such as protons, neutrons, and mesons.Leptons participate only in the electromagnetic and weak interactions. The electromagneticforce is mediated by photons, and the weak force is mediated by W and Zbosons. Note that all fundamental forces are mediated by bosons (particles with spin1) whose properties are given, to a large extent, by symmetries involved in the theories.However, the Standard Model does not answer all questions. A major questionis why the photon has no mass while the W and Z bosons do. Because of this massdifference, the electromagnetic and weak forces are quite distinct at low energies,but become similar in nature at very high energies, where the rest energies of theW and Z bosons are insignificant fractions of their total energies. This behaviorduring the transition from high to low energies, called symmetry breaking, doesn’tanswer the question of the origin of particle masses. To resolve that problem, ahypothetical particle called the Higgs boson has been proposed which provides amechanism for breaking the electroweak symmetry and bestowing different particlemasses on different particles. The Standard Model, including the Higgs mechanism,provides a logically consistent explanation of the massive nature of the Wand Z bosons. Unfortunately, the Higgs boson has not yet been found, but physicistsknow that its mass should be less than 1 TeV/c 2 (10 12 eV).In order to determine whether the Higgs boson exists, two quarks of at least 1 TeVof energy must collide, but calculations show that this requires injecting 40 TeV ofenergy within the volume of a proton. Scientists are convinced that because of thelimited energy available in conventional accelerators using fixed targets, it is necessaryto build colliding-beam accelerators called colliders. The concept of a collideris straightforward. In such a device, particles with equal masses and kinetic energies,traveling in opposite directions in an accelerator ring, collide head-on to producethe required reaction and the formation of new particles. Because the totalmomentum of the interacting particles is zero, all of their kinetic energy is availablefor the reaction. The Large Electron–Positron (LEP) collider at CERN, nearGeneva, Switzerland, and the Stanford Linear Collider in California collide bothelectrons and positrons. The Super Proton Synchrotron at CERN accelerates

30.15 The Cosmic Connection 999protons and antiprotons to energies of 270 GeV, and the world’s highest-energyproton acclerator, the Tevatron, at the Fermi National Laboratory in Illinois, producesprotons at almost 1 000 GeV (or 1 TeV). CERN has started construction ofthe Large Hadron Collider (LHC), a proton–proton collider that will provide acenter-of-mass energy of 14 TeV and allow an exploration of Higgs-boson physics.The accelerator is being constructed in the same 27-km circumference tunnel asCERN’s LEP collider, and construction is expected to be completed in 2005.Following the success of the electroweak theory, scientists attempted to combineit with QCD in a grand unification theory (GUT). In this model, the electroweakforce was merged with the strong color force to form a grand unifiedforce. One version of the theory considers leptons and quarks as members of thesame family that are able to change into each other by exchanging an appropriateparticle. Many GUT theories predict that protons are unstable and will decay witha lifetime of about 10 31 years, a period far greater than the age of the Universe. Asyet, proton decays have not been observed.Applying Physics 30.4Head-on CollisionsConsider a car making a head-on collision with anidentical car moving in the opposite direction at thesame speed. Compare that collision to one in whichone of the cars collides with a second car that is atrest. In which collision is there a larger transformationof kinetic energy to other forms? How does this idearelate to producing exotic particles in collisions?Explanation In the head-on collision with both carsmoving, conservation of momentum causes most, ifnot all, of the kinetic energy to be transformed toother forms. In the collision between a moving car anda stationary car, the cars are still moving after the collisionin the direction of the moving car, but with reducedspeed. Thus, only part of the kinetic energy istransformed to other forms. This suggests the advantageof using colliding beams to produce exotic particles,as opposed to firing a beam into a stationary target.When particles moving in opposite directionscollide, all of the kinetic energy is available for transformationinto other forms—in this case, the creationof new particles. When a beam is fired into a stationarytarget, only part of the energy is available for transformation,so particles of higher mass cannot be created.30.15 THE COSMIC CONNECTIONIn this section we describe one of the most fascinating theories in all of science—the Big Bang theory of the creation of the Universe—and the experimentalevidence that supports it. This theory of cosmology states that the Universe had a beginningand that this beginning was so cataclysmic that it is impossible to look backbeyond it. According to the theory, the Universe erupted from an infinitely densesingularity about 15 to 20 billion years ago. The first few minutes after the Big Bangsaw such extremes of energy that it is believed that all four interactions of physicswere unified and all matter was contained in an undifferentiated “quark soup.”The evolution of the four fundamental forces from the Big Bang to thepresent is shown in Figure 30.14. During the first 10 43 s (the ultrahot epoch, withBig BangUnifiedforceGravitationalStrong andelectroweakTwoforcesQuarks and leptonsElectroweakThreeforcesProtons andneutronscan formNuclei can formAtoms can formElectromagnetic forceFour forces10 –40 10 –30 10 –20 10 –10 10 0 10 10 10 20Age of the Universe (s)GravitationStrong forceWeak forcePresent ageof the UniverseFigure 30.14 A brief history of theUniverse from the Big Bang to thepresent. The four forces becamedistinguishable during the firstmicrosecond. Following this, all thequarks combined to form particlesthat interact via the strong force. Theleptons remained separate, however,and exist as individually observableparticles to this day.

1000 Chapter 30 Nuclear Energy and Elementary ParticlesGEORGE GAMOW(1904 – 1968)Gamow and two of his students, RalphAlpher and Robert Herman, were the firstto take the first half hour of the Universeseriously. In a mostly overlooked paperpublished in 1948, they made trulyremarkable cosmological predictions. Theycorrectly calculated the abundances ofhydrogen and helium after the first halfhour (75% H and 25% He) and predictedthat radiation from the Big Bang shouldstill be present and have an apparenttemperature of about 5 K.Figure 30.15 Robert W. Wilson(left) and Arno A. Penzias (right), withBell Telephone Laboratories’ hornreflectorantenna.Courtesy of AIP Emilio Segre Visual ArchivesAT&T Bell LaboratoriesT 10 32 K), it is presumed that the strong, electroweak, and gravitational forceswere joined to form a completely unified force. In the first 10 35 s following theBig Bang (the hot epoch, with T 10 29 K), gravity broke free of this unificationand the strong and electroweak forces remained as one, described by a grand unificationtheory. This was a period when particle energies were so great( 10 16 GeV) that very massive particles as well as quarks, leptons, and their antiparticles,existed. Then, after 10 35 s, the Universe rapidly expanded and cooled(the warm epoch, with T 10 29 to 10 15 K), the strong and electroweak forcesparted company, and the grand unification scheme was broken. As the Universecontinued to cool, the electroweak force split into the weak force and the electromagneticforce about 10 10 s after the Big Bang.After a few minutes, protons condensed out of the hot soup. For half an hourthe Universe underwent thermonuclear detonation, exploding like a hydrogenbomb and producing most of the helium nuclei now present. The Universe continuedto expand, and its temperature dropped. Until about 700 000 years afterthe Big Bang, the Universe was dominated by radiation. Energetic radiationprevented matter from forming single hydrogen atoms because collisions wouldinstantly ionize any atoms that might form. Photons underwent continuousCompton scattering from the vast number of free electrons, resulting in aUniverse that was opaque to radiation. By the time the Universe was about 700 000years old, it had expanded and cooled to about 3 000 K, and protons could bind toelectrons to form neutral hydrogen atoms. Because the energies of the atoms werequantized, far more wavelengths of radiation were not absorbed by atoms thanwere, and the Universe suddenly became transparent to photons. Radiation nolonger dominated the Universe, and clumps of neutral matter grew steadily—firstatoms, followed by molecules, gas clouds, stars, and finally galaxies.Observation of Radiation from the Primordial FireballIn 1965 Arno A. Penzias (b. 1933) and Robert W. Wilson (b. 1936) of Bell Laboratoriesmade an amazing discovery while testing a sensitive microwave receiver. A peskysignal producing a faint background hiss was interfering with their satellite communicationsexperiments. In spite of their valiant efforts, the signal remained. Ultimatelyit became clear that they were observing microwave background radiation(at a wavelength of 7.35 cm) representing the leftover “glow” from the Big Bang.The microwave horn that served as their receiving antenna is shown in Figure30.15. The intensity of the detected signal remained unchanged as the antennawas pointed in different directions. The fact that the radiation had equal strengthsin all directions suggested that the entire Universe was the source of this radiation.Evicting a flock of pigeons from the 20-foot horn and cooling the microwavedetector both failed to remove the signal. Through a casual conversation, Penziasand Wilson discovered that a group at Princeton had predicted the residual radiationfrom the Big Bang and were planning an experiment to confirm the theory.The excitement in the scientific community was high when Penzias and Wilsonannounced that they had already observed an excess microwave backgroundcompatible with a 3-K blackbody source.Because Penzias and Wilson made their measurements at a single wavelength,they did not completely confirm the radiation as 3-K blackbody radiation. Subsequentexperiments by other groups added intensity data at different wavelengths,as shown in Figure 30.16. The results confirm that the radiation is that of a blackbodyat 2.9 K. This figure is perhaps the most clear-cut evidence for the Big Bangtheory. The 1978 Nobel Prize in physics was awarded to Penzias and Wilson fortheir important discovery.The discovery of the cosmic background radiation produced a problem, however:the radiation was too uniform. Scientists believed there had to be slight fluctuationsin this background in order for such objects as galaxies to form. In 1989, NASAlaunched a satellite called the Cosmic Background Explorer (COBE, pronouncedKOH-bee) to study this radiation in greater detail. In 1992, George Smoot

30.16 Problems and Perspectives 1001Radiant energy density per wavelength interval (eV/m 3 /m)10 810 610 410 210 010 –2PenziasandWilson0.01 0.1 1 10 100Wavelength (cm)Figure 30.16 Theoretical blackbody (browncurve) and measured radiation spectra (bluepoints) of the Big Bang. Most of the data werecollected from the Cosmic Background Explorer(COBE) satellite. The datum of Wilson andPenzias is indicated.(b. 1945) at the Lawrence Berkeley Laboratory found that the background was notperfectly uniform, but instead contained irregularities corresponding to temperaturevariations of 0.000 3 K. It is these small variations that provided nucleationsites for the formation of the galaxies and other objects we now see in the sky.30.16 PROBLEMS AND PERSPECTIVESWhile particle physicists have been exploring the realm of the very small, cosmologistshave been exploring cosmic history back to the first microsecond of the BigBang. Observation of the events that occur when two particles collide in an acceleratoris essential in reconstructing the early moments in cosmic history. Perhapsthe key to understanding the early Universe is first to understand the world ofelementary particles. Cosmologists and particle physicists find that they have manycommon goals and are joining efforts to study the physical world at its most fundamentallevel.Our understanding of physics at short and long distances is far from complete.Particle physics is faced with many questions: why is there so little antimatter in theUniverse? Do neutrinos have a small mass, and if so, how much do they contributeto the “dark matter” holding the universe together gravitationally? How can weunderstand the latest astronomical measurements, which show that the expansionof the universe is accelerating and that there may be a kind of “antigravity force”acting between widely separated galaxies? Is it possible to unify the strong andelectroweak theories in a logical and consistent manner? Why do quarks and leptonsform three similar but distinct families? Are muons the same as electrons(apart from their different masses), or do they have subtle differences that havenot been detected? Why are some particles charged and others neutral? Why doquarks carry a fractional charge? What determines the masses of the fundamentalparticles? The questions go on and on. Because of the rapid advances and new discoveriesin the related fields of particle physics and cosmology, by the time youread this book some of these questions may have been resolved and others mayhave emerged.An important question that remains is whether leptons and quarks have a substructure.If they do, one could envision an infinite number of deeper structurelevels. However, if leptons and quarks are indeed the ultimate constituents of matter,as physicists today tend to believe, we should be able to construct a final theoryof the structure of matter, as Einstein dreamed of doing. In the view of many physicists,the end of the road is in sight, but how long it will take to reach that goal isanyone’s guess.

1002 Chapter 30 Nuclear Energy and Elementary ParticlesSUMMARYTake a practice test by logging intoPhysicsNow at www.cp7e.com and clicking on the Pre-Test link for this chapter.30.1 Nuclear Fission &30.2 Nuclear ReactorsIn nuclear fission, the total mass of the products is alwaysless than the original mass of the reactants. Nuclearfission occurs when a heavy nucleus splits, or fissions,into two smaller nuclei. The lost mass is transformedinto energy, electromagnetic radiation, and the kineticenergy of daughter particles.A nuclear reactor is a system designed to maintain aself-sustaining chain reaction. Nuclear reactors usingcontrolled fission events are currently being used togenerate electric power. A useful parameter for describingthe level of reactor operation is the reproductionconstant K, which is the average number ofneutrons from each fission event that will cause anotherevent. A self-sustaining reaction is achievedwhen K 1.30.3 Nuclear FusionIn nuclear fusion, two light nuclei combine to forma heavier nucleus. This type of nuclear reactionoccurs in the Sun, assisted by a quantum tunnelingprocess that helps particles get through the Coulombbarrier.Controlled fusion events offer the hope of plentifulsupplies of energy in the future. The nuclear fusion reactoris considered by many scientists to be the ultimateenergy source because its fuel is water. Lawson’s criterionstates that a fusion reactor will provide a net outputpower if the product of the plasma ion density n andthe plasma confinement time satisfies the followingrelationships:n 10 14 s/cm 3 Deuterium–tritium interaction [30.5]n 10 16 s/cm 3 Deuterium–deuterium interaction30.5 The Fundamental Forces ofNatureThere are four fundamental forces of nature: the strong(hadronic), electromagnetic, weak, and gravitationalforces. The strong force is the force between nucleonsthat keeps the nucleus together. The weak force is responsiblefor beta decay. The electromagnetic and weakforces are now considered to be manifestations of a singleforce called the electroweak force.Every fundamental interaction is said to be mediatedby the exchange of field particles. The electromagneticinteraction is mediated by the photon, the weak interactionby the W and Z 0 bosons, the gravitationalinteraction by gravitons, and the strong interaction bygluons.30.6 Positrons and OtherAntiparticlesAn antiparticle and a particle have the same mass, butopposite charge, and may also have other propertieswith opposite values, such as lepton number and baryonnumber. It is possible to produce particle–antiparticlepairs in nuclear reactions if the available energy isgreater than 2mc 2 , where m is the mass of the particle(or antiparticle).30.8 Classification of ParticlesParticles other than photons are classified as hadronsor leptons. Hadrons interact primarily through thestrong force. They have size and structure and henceare not elementary particles. There are two types ofhadrons: baryons and mesons. Mesons have a baryonnumber of zero and have either zero or integer spin.Baryons, which generally are the most massive particles,have nonzero baryon numbers and spins of 1/2or 3/2. The neutron and proton are examples ofbaryons.Leptons have no known structure, down to the limitsof current resolution (about 10 19 m). Leptons interactonly through the weak and electromagnetic forces.There are six leptons: the electron, e ; the muon, ;the tau, ; and their associated neutrinos, e , ,and .30.9 Conservation Laws &30.10 Strange Particles andStrangenessIn all reactions and decays, quantities such as energy,linear momentum, angular momentum, electric charge,baryon number, and lepton number are strictly conserved.Certain particles have properties called strangenessand charm. These unusual properties are conservedonly in those reactions and decays that occur viathe strong force.

Problems 100330.12 Quarks &30.13 Colored QuarksRecent theories postulate that all hadrons are composedof smaller units known as quarks which have fractionalelectric charges and baryon numbers of 1/3 and comein six “flavors”: up, down, strange, charmed, top, andbottom. Each baryon contains three quarks, and eachmeson contains one quark and one antiquark.According to the theory of quantumchromodynamics, quarks have a property called color,and the strong force between quarks is referred to as thecolor force. The color force increases as the distancebetween particles increases, so quarks are confinedand are never observed in isolation. When two boundquarks are widely separated, a new quark–antiquarkpair forms between them, and the single particle breaksinto two new particles, each composed of a quark–antiquark pair.30.15 The Cosmic ConnectionObservation of background microwave radiation by Penziasand Wilson strongly confirmed that the Universestarted with a Big Bang about 15 billion years ago andhas been expanding ever since. The background radiationis equivalent to that of a blackbody at a temperatureof about 3 K.The cosmic microwave background has very small irregularities,corresponding to temperature variations of0.000 3 K. Without these irregularities acting as nucleationsites, particles would never have clumped together to formgalaxies and stars.CONCEPTUAL QUESTIONS1. If high-energy electrons with de Broglie wavelengthssmaller than the size of the nucleus are scatteredfrom nuclei, the behavior of the electrons is consistentwith scattering from very massive structuresmuch smaller in size than the nucleus, namely,quarks. How is this similar to a classic experimentthat detected small structures in an atom?2. What factors make a fusion reaction difficult toachieve?3. Doubly charged baryons are known to exist. Whyare there no doubly charged mesons?4. Why would a fusion reactor produce less radioactivewaste than a fission reactor?5. Atoms didn’t exist until hundreds of thousands ofyears after the Big Bang. Why?6. Particles known as resonances have very short halflives,on the order of 10 23 s. Would you guess theyare hadrons or leptons?7. Describe the quark model of hadrons, including theproperties of quarks.8. In the theory of quantum chromodynamics,quarks come in three colors. How would you justifythe statement “All baryons and mesons arecolorless?”9. Describe the properties of baryons and mesons andthe important differences between them.10. Identify the particle decays in Table 30.2 that occurby the electromagnetic interaction. Justify youranswer.11. Kaons all decay into final states that contain no protonsor neutrons. What is the baryon number ofkaons?12. When an electron and a positron meet at low speedsin free space, why are two 0.511-MeV gamma raysproduced, rather than one gamma ray with an energyof 1.02 MeV ?13. Two protons in a nucleus interact via the strong interaction.Are they also subject to a weak interaction?14. Why is a neutron stable inside the nucleus? (In freespace, the neutron decays in 900 s.)15. An antibaryon interacts with a meson. Can a baryonbe produced in such an interaction? Explain.16. Why is water a better shield against neutrons thanlead or steel is?17. How many quarks are there in (a) a baryon, (b) an antibaryon,(c) a meson, and (d) an antimeson? How doyou account for the fact that baryons have half-integralspins and mesons have spins of 0 or 1? [Hint :1quarks have spin .]2

1004 Chapter 30 Nuclear Energy and Elementary Particles18. A typical chemical reaction is one in which a water 19. The neutral meson decays by the strong interaction1event such as 0 n 23592 U : 13653 I 9839 Y 21 0 n . K 0 : , but with a much longer half-lifeWould you expect the energy released in this nuclearof about 10 10 s. How do you explain these obser-molecule is formed by combining hydrogen andinto two pions according to : ,oxygen. In such a reaction, about 2.5 eV of energy with a half-life of about 10 23 s. The neutral Kis released. Compare this reaction to a nuclear meson also decays into two pions according toevent to be much greater, much less, or vations?about the same as that released in the chemical reaction?Explain.PROBLEMSthat each fission produces 200 MeV of thermal1, 2, 3 straightforward, intermediate, challenging full solution available in Student Solutions Manual/Study Guide coached problem with hints available at www.cp7e.com = biomedical applicationSection 30.1 Nuclear Fissionn 23592 U : 98 13540Zr 52 Te 3n engine has an efficiency of 20%? (AssumeSection 30.2 Nuclear Reactors1. If the average energy released in a fission event isenergy. Calculate the mass of 235 U consumedeach day.208 MeV, find the total number of fission events requiredto operate a 100-W lightbulb for 1.0 h.7. Suppose that the water exerts an average frictionaldrag of 1.0 10 5 N on a nuclear-powered ship.2. Find the energy released in the fission reactionThe atomic masses of the fission products are9897.912 0 u for Zr and 134.908 7 u for1354052 Te.How far can the ship travel per kilogram of fuel ifthe fuel consists of enriched uranium containing1.7% of the fissionable isotope 235 U and the ship’s208 MeV is released per fission event.)8. It has been estimated that the Earth contains1.0 10 9 tons of natural uranium that can be3. Find the energy released in the following fission reaction:mined economically. If all the world’s energyneeds (7.0 10 12 J/s) were supplied by 235 U10 n 23592 U : 8838Sr 13654 Xe 121 0 nfission, how long would this supply last? [Hint :See Appendix B for the percent abundance4. Strontium-90 is a particularly dangerous fissionproduct of 235 U because it is radioactive and it substitutesfor calcium in bones. What other direct fissionproducts would accompany it in the neutroninducedfission of 235 U? [Note : This reaction mayrelease two, three, or four free neutrons.]5. Assume that ordinary soil contains natural uranium inamounts of 1 part per million by mass. (a) How muchuranium is in the top 1.00 meter of soil on a 1-acre(43 560-ft 2 ) plot of ground, assuming the specificgravity of soil is 4.00? (b) How much of the isotope235 U, appropriate for nuclear reactor fuel, is in thissoil? [Hint : See Appendix B for the percent abundanceof235.]92 U6. A typical nuclear fission power plant producesabout 1.00 GW of electrical power. Assume thatthe plant has an overall efficiency of 40.0% and9.of23592U.]An all-electric home uses approximately2 000 kWh of electric energy per month.How much uranium-235 would be required to providethis house with its energy needs for 1 year? (Assume100% conversion efficiency and 208 MeV releasedper fission.)Section 30.3 Nuclear Fusion10. Find the energy released in the fusion reaction1121H H : He 11. When a star has exhausted its hydrogen fuel, it mayfuse other nuclear fuels. At temperatures above1.0 10 8 K, helium fusion can occur. Write theequations for the following processes: (a) Two alpha32

Problems 1005particles fuse to produce a nucleus A and a gammaray. What is nucleus A? (b) Nucleus A absorbs an alphaparticle to produce a nucleus B and a gammaray. What is nucleus B? (c) Find the total energy releasedin the reactions given in (a) and (b). [Note :8The mass of Be 8.005 305 u.]412. Another series of nuclear reactions that can produceenergy in the interior of stars is the cycle describedbelow. This cycle is most efficient when thecentral temperature in a star is above 1.6 10 7 K.Because the temperature at the center of the Sun isonly 1.5 10 7 K, the following cycle produces lessthan 10% of the Sun’s energy. (a) A high-energyproton is absorbed by 12 C. Another nucleus, A, isproduced in the reaction, along with a gamma ray.Identify nucleus A. (b) Nucleus A decays throughpositron emission to form nucleus B. Identify nucleusB. (c) Nucleus B absorbs a proton to produce nucleusC and a gamma ray. Identify nucleus C.(d) Nucleus C absorbs a proton to produce nucleusD and a gamma ray. Identify nucleus D. (e) NucleusD decays through positron emission to produce nucleusE. Identify nucleus E. (f) Nucleus E absorbs aproton to produce nucleus F plus an alpha particle.What is nucleus F ? [Note : If nucleus F is not 12 C—that is, the nucleus you started with—you havemade an error and should review the sequence ofevents.]13. If an all-electric home uses approximately2 000 kWh of electric energy per month, how manyfusion events described by the reaction21 H 3 1 H : 4 2 He 1 0 n would be required to keepthis home running for one year?14. To understand why plasma containment is necessary,consider the rate at which an unconfinedplasma would be lost. (a) Estimate the rms speed ofdeuterons in a plasma at 4.00 10 8 K. (b) Estimatethe order of magnitude of the time such a plasmawould remain in a 10-cm cube if no steps weretaken to contain it.15. The oceans have a volume of 317 million cubic milesand contain 1.32 10 21 kg of water. Of all the hydrogennuclei in this water, 0.030 0% of the mass isdeuterium. (a) If all of these deuterium nucleiwere fused to helium via the first reaction in Equation30.4, determine the total amount of energythat could be released. (b) The present world electricpower consumption is about 7.00 10 12 W.If consumption were 100 times greater, howmany years would the energy supply calculated inpart (a) last?Section 30.6 Positrons and OtherAntiparticles16. Two photons are produced when a proton and anantiproton annihilate each other. What is the minimumfrequency and corresponding wavelength ofeach photon?17.A photon produces a proton–antiproton pair according to the reaction. What is the minimum possible frequencyof the photon? What is its wavelength?18. A photon with an energy of 2.09 GeV creates aproton–antiproton pair in which the proton has akinetic energy of 95.0 MeV. What is the kinetic energyof the antiproton?Section 30.7 Mesons and the Beginning of ParticlePhysics19. When a high-energy proton or pion traveling nearthe speed of light collides with a nucleus, it travelsan average distance of 3.0 10 15 m before interactingwith another particle. From this information,estimate the time for the strong interaction tooccur.20. : p pCalculate the order of magnitude of the range of theforce that might be produced by the virtual exchangeof a proton.21. One of the mediators of the weak interaction is theZ 0 boson, which has a mass of 96 GeV/c 2 . Use thisinformation to find an approximate value for therange of the weak interaction.22. If a 0 at rest decays into two ’s, what is the energyof each of the ’s?Section 30.9 Conservation LawsSection 30.10 Strange Particles andStrangeness23. Each of the following reactions is forbidden. Determinea conservation law that is violated for eachreaction.(a) p p : e (b) p : p (c) p p : p (d) p p : p p n(e) p : n 0

1006 Chapter 30 Nuclear Energy and Elementary Particles24. For the following two reactions, the first may occurbut the second cannot. Explain.25.K 0 : (can occur) 0 : (cannot occur)Identify the unknown particle onthe left side of the reaction? p : n 26. Determine the type of neutrino or antineutrino involvedin each of the following processes:(a) : 0 e ?(b) ? p : p (c) 0 : p ?(d) : ? ?27. The following reactions or decays involve one ormore neutrinos. Supply the missing neutrinos.(a) : ?(b) K : ?(c) ? p : n e (d) ? n : p e (e) ? n : p (f) : e ? ?28. Determine which of the reactions below can occur.For those that cannot occur, determine the conservationlaw (or laws) that each violates:(a) p : 0(b) p p : p p 0(c) p p : p (d) : (e) n : p e e(f) : n29. Which of the following processes are allowed by thestrong interaction, the electromagnetic interaction,the weak interaction, or no interaction at all?(a) p : 2 0(b) K n : 0 (c) K : 0(d) : 0(e) 0 : 230.A K 0 particle at rest decays into a and a . Whatwill be the speed of each of the pions? The mass ofthe K 0 is 497.7 MeV/c 2 and the mass of each pion is139.6 MeV/c 2 .31. Determine whether or not strangeness is conservedin the following decays and reactions:(a) 0 : p (b) p : 0 K 0(c) p p : 0 0(d) p : (e) : 0 (f) 0 : p 32. Fill in the missing particle. Assume that (a) occursvia the strong interaction while (b) and (c) involvethe weak interaction.(a) K p : ____ p(b) : ____ (c) K : ____ 33. Identify the conserved quantities in the followingprocesses:(a) : 0 (b) K 0 : 2 0(c) K p : 0 n(d) 0 : 0 (e) e e : (f) p n : 0 Section 30.12 QuarksSection 30.13 Colored Quarks34. The quark composition of the proton is uud, whilethat of the neutron in udd. Show that the charge,baryon number, and strangeness of these particlesequal the sums of these numbers for their quarkconstituents.35.Find the number of electrons, and of each speciesof quark, in 1 L of water.36. The quark compositions of the K 0 and 0 particlesare ds and uds, respectively. Show that the charge,baryon number, and strangeness of these particlesequal the sums of these numbers for the quark constituents.37. Identify the particles corresponding to the followingquark states: (a) suu; (b) ud; (c) sd; (d) ssd.38. What is the electrical charge of the baryons with thequark compositions (a) u u d and (b) ud d? What arethese baryons called?39. Analyze the first three of the following reactions atthe quark level, and show that each conserves thenet number of each type of quark; then, in the lastreaction, identify the mystery particle:(a) p : K 0 0(b) p : K (c) K p : K K 0 (d) p p : K 0 p ?

Problems 100740. Assume binding energies can be neglected. Find themasses of the u and d quarks from the masses of theproton and neutron.ADDITIONAL PROBLEMS41. A 0 particle traveling through matter strikes a protonand a , and a gamma ray, as well as a thirdparticle, emerges. Use the quark model of each todetermine the identity of the third particle.42. It was stated in the text that the reactionoccurs with high probability,whereas the reactionnever occurs.Analyze these reactions at the quark level andshow that the first conserves the net number of eachtype of quark while the second does not. p : K 0 043. Two protons approach each other with equal andopposite velocities. Find the minimum kinetic energyof each of the protons if they are to produce a meson at rest in the reactionp p : p n 44. Name at least one conservation law that preventseach of the following reactions from occurring:(a) p : 0(b) : e(c) p : Find the energy released in the fu-45.sion reaction p : K 0 nH 3 2He : 4 2He e 46. Occasionally, high-energy muons collide with electronsand produce two neutrinos according to thereaction e : 2. What kind of neutrinos arethese?47. Each of the following decays is forbidden. For eachprocess, determine a conservation law that is violated:(a) : e (b) n : p e e(c) 0 : p 0(d) p : e 0(e) 0 : n 048. Two protons approach each other with 70.4 MeV ofkinetic energy and engage in a reaction in which aproton and a positive pion emerge at rest. Whatthird particle, obviously uncharged and thereforedifficult to detect, must have been created?49. The atomic bomb dropped on Hiroshima on August6, 1945, released 5 10 13 J of energy (equivalent tothat from 12 000 tons of TNT). Estimate (a) thenumber of23592U nuclei fissioned and (b) the mass ofthis235U.50. A 0 particle at rest decays according to 0 : 0 . Find the gamma-ray energy. [Hint: remember toconserve momentum.]51. If baryon number is not conserved, then one possiblemechanism by which a proton can decay isp : e Show that this reaction violates the conservation ofbaryon number. (b) Assuming that the reaction occursand that the proton is initially at rest, determinethe energy and momentum of the photon afterthe reaction. [Hint: recall that energy andmomentum must be conserved in the reaction.](c) Determine the speed of the positron after thereaction.52. Classical general relativity views the space–timemanifold as a deterministic structure completelywell defined down to arbitrarily small distances.On the other hand, quantum general relativity forbidsdistances smaller than the Planck lengthL (G/c 3 ) 1/2 . (a) Calculate the value of L. Theanswer suggests that, after the Big Bang (when allthe known Universe was reduced to a singularity),nothing could be observed until that singularitygrew larger than the Planck length, L. Since the sizeof the singularity grew at the speed of light, we caninfer that during the time it took for light to travelthe Planck length, no observations were possible.(b) Determine this time (known as the Planck timeT), and compare it to the ultra-hot epoch discussedin the text. (c) Does your answer to part (b) suggestthat we may never know what happened between thetime t 0 and the time t T ?53.92(a) Show that about 1.0 10 10 J would be releasedby the fusion of the deuterons in 1.0 gal of water.Note that 1 out of every 6 500 hydrogen atoms is adeuteron. (b) The average energy consumption rateof a person living in the United States is about 1.0 10 4 J/s (an average power of 10 kW). At this rate,how long would the energy needs of one person besupplied by the fusion of the deuterons in 1.0 gal ofwater? Assume that the energy released perdeuteron is 1.64 MeV.54. Calculate the mass of 235 U required to provide thetotal energy requirements of a nuclear submarine

1008 Chapter 30 Nuclear Energy and Elementary Particlesduring a 100-day patrol, assuming a constant powerdemand of 100 000 kW, a conversion efficiency of30%, and an average energy released per fission of208 MeV.55. A 2.0-MeV neutron is emitted in a fission reactor. Ifit loses one-half of its kinetic energy in each collisionwith a moderator atom, how many collisionsmust it undergo in order to achieve thermal energy(0.039 eV)?© 2005 Sidney Harris

A.1 MATHEMATICAL NOTATIONMany mathematical symbols are used throughout this book. You are no doubtfamiliar with some, such as the symbol to denote the equality of two quantities.The symbol denotes a proportionality. For example, y x 2 means that y isproportional to the square of x.The symbol means is less than, and means is greater than. For example,x y means x is greater than y.The symbol means is much less than, and means is much greater than.The symbol indicates that two quantities are approximately equal to each other.The symbol means is defined as. This is a stronger statement than a simple .It is convenient to use the notation x (read as “delta x”) to indicate the changein the quantity x. (Note that x does not mean “the product of and x.”) For example,suppose that a person out for a morning stroll starts measuring her distanceaway from home when she is 10 m from her doorway. She then moves along astraight-line path and stops strolling 50 m from the door. Her change in positionduring the walk is x 50 m 10 m 40 m or, in symbolic form,x x f x iIn this equation x f is the final position and x i is the initial position.We often have occasion to add several quantities. A useful abbreviation for representingsuch a sum is the Greek letter (capital sigma). Suppose we wish to adda set of five numbers represented by x 1 , x 2 , x 3 , x 4 , and x 5 . In the abbreviated notation,we would write the sum aswhere the subscript i on x represents any one of the numbers in the set. For example,if there are five masses in a system, m 1 , m 2 , m 3 , m 4 , and m 5 , the total mass ofthe system M m 1 m 2 m 3 m 4 m 5 could be expressed asFinally, the magnitude of a quantity x, written x , is simply the absolute value ofthat quantity. The sign of x is always positive, regardless of the sign of x. Forexample, if x 5, x 5; if x 8, x 8.A.2 SCIENTIFIC NOTATIONMany quantities that scientists deal with often have very large or very small values.For example, the speed of light is about 300 000 000 m/s and the ink required tomake the dot over an i in this textbook has a mass of about 0.000 000 001 kg. Obviously,it is cumbersome to read, write, and keep track of numbers such as these.We avoid this problem by using a method dealing with powers of the number 10:10 1 10x 1 x 2 x 3 x 4 x 5 5M 5m ii 110 0 110 2 10 10 10010 3 10 10 10 1 000x ii 110 4 10 10 10 10 10 00010 5 10 10 10 10 10 100 000APPENDIX AMathematical ReviewA.1

A.2 Appendix A Mathematical Reviewand so on. The number of zeros corresponds to the power to which 10 is raised,called the exponent of 10. For example, the speed of light, 300 000 000 m/s, canbe expressed as 3 10 8 m/s.For numbers less than one, we note the following:10 1 1 10 0.110 2 10 3 10 4 10 5 110 10 0.01110 10 10 0.001110 10 10 10 0.000 1110 10 10 10 10In these cases, the number of places the decimal point is to the left of the digit 1equals the value of the (negative) exponent. Numbers that are expressed as somepower of 10 multiplied by another number between 1 and 10 are said to be inscientific notation. For example, the scientific notation for 5 943 000 000 is5.943 10 9 and that for 0.000 083 2 is 8.32 10 5 .When numbers expressed in scientific notation are being multiplied, the followinggeneral rule is very useful:10 n 10 m 10 nm [A.1]where n and m can be any numbers (not necessarily integers). For example,10 2 10 5 10 7 . The rule also applies if one of the exponents is negative. For example,10 3 10 8 10 5 .When dividing numbers expressed in scientific notation, note that10 n10 m 10n 10 m 10 nm 0.000 01EXERCISESWith help from the above rules, verify the answers to the following:1. 86 400 8.64 10 42. 9 816 762.5 9.816 762 5 10 63. 0.000 000 039 8 3.98 10 84. (4.0 10 8 )(9.0 10 9 ) 3.6 10 185. (3.0 10 7 )(6.0 10 12 ) 1.8 10 475 10 116.5.0 103 1.5 107(3 10 6 )(8 10 2 )7.(2 10 17 )(6 10 5 ) 2 1018[A.2]A.3 ALGEBRAA. Some Basic RulesWhen algebraic operations are performed, the laws of arithmetic apply. Symbolssuch as x, y, and z are frequently used to represent quantities that are not specified,what are called the unknowns.First, consider the equation8x 32If we wish to solve for x, we can divide (or multiply) each side of the equation bythe same factor without destroying the equality. In this case, if we divide both sides

A.3 Algebra A.3by 8, we have8x8 32 8x 4Next consider the equationx 2 8In this type of expression, we can add or subtract the same quantity from eachside. If we subtract 2 from each side, we getIn general, if x a b, then x b a.Now consider the equationx 2 2 8 2x 6x5 9If we multiply each side by 5, we are left with x on the left by itself and 45 on theright: x 5 (5) 9 5x 45In all cases, whatever operation is performed on the left side of the equality mustalso be performed on the right side.The following rules for multiplying, dividing, adding, and subtracting fractionsshould be recalled, where a, b, and c are three numbers:RuleExampleMultiplyingDividingAdding a b c d acbd(a/b)(c/d) adbcab c dad bcbd 2 3 4 5 8152/34/5 (2)(5)(4)(3) 101223 4 5(2)(5) (4)(3)(3)(5) 215EXERCISESIn the following exercises, solve for x:ANSWERS1.2.3.4.a 11 x3x 5 13ax 5 bx 252x 6 34x 8x 1 aax 6x 7a bx 11 7B. PowersWhen powers of a given quantity x are multiplied, the following rule applies:x n x m x nmFor example, x 2 x 4 x 24 x 6 .[A.3]

A.4 Appendix A Mathematical ReviewTABLE A.1Rules of Exponentsx 0 1x 1 xx n x m x nmx n /x m x nmx 1/n n √x(x n ) m x nmWhen dividing the powers of a given quantity, note thatFor example, x 8 /x 2 x 8 2 x 6 .A power that is a fraction, such asx nx m xnm13 ,corresponds to a root as follows:[A.4]x 1/n n √x[A.5]For example, 4 1/3 √ 3 4 1.587 4. (A scientific calculator is useful for suchcalculations.)Finally, any quantity x n that is raised to the mth power is(x n ) m x nm [A.6]Table A.1 summarizes the rules of exponents.EXERCISESVerify the following:1. 3 2 3 3 2432. x 5 x 8 x 33. x 10 /x 5 x 154. 5 1/3 1.709 975 (Use your calculator.)5. 60 1/4 2.783 158 (Use your calculator.)6. (x 4 ) 3 x 12C. FactoringSome useful formulas for factoring an equation areax ay az a(x y z) common factora 2 2ab b 2 (a b) 2 perfect squarea 2 b 2 (a b)(a b)differences of squaresD. Quadratic EquationsThe general form of a quadratic equation isax 2 bx c 0[A.7]where x is the unknown quantity and a, b, and c are numerical factors referred toas coefficients of the equation. This equation has two roots, given byx b √b 2 4ac[A.8]2aIf b 2 4ac, the roots will be real.EXAMPLEThe equation x 2 5x 4 0 has the following roots corresponding to the two signs of the square root term:x 5 √52 (4)(1)(4)2(1)5 √92 5 32that is,x 5 321x 5 32where x refers to the root corresponding to the positive sign and x refers to the root corresponding to thenegative sign.4

A.3 Algebra A.5EXERCISESSolve the following quadratic equations:ANSWERS1.2.3.x 2 2x 3 02x 2 5x 2 02x 2 4x 9 0x 1x 2x 1 √22/2x 3x 1/2x 1 √22/2E. Linear EquationsA linear equation has the general formy ax b[A.9]where a and b are constants. This equation is referred to as being linear becausethe graph of y versus x is a straight line, as shown in Figure A.1. The constant b,called the intercept, represents the value of y at which the straight line intersectsthe y axis. The constant a is equal to the slope of the straight line. If any two pointson the straight line are specified by the coordinates (x 1 , y 1 ) and (x 2 , y 2 ), as in FigureA.1, then the slope of the straight line can be expressedSlope y 2 y 1 y[A.10]x 2 x 1 xNote that a and b can have either positive or negative values. If a 0, thestraight line has a positive slope, as in Figure A.1. If a 0, the straight line has anegative slope. In Figure A.1, both a and b are positive. Three other possible situationsare shown in Figure A.2: a 0, b 0; a 0, b 0; and a 0, b 0.EXERCISES1. Draw graphs of the following straight lines:(a) y 5x 3 (b) y 2x 4 (c) y 3x 62. Find the slopes of the straight lines described in Exercise 1.Answers: (a) 5 (b) 2 (c) 33. Find the slopes of the straight lines that pass through the following sets ofpoints: (a) (0, 4) and (4, 2), (b) (0, 0) and (2, 5), and (c) ( 5, 2)and (4, 2)3Answers: (a) (b) (c)2 5 2 4 9y(0, b)Figure A.1Figure A.2(x 1 , y 1 )u∆x(x 2 , y 2 )∆yu(0, 0) xy(3)(1)(2)a < 0b < 0a > 0b < 0a < 0b > 0xF. Solving Simultaneous Linear EquationsConsider an equation such as 3x 5y 15, which has two unknowns, x and y.Such an equation does not have a unique solution. That is, (x 0, y 3),9(x 5, y 0) and (x 2, y ) are all solutions to this equation.5If a problem has two unknowns, a unique solution is possible only if we have twoindependent equations. In general, if a problem has n unknowns, its solution requiresn independent equations. In order to solve two simultaneous equations involvingtwo unknowns, x and y, we solve one of the equations for x in terms of yand substitute this expression into the other equation.EXAMPLESolve the following two simultaneous equations:(1) 5x y 8 (2)2x 2y 4Solution From (2), we find that x y 2. Substitution of this into (1) gives

A.6 Appendix A Mathematical Review5(y 2) y 86y 18y 3x y 2 1Alternate Solution Multiply each term in (1) by the factor 2 and add the result to (2):10x 2y 162x 2y 412x 12x 1y x 2 3y543x – 2y = –1(5, 3)211 2 3 4 5 6 xx – y = 2Figure A.3Two linear equations with two unknowns can also be solved by a graphicalmethod. If the straight lines corresponding to the two equations are plotted in aconventional coordinate system, the intersection of the two lines represents the solution.For example, consider the two equationsx y 2x 2y 1These are plotted in Figure A.3. The intersection of the two lines has the coordinatesx 5, y 3. This represents the solution to the equations. You should checkthis solution by the analytical technique discussed above.EXERCISESSolve the following pairs of simultaneous equations involving two unknowns:1.2.3.x y 8x y 298 T 10aT 49 5a6x 2y 68x 4y 28ANSWERSx 5, y 3T 65, a 3.27x 2, y 3G. LogarithmsSuppose that a quantity x is expressed as a power of some quantity a:x a y[A.11]The number a is called the base number. The logarithm of x with respect to thebase a is equal to the exponent to which the base must be raised in order to satisfythe expression x a y :y log a x[A.12]Conversely, the antilogarithm of y is the number x:x antilog a y[A.13]In practice, the two bases most often used are base 10, called the common logarithmbase, and base e 2.718 ..., called the natural logarithm base. Whencommon logarithms are used,y log 10 x (or x 10 y )[A.14]

A.5 Trigonometry A.7When natural logarithms are used,y ln e x (or x e y )[A.15]For example, log 10 52 1.716, so that antilog 10 1.716 10 1.716 52. Likewise,ln e 52 3.951, so antiln e 3.951 e 3.951 52.In general, note that you can convert between base 10 and base e with theequalityln e x (2.302 585)log 10 x[A.16]Finally, some useful properties of logarithms arelog (ab) log a log blog (a/b) log a log blog (an ) n log aln e 1ln e a aln 1a ln aA.4 GEOMETRYTable A.2 gives the areas and volumes for several geometric shapes used throughoutthis text:TABLE A.2Useful Information for GeometrywArea = wrSurface area = 4pr 2Volume = 4 3 pr 3RectangleSphererArea = pr 2Circumference = 2prrSurface area = 2pr 2 + 2prVolume = pr 2 CircleCylinderbTriangleh1 Area = bh 2hRectangular boxwSurface area =2(h + w + hw)Volume = whA.5 TRIGONOMETRYSome of the most basic facts concerning trigonometry are presented in Chapter 1,and we encourage you to study the material presented there if you are having troublewith this branch of mathematics. In addition to the discussion of Chapter 1,certain useful trig identities that can be of value to you follow.sin 2 cos 2 1sin cos(90 )cos sin(90 )sin 2 2 sin cos cos 2 cos 2 sin 2 sin( ) sin cos cos sin cos( ) cos cos sin sin

A.8 Appendix A Mathematical ReviewThe following relationships apply to any triangle, as shown in Figure A.4:g 180°abLaw of cosinesa 2 b 2 c 2 2bc cos b 2 a 2 c 2 2ac cos c 2 a 2 b 2 2ab cos bFigure A.4caLaw of sinesasin bsin csin

APPENDIX BAn Abbreviated Table of IsotopesMassNumberAtomic Chemical (* Indicates Half-LifeNumber Atomic Radioactive) Atomic Percent (If Radioactive)Z Element Symbol Mass (u) A Mass (u) Abundance T 1/20 (Neutron) n 1* 1.008 665 10.4 min1 Hydrogen H 1.007 94 1 1.007 825 99.988 5Deuterium D 2 2.014 102 0.011 5Tritium T 3* 3.016 049 12.33 yr2 Helium He 4.002 602 3 3.016 029 0.000 1374 4.002 603 99.999 8633 Lithium Li 6.941 6 6.015 122 7.57 7.016 004 92.54 Beryllium Be 9.012 182 7* 7.016 929 53.3 days9 9.012 182 1005 Boron B 10.811 10 10.012 937 19.911 11.009 306 80.16 Carbon C 12.010 7 10* 10.016 853 19.3 s11* 11.011 434 20.4 min12 12.000 000 98.9313 13.003 355 1.0714* 14.003 242 5 730 yr7 Nitrogen N 14.006 7 13* 13.005 739 9.96 min14 14.003 074 99.63215 15.000 109 0.3688 Oxygen O 15.999 4 15* 15.003 065 122 s16 15.994 915 99.75718 17.999 160 0.2059 Fluorine F 18.998 403 2 19 18.998 403 10010 Neon Ne 20.179 7 20 19.992 440 90.4822 21.991 385 9.2511 Sodium Na 22.989 77 22* 21.994 437 2.61 yr23 22.989 770 10024* 23.990 963 14.96 h12 Magnesium Mg 24.305 0 24 23.985 042 78.9925 24.985 837 10.0026 25.982 593 11.0113 Aluminum Al 26.981 538 27 26.981 539 10014 Silicon Si 28.085 5 28 27.976 926 92.229 715 Phosphorus P 30.973 761 31 30.973 762 10032* 31.973 907 14.26 days16 Sulfur S 32.066 32 31.972 071 94.9335* 34.969 032 87.5 days17 Chlorine Cl 35.452 7 35 34.968 853 75.7837 36.965 903 24.2218 Argon Ar 39.948 40 39.962 383 99.600 319 Potassium K 39.098 3 39 38.963 707 93.258 140* 39.963 999 0.011 7 1.28 10 9 yr(Continued)A.9

A.10 Appendix B An Abbreviated Table of IsotopesMassNumberAtomic Chemical (* Indicates Half-LifeNumber Atomic Radioactive) Atomic Percent (If Radioactive)Z Element Symbol Mass (u) A Mass (u) Abundance T 1/220 Calcium Ca 40.078 40 39.962 591 96.94121 Scandium Sc 44.955 910 45 44.955 910 10022 Titanium Ti 47.867 48 47.947 947 73.7223 Vanadium V 50.941 5 51 50.943 964 99.75024 Chromium Cr 51.996 1 52 51.940 512 83.78925 Manganese Mn 54.938 049 55 54.938 050 10026 Iron Fe 55.845 56 55.934 942 91.75427 Cobalt Co 58.933 200 59 58.933 200 10060* 59.933 822 5.27 yr28 Nickel Ni 58.693 4 58 57.935 348 68.076 960 59.930 790 26.223 129 Copper Cu 63.546 63 62.929 601 69.1765 64.927 794 30.8330 Zinc Zn 65.39 64 63.929 147 48.6366 65.926 037 27.9068 67.924 848 18.7531 Gallium Ga 69.723 69 68.925 581 60.10871 70.924 705 39.89232 Germanium Ge 72.61 70 69.924 250 20.8472 71.922 076 27.5474 73.921 178 36.2833 Arsenic As 74.921 60 75 74.921 596 10034 Selenium Se 78.96 78 77.917 310 23.7780 79.916 522 49.6135 Bromine Br 79.904 79 78.918 338 50.6981 80.916 291 49.3136 Krypton Kr 83.80 82 81.913 485 11.5883 82.914 136 11.4984 83.911 507 57.0086 85.910 610 17.3037 Rubidium Rb 85.467 8 85 84.911 789 72.1787* 86.909 184 27.83 4.75 10 10 yr38 Strontium Sr 87.62 86 85.909 262 9.8688 87.905 614 82.5890* 89.907 738 29.1 yr39 Yttrium Y 88.905 85 89 88.905 848 10040 Zirconium Zr 91.224 90 89.904 704 51.4591 90.905 645 11.2292 91.905 040 17.1594 93.906 316 17.3841 Niobium Nb 92.906 38 93 92.906 378 10042 Molybdenum Mo 95.94 92 91.906 810 14.8495 94.905 842 15.9296 95.904 679 16.6898 97.905 408 24.1343 Technetium Tc 98* 97.907 216 4.2 10 6 yr99* 98.906 255 2.1 10 5 yr

An Abbreviated Table of Isotopes A.11MassNumberAtomic Chemical (* Indicates Half-LifeNumber Atomic Radioactive) Atomic Percent (If Radioactive)Z Element Symbol Mass (u) A Mass (u) Abundance T 1/244 Ruthenium Ru 101.07 99 98.905 939 12.76100 99.904 220 12.60101 100.905 582 17.06102 101.904 350 31.55104 103.905 430 18.6245 Rhodium Rh 102.905 50 103 102.905 504 10046 Palladium Pd 106.42 104 103.904 035 11.14105 104.905 084 22.33106 105.903 483 27.33108 107.903 894 26.46110 109.905 152 11.7247 Silver Ag 107.868 2 107 106.905 093 51.839109 108.904 756 48.16148 Cadmium Cd 112.411 110 109.903 006 12.49111 110.904 182 12.80112 111.902 757 24.13113* 112.904 401 12.22 9.3 10 15 yr114 113.903 358 28.7349 Indium In 114.818 115* 114.903 878 95.71 4.4 10 14 yr50 Tin Sn 118.710 116 115.901 744 14.54118 117.901 606 24.22120 119.902 197 32.5851 Antimony Sb 121.760 121 120.903 818 57.21123 122.904 216 42.7952 Tellurium Te 127.60 126 125.903 306 18.84128* 127.904 461 31.74 8 10 24 yr130* 129.906 223 34.08 1.25 10 21 yr53 Iodine I 126.904 47 127 126.904 468 100129* 128.904 988 1.6 10 7 yr54 Xenon Xe 131.29 129 128.904 780 26.44131 130.905 082 21.18132 131.904 145 26.89134 133.905 394 10.44136* 135.907 220 8.87 2.36 10 21 yr55 Cesium Cs 132.905 45 133 132.905 447 10056 Barium Ba 137.327 137 136.905 821 11.232138 137.905 241 71.69857 Lanthanum La 138.905 5 139 138.906 349 99.91058 Cerium Ce 140.116 140 139.905 434 88.450142* 141.909 240 11.114 5 10 16 yr59 Praseodymium Pr 140.907 65 141 140.907 648 10060 Neodymium Nd 144.24 142 141.907 719 27.2144* 143.910 083 23.8 2.3 10 15 yr146 145.913 112 17.261 Promethium Pm 145* 144.912 744 17.7 yr62 Samarium Sm 150.36 147* 146.914 893 14.99 1.06 10 11 yr149* 148.917 180 13.82 2 10 15 yr152 151.919 728 26.75154 153.922 205 22.75(Continued)

A.12 Appendix B An Abbreviated Table of IsotopesMassNumberAtomic Chemical (* Indicates Half-LifeNumber Atomic Radioactive) Atomic Percent (If Radioactive)Z Element Symbol Mass (u) A Mass (u) Abundance T 1/263 Europium Eu 151.964 151 150.919 846 47.81153 152.921 226 52.1964 Gadolinium Gd 157.25 156 155.922 120 20.47158 157.924 100 24.84160 159.927 051 21.8665 Terbium Tb 158.925 34 159 158.925 343 10066 Dysprosium Dy 162.50 162 161.926 796 25.51163 162.928 728 24.90164 163.929 171 28.1867 Holmium Ho 164.930 32 165 164.930 320 10068 Erbium Er 167.6 166 165.930 290 33.61167 166.932 045 22.93168 167.932 368 26.7869 Thulium Tm 168.934 21 169 168.934 211 10070 Ytterbium Yb 173.04 172 171.936 378 21.83173 172.938 207 16.13174 173.938 858 31.8371 Lutecium Lu 174.967 175 174.940 768 97.4172 Hafnium Hf 178.49 177 176.943 220 18.60178 177.943 698 27.28179 178.945 815 13.62180 179.946 549 35.0873 Tantalum Ta 180.947 9 181 180.947 996 99.98874 Tungsten W 183.84 182 181.948 206 26.50(Wolfram) 183 182.950 224 14.31184* 183.950 933 30.64 3 10 17 yr186 185.954 362 28.4375 Rhenium Re 186.207 185 184.952 956 37.40187* 186.955 751 62.60 4.4 10 10 yr76 Osmium Os 190.23 188 187.955 836 13.24189 188.958 145 16.15190 189.958 445 26.26192 191.961 479 40.7877 Iridium Ir 192.217 191 190.960 591 37.3193 192.962 924 62.778 Platinum Pt 195.078 194 193.962 664 32.967195 194.964 774 33.832196 195.964 935 25.24279 Gold Au 196.966 55 197 196.966 552 10080 Mercury Hg 200.59 199 198.968 262 16.87200 199.968 309 23.10201 200.970 285 13.18202 201.970 626 29.8681 Thallium Tl 204.383 3 203 202.972 329 29.524205 204.974 412 70.476(Th C) 208* 207.982 005 3.053 min(Ra C) 210* 209.990 066 1.30 min

An Abbreviated Table of Isotopes A.13MassNumberAtomic Chemical (* Indicates Half-LifeNumber Atomic Radioactive) Atomic Percent (If Radioactive)Z Element Symbol Mass (u) A Mass (u) Abundance T 1/282 Lead Pb 207.2 204* 203.973 029 1.4 1.4 10 17 yr206 205.974 449 24.1207 206.975 881 22.1208 207.976 636 52.4(Ra D) 210* 209.984 173 22.3 yr(Ac B) 211* 210.988 732 36.1 min(Th B) 212* 211.991 888 10.64 h(Ra B) 214* 213.999 798 26.8 min83 Bismuth Bi 208.980 38 209 208.980 383 100(Th C) 211* 210.987 258 2.14 min84 Polonium Po(Ra F) 210* 209.982 857 138.38 days(Ra C) 214* 213.995 186 164 s85 Astatine At 218* 218.008 682 1.6 s86 Radon Rn 222* 222.017 570 3.823 days87 Francium Fr(Ac K) 223* 223.019 731 22 min88 Radium Ra 226* 226.025 403 1 600 yr(Ms Th 1 ) 228* 228.031 064 5.75 yr89 Actinium Ac 227* 227.027 747 21.77 yr90 Thorium Th 232.038 1(Rd Th) 228* 228.028 731 1.913 yr(Th) 232* 232.038 050 100 1.40 10 10 yr91 Protactinium Pa 231.035 88 231* 231.035 879 32.760 yr92 Uranium U 238.028 9 232* 232.037 146 69 yr233* 233.039 628 1.59 10 5 yr(Ac U) 235* 235.043 923 0.720 0 7.04 10 8 yr236* 236.045 562 2.34 10 7 yr(UI) 238* 238.050 783 99.274 5 4.47 10 9 yr93 Neptunium Np 237* 237.048 167 2.14 10 6 yr94 Plutonium Pu 239* 239.052 156 2.412 10 4 yr242* 242.058 737 3.73 10 6 yr244* 244.064 198 8.1 10 7 yra Chemical atomic masses are from T. B. Coplen, “Atomic Weights of the Elements 1999,” a technical report to the International Union of Pure and Applied Chemistry,and published in Pure and Applied Chemistry, 73(4), 667–683, 2001. Atomic masses of the isotopes are from G. Audi and A. H. Wapstra, “The 1995 Update to the AtomicMass Evaluation,” Nuclear Physics, A595, vol. 4, 409–480, December 25, 1995. Percent abundance values are from K. J. R. Rosman and P. D. P. Taylor, “Isotopic Compositionsof the Elements 1999”, a technical report to the International Union of Pure and Applied Chemistry, and published in Pure and Applied Chemistry, 70(1), 217–236,1998.

APPENDIX CSome Useful TablesTABLE C.1Mathematical Symbols Used in the Text and Their MeaningSymbolxx ixMeaningis equal tois not equal tois defined asis proportional tois greater thanis less thanis much greater thanis much less thanis approximately equal tois on the order of magnitude ofchange in x or uncertainty in xsum of all quantities x iabsolute value of x (always a positive quantity)TABLE C.2Standard Symbols for UnitsSymbol Unit Symbol UnitA ampere kcal kilocalorieÅ angstrom kg kilogramatm atmosphere km kilometerBq bequerel kmol kilomoleBtu British thermal unit L literC coulomb lb pound°C degree Celsius ly light yearcal calorie m metercm centimeter min minuteCi curie mol moled day N newtondeg degree (angle) nm nanometereV electronvolt Pa pascal°F degree Fahrenheit rad radianF farad rev revolutionft foot s secondG Gauss T teslag gram u atomic mass unitH henry V volth hour W watthp horsepower Wb weberHz hertz yr yearin. inch m micrometerJ joule ohmK kelvinA.14

Some Useful Tables A.15TABLE C.3The Greek AlphabetAlpha Nu Beta Xi Gamma Omicron Delta Pi Epsilon Rho Zeta Sigma Eta Tau Theta Upsilon Iota Phi Kappa Chi Lambda Psi Mu Omega TABLE C.4Physical Data Often Used aAverage Earth-Moon distance 3.84 10 8 mAverage Earth-Sun distance1.496 10 11 mEquatorial radius of Earth6.38 10 6 mDensity of air (20°C and 1 atm) 1.20 kg/m 3Density of water (20°C and 1 atm) 1.00 10 3 kg/m 3Free-fall acceleration 9.80 m/s 2Mass of Earth5.98 10 24 kgMass of Moon7.36 10 22 kgMass of Sun1.99 10 30 kgStandard atmospheric pressure 1.013 10 5 Paa These are the values of the constants as used in the text.TABLE C.5Some Fundamental Constants aQuantity Symbol Value bAtomic mass unit u 1.660 540 2(10) 10 27 kg931.494 32(28) MeV/c 2Avogadro’s number N A 6.022 136 7(36) 10 23 (mol) 1Bohr radiusa 0.529 177 249(24) 10 10 0 mm e e 2 k eBoltzmann’s constant k B R/N A 1.380 658(12) 10 23 J/KCompton wavelengthm e c2.426 310 58(22) 10 12 mCoulomb constantk e 1408.987 551 787 10 9 N m 2 /C 2 (exact)Electron mass m e 9.109 389 7(54) 10 31 kg5.485 799 03(13) 10 4 u0.510 999 06(15) MeV/c 2Electron volt eV 1.602 177 33(49) 10 19 JElementary charge e 1.602 177 33(49) 10 19 CGas constant R 8.314 510(70) J/K molGravitational constant G 6.672 59(85) 10 11 N m 2 /kg 2Hydrogen ionization energy C h2E 1 m ee 4 k2 e2 2 e 2 k e2a 013.605 698(40) eVNeutron mass m n 1.674 928 6(10) 10 27 kg1.008 664 904(14) u939.565 63(28) MeV/c 2Permeability of free space 0 4 10 7 T m/A (exact)Permittivity of free space 0 1/ 0 c 2 8.854 187 817 10 12 C 2 /N m 2 (exact)Planck’s constant h 6.626 075(40) 10 34 J sh/21.054 572 66(63) 10 34 J sProton mass m p 1.672 623(10) 10 27 kg1.007 276 470(12) u938.272 3(28) MeV/c 2Rydberg constant R H 1.097 373 153 4(13) 10 7 m 1Speed of light in vacuum c 2.997 924 58 10 8 m/s (exact)a These constants are the values recommended in 1986 by CODATA, based on a least-squares adjustment of data from differentmeasurements. For a more complete list, see Cohen, E. Richard, and Barry N. Taylor, Rev. Mod. Phys. 59:1121, 1987.b The numbers in parentheses for the values below represent the uncertainties in the last two digits.

APPENDIX DSI UnitsTABLE D.1SI Base UnitsSI Base UnitBase Quantity Name SymbolLength meter mMass kilogram kgTime second sElectric current ampere ATemperature kelvin KAmount of substance mole molLuminous intensity candela cdTABLE D.2Derived SI UnitsExpression in Expression inTerms of Base Terms ofQuantity Name Symbol Units Other SI UnitsPlane angle radian rad m/mFrequency hertz Hz s 1Force newton N kg m/s 2 J/mPressure pascal Pa kg/m s 2 N/m 2Energy: work joule J kg m 2 /s 2 N mPower watt W kg m 2 /s 3 J/sElectric charge coulomb C A sElectric potential (emf) volt V kg m 2 /A s 3 W/A, J/CCapacitance farad F A 2 s 4 /kg m 2 C/VElectric resistance ohm kg m 2 /A 2 s 3 V/AMagnetic flux weber Wb kg m 2 /A s 2 V s, T m 2Magnetic field intensity tesla T kg/A s 2 Wb/m 2Inductance henry H kg m 2 /A 2 s 2 Wb/AA.16

Answers to Quick Quizzes,Odd-Numbered Conceptual Questionsand ProblemsChapter 15QUICK QUIZZES1. (b)2. (b)3. (c)4. (a)5. (c) and (d)6. (a)7. (c)8. (b)9. (d)10. (b) and (d)CONCEPTUAL QUESTIONS1. Electrons have been removed from the object.3. The configuration shown is inherently unstable. The negativecharges repel each other. If there is any slight rotationof one of the rods, the repulsion can result in furtherrotation away from this configuration. There are threeconceivable final configurations shown below. Configuration(a) is stable: If the positive upper ends are pushedtowards each other, their mutual repulsion will move thesystem back to the original configuration. Configuration(b) is an equilibrium configuration, but it is unstable: Ifthe lower ends are moved towards each other, theirmutual attraction will be larger than that of the upperends, and the configuration will shift to (c), another possiblestable configuration.+–(a)+–+ –(c)+–(b)+ –Figure Q15.35. Move an object A with a net positive charge so it is near,but not touching, a neutral metallic object B that is insulatedfrom the ground. The presence of A will polarize B,causing an excess negative charge to exist on the sidenearest A and an excess positive charge of equal magnitudeto exist on the side farthest from A. While A is stillnear B, touch B with your hand. Additional electrons willthen flow from ground, through your body and onto B.With A continuing to be near but not in contact with B,remove your hand from B, thus trapping the excess electronson B. When A is now removed, B is left with excesselectrons, or a net negative charge. By means of mutualrepulsion, this negative charge will now spread uniformlyover the entire surface of B.–+7. An object’s mass decreases very slightly (immeasurably)when it is given a positive charge, because it loses electrons.When the object is given a negative charge, its massincreases slightly because it gains electrons.9. Electric field lines start on positive charges and end onnegative charges. Thus, if the fair-weather field is directedinto the ground, the ground must have a negative charge.11. The two charged plates create a region with a uniformelectric field between them, directed from the positive towardthe negative plate. Once the ball is disturbed so as totouch one plate (say, the negative one), some negativecharge will be transferred to the ball and it will be actedupon by an electric force that will accelerate it to the positiveplate. Once the ball touches the positive plate, it willrelease its negative charge, acquire a positive charge, andaccelerate back to the negative plate. The ball will continueto move back and forth between the plates until ithas transferred all their net charge, thereby making bothplates neutral.13. The electric shielding effect of conductors depends onthe fact that there are two kinds of charge: positive andnegative. As a result, charges can move within the conductorso that the combination of positive and negativecharges establishes an electric field that exactly cancelsthe external field within the conductor and any cavitiesinside the conductor. There is only one type of gravitationcharge, however, because there is no negative mass. As aresult, gravitational shielding is not possible.15. The electric field patterns of each of these three configurationsdo not have sufficient symmetry to make the calculationspractical. Gauss’s law is useful only for calculatingthe electric fields of highly symmetric charge distributions,such as uniformly charged spheres, cylinders, andsheets.17. No, the wall is not positively charged. The balloon inducesa charge of opposite sign in the wall, causing theballoon and the wall to be attracted to each other. Theballoon eventually falls because its charge slowly diminishesafter leaking to ground. Some of the balloon’scharge could also be lost due to positive ions in the surroundingatmosphere, which would tend to neutralize thenegative charges on the balloon.19. When the comb is nearby, charges separate on the paper,and the paper is attracted to the comb. After contact,charges from the comb are transferred to the paper, sothat it has the same type of charge as the comb. The paperis thus repelled.21. The attraction between the ball and the object could bean attraction of unlike charges, or it could be an attractionbetween a charged object and a neutral object as a resultof polarization of the molecules of the neutral object.Two additional experiments could help us determinewhether the object is charged. First, a known neutral ballcould be brought near the object, and if there is anattraction, the object is negatively charged. Another possibilityis to bring a known negatively charged ball near theA.17

A.18 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problemsobject. In that case, if there is a repulsion, then the objectis negatively charged. If there is an attraction, then theobject is neutral.PROBLEMS1. 1.1 10 8 N (attractive)3. 91 N (repulsion)5. (a) 36.8 N (b) 5.54 10 27 m/s 27. 5.12 10 5 N9. (a) 2.2 10 5 N (attraction)(b) 9.0 10 7 N (repulsion)11. 1.38 10 5 N at 77.5° below the negative x -axis13. 0.872 N at 30.0° below the positive x -axis15. 7.2 nC17. 1.5 10 3 C19. 7.20 10 5 N/C (downward)21. 1.2 10 4 N/C23. (a) 6.12 10 10 m/s 2 (b) 19.6 s (c) 11.8 m(d) 1.20 10 15 J25. zero27. 1.8 m to the left of the 2.5-C charge33. (a) 0 (b) 5 C inside, 5 C outside (c) 0 inside, 5 C outside (d) 0 inside, 5 C outside35. 1.3 10 3 C37. (a) 4.8 10 15 N (b) 2.9 10 12 m/s 239. (a) 858 N m 2 /C (b) 0 (c) 657 N m 2 /C41. 4.1 10 6 N/C43. (a) 0 (b) k e q/r 2 outward47. 57.5 N49. 24 N/C in the positive x -direction51. (a) E 2k e qb (a 2 b 2 ) 3/2 in the positive x-direction(b) E k e Qb(a 2 b 2 ) 3/2 in the positive x-direction53. (a) 0 (b) 7.99 10 7 N/C (outward)(c) 0 (d) 7.34 10 6 N/C (outward)55. 3.55 10 5 N m 2 /C57. 4.4 10 5 N/C59. (a) 10.9 nC (b) 5.44 10 3 N61. 10 7 C63. (a) 1.00 10 3 N/C (b) 3.37 10 8 s (c) accelerate at1.76 10 14 m/s 2 in the direction opposite that of theelectric fieldChapter 16QUICK QUIZZES1. (b)2. (b), (d)3. (d)4. (c)5. (a)6. (c)7. (a) C decreases. (b) Q stays the same. (c) E staysthe same. (d) V increases. (e) The energy storedincreases.8. (a) C increases. (b) Q increases. (c) E stays the same.(d) V remains the same. (e) The energy stored increases.9. (a)CONCEPTUAL QUESTIONS1. (a) The proton moves in a straight line with constant accelerationin the direction of the electric field. (b) As itsvelocity increases, its kinetic energy increases and the electricpotential energy associated with the proton decreases.3. The work done in pulling the capacitor plates fartherapart is transferred into additional electric energy storedin the capacitor. The charge is constant and the capacitancedecreases, but the potential difference between theplates increases, which results in an increase in the storedelectric energy.5. If the power line makes electrical contact with the metalof the car, it will raise the potential of the car to 20 kV. Itwill also raise the potential of your body to 20 kV, becauseyou are in contact with the car. In itself, this is not a problem.If you step out of the car, however, your body at20 kV will make contact with the ground, which is at zerovolts. As a result, a current will pass through your bodyand you will likely be injured. Thus, it is best to stay in thecar until help arrives.7. If two points on a conducting object were at different potentials,then free charges in the object would move andwe would not have static conditions, in contradiction tothe initial assumption. (Free positive charges would migratefrom locations of higher to locations of lower potential.Free electrons would rapidly move from locations oflower to locations of higher potential.) All of the chargeswould continue to move until the potential became equaleverywhere in the conductor.9. The capacitor often remains charged long after the voltagesource is disconnected. This residual charge can belethal. The capacitor can be safely handled after dischargingthe plates by short-circuiting the device with a conductor,such as a screwdriver with an insulating handle.11. Field lines represent the direction of the electric force ona positive test charge. If electric field lines were to cross,then, at the point of crossing, there would be an ambiguityregarding the direction of the force on the test charge,because there would be two possible forces there. Thus,electric field lines cannot cross. It is possible for equipotentialsurfaces to cross. (However, equipotential surfacesat different potentials cannot intersect.) For example,suppose two identical positive charges are at diagonallyopposite corners of a square and two negative charges ofequal magnitude are at the other two corners. Then theplanes perpendicular to the sides of the square at theirmidpoints are equipotential surfaces. These two planescross each other at the line perpendicular to the square atits center.13. You should use a dielectric-filled capacitor whose dielectricconstant is very large. Further, you should make thedielectric as thin as possible, keeping in mind that dielectricbreakdown must also be considered.15. (a) ii (b) i17. It would make no difference at all. An electron volt is thekinetic energy gained by an electron in being acceleratedthrough a potential difference of 1 V. A proton acceleratedthrough 1 V would have the same kinetic energy, becauseit carries the same charge as the electron (exceptfor the sign). The proton would be moving in the oppositedirection and more slowly after accelerating through1 V, due to its opposite charge and its larger mass, but itwould still gain 1 electron volt, or 1 proton volt, of kineticenergy.PROBLEMS1. (a) 6.40 10 19 J (b) 6.40 10 19 J (c) 4.00 V3. 1.4 10 20 J

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.195. 1.7 10 6 N/C7. (a) 1.13 10 5 N/C (b) 1.80 10 14 N(c) 4.38 10 17 J9. (a) 0.500 m (b) 0.250 m11. (a) 1.44 10 7 V (b) 7.19 10 8 V13. (a) 2.67 10 6 V (b) 2.13 10 6 V15. (a) 103 V (b) 3.85 10 7 J; positive work must be doneto separate the charges.17. 11.0 kV19. 2.74 10 14 m21. 0.719 m, 1.44 m, 2.88 m. No. The equipotentials are notuniformly spaced. Instead, the radius of an equipotenial isinversely proportional to the potential.23. (a) 1.1 10 8 F (b) 27 C25. (a) 11.1 kV/m toward the negative plate (b) 3.74 pF(c) 74.7 pC and 74.7 pC27. (a) 90.4 V (b) 9.04 10 4 V/m29. (a) 13.3 C on each (b) 20.0 C, 40.0 C31. (a) 2.00 F (b) Q 3 24.0 C, Q 4 16.0 C,Q 2 8.00 C, (V ) 2 (V ) 4 4.00 V, (V ) 3 8.00 V33. (a) 5.96 F (b) Q 20 89.5 C, Q 6 63.2 C,Q 3 Q 15 26.3 C35. Q 1 16.0 C, Q 5 80.0 C, Q 8 64.0 C,Q 4 32.0 C37. (a) Q 25 1.25 mC, Q 40 2.00 mC (b) Q 25 288 C,Q 40 462 C, V 11.5 V39. Q 1 3.33 C, Q 2 6.67 C41. 83.6 C43. 2.55 10 11 J45. 3.2 10 10 J47. 4.049. (a) 8.13 nF (b) 2.40 kV51. (a) volume 9.09 10 16 m 3 , area 4.54 10 10 m 2(b) 2.01 10 13 F (c) 2.01 10 14 C,1.26 10 5 electronic charges55. 4.29 F57. 6.25 F59. 4.47 kV61. 0.75 mC on C 1 , 0.25 mC on C 265. 50 NChapter 17QUICK QUIZZES1. (d)2. (b)3. (c), (d)4. (b)5. (b)6. (a)7. (b)8. (a)CONCEPTUAL QUESTIONS1. Charge. Because an ampere is a unit of current (1 A 1 C/s) and an hour is a unit of time (1 h 3 600 s), then1A h 3 600 C.3. The gravitational force pulling the electron to the bottomof a piece of metal is much smaller than the electrical repulsionpushing the electrons apart. Thus, free electronsstay distributed throughout the metal. The concept ofcharges residing on the surface of a metal is true for ametal with an excess charge. The number of free electronsin an electrically neutral piece of metal is the same as thenumber of positive ions—the metal has zero net charge.5. A voltage is not something that “surges through” a completedcircuit. A voltage is a potential difference that is appliedacross a device or a circuit. It would be more correctto say “1 ampere of electricity surged through the victim’sbody.” Although this amount of current would have disastrousresults on the human body, a value of 1 (ampere)doesn’t sound as exciting for a newspaper article as 10 000(volts). Another possibility is to write “10 000 volts of electricitywere applied across the victim’s body,” which stilldoesn’t sound quite as exciting.7. We would conclude that the conductor is nonohmic.9. The shape, dimensions, and the resistivity affect the resistanceof a conductor. Because temperature and impuritiesaffect the conductor’s resistivity, these factors also affectresistance.11. The radius of wire B is the square root of three times theradius of wire A. Therefore the cross-sectional area of Bthree times larger than that of A.13. The drift velocity might increase steadily as time goes on,because collisions between electrons and atoms in thewire would be essentially nonexistent and the conductionelectrons would move with constant acceleration. The currentwould rise steadily without bound also, because I isproportional to the drift velocity.15. Once the switch is closed, the line voltage is applied acrossthe bulb. As the voltage is applied across the cold filamentwhen it is first turned on, the resistance of the filament islow, the current is high, and a relatively large amount ofpower is delivered to the bulb. As the filament warms, itsresistance rises and the current decreases. As a result, thepower delivered to the bulb decreases. The large currentspike at the beginning of the bulb’s operation is the reasonthat lightbulbs often fail just after they are turned on.PROBLEMS1. 3.00 10 20 electrons move past in the direction oppositeto the current.3. 2.00 C5. 1.05 mA7. 27 yr9. (a) n is unaffected (b) v d is doubled11. 32 V is 200 times larger than 0.16 V13. 0.17 mm15. (a) 30 (b) 4.7 10 4 m17. silver ( 1.59 10 8 m)19. 256 21. 1.98 A23. 26 mA25. (a) 5.89 10 2 (b) 5.45 10 2 27. (a) 3.0 A (b) 2.9 A29. (a) 1.2 (b) 8.0 10 4 (a 0.080% increase)31. 5.00 A, 24.0 33. 18 bulbs35. 11.2 min37. 34.4 39. 1.6 cm41. 295 metric tons/h43. 26 cents45. 23 cents47. $1.2

A.20 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems49. 1.1 km51. 1.47 10 6 m; differs by 2.0% from value inTable 17.153. (a) $3.06 (b) No. The circuit must be able to handle atleast 26 A.55. (a) 667 A (b) 50.0 km57. 3.77 10 28 /m 359. (a) 144 (b) 26 m (c) To fit the required length into asmall space. (d) 25 m61. 37 M63. 0.48 kg/s65. (a) 2.6 10 5 (b) 76 kg67. (a) 470 W (b) 1.60 mm or more (c) 2.93 mm or moreChapter 18QUICK QUIZZES1. (a), (d)2. (b)3. (a)4. Parallel : (a) unchanged (b) unchanged (c) increase(d) decrease5. Series: (a) decrease (b) decrease (c) decrease(d) increase6. (c)CONCEPTUAL QUESTIONS1. No. When a battery serves as a source and supplies currentto a circuit, the conventional current flows throughthe battery from the negative terminal to the positive one.However, when a source having a larger emf than thebattery is used to charge the battery, the conventionalcurrent is forced to flow through the battery from the positiveterminal to the negative one.3. The total amount of energy delivered by the battery willbe less than W. Recall that a battery can be considered anideal, resistanceless battery in series with the internal resistance.When the battery is being charged, the energydelivered to it includes the energy necessary to charge theideal battery, plus the energy that goes into raising thetemperature of the battery due to I 2 r heating in the internalresistance. This latter energy is not available duringdischarge of the battery, when part of the reduced availableenergy again transforms into internal energy in theinternal resistance, further reducing the available energybelow W.5. The starter in the automobile draws a relatively large currentfrom the battery. This large current causes a significantvoltage drop across the internal resistance of the battery.As a result, the terminal voltage of the battery isreduced, and the headlights dim accordingly.7. An electrical appliance has a given resistance. Thus, whenit is attached to a power source with a known potential difference,a definite current will be drawn, and the devicecan therefore be labeled with both the voltage and thecurrent. Batteries, however, can be applied to a number ofdevices. Each device will have a different resistance, so thecurrent will vary with the device. As a result, only the voltageof the battery can be specified.9. Connecting batteries in parallel does not increase theemf. A high-current device connected to two batteriesin parallel can draw currents from both batteries. Thus,connecting the batteries in parallel increases the possiblecurrent output and, therefore, the possible power output.11. The lightbulb will glow for a very short while as the capacitoris being charged. Once the capacitor is almost totallycharged, the current in the circuit will be nearly zero andthe bulb will not glow.13. The bird is resting on a wire of fixed potential. In order tobe electrocuted, a large potential difference is requiredbetween the bird’s feet. The potential difference betweenthe bird’s feet is too small to harm the bird.15. The junction rule is a statement of conservation ofcharge. It says that the amount of charge that enters ajunction in some time interval must equal the charge thatleaves the junction in that time interval. The loop rule is astatement of conservation of energy. It says that the increasesand decreases in potential around a closed loop ina circuit must add to zero.17. A few of the factors involved are as follows: the conductivityof the string (is it wet or dry?); how well you are insulatedfrom ground (are you wearing thick rubber- orleather-soled shoes?); the magnitude of the potential differencebetween you and the kite; and the type and conditionof the soil under your feet.19. She will not be electrocuted if she holds onto only onehigh-voltage wire, because she is not completing a circuit.There is no potential difference across her body as long asshe clings to only one wire. However, she should releasethe wire immediately once it breaks, because she will becomepart of a closed circuit when she reaches theground or comes into contact with another object.21. (a) The intensity of each lamp increases because lamp C isshort circuited and there is current (which increases) onlyin lamps A and B. (b) The intensity of lamp C goes to zerobecause the current in this branch goes to zero. (c) Thecurrent in the circuit increases because the total resistancedecreases from 3R (with the switch open) to 2R (after theswitch is closed). (d) The voltage drop across lamps A andB increases, while the voltage drop across lamp C becomeszero. (e) The power dissipated increases from 2 /3R (withthe switch open) to 2 /2R (after the switch is closed).23. The statement is false. The current in each bulb is thesame, because they are connected in series. The bulb thatglows brightest has the larger resistance and hence dissipatesmore powerPROBLEMS1. 4.92 3. 73.8 W. Your circuit diagram will consist of two 0.800-resistors in series with the 192- resistance of the bulb.5. (a) 17.1 (b) 1.99 A for 4.00 and 9.00 , 1.17 A for7.00 , 0.818 A for 10.0 7. 2.5R9. (a) 0.227 A (b) 5.68 V11. 55 13. 0.43 A15. (a) Connect two 50- resistors in parallel, and then connectthis combination in series with a 20- resistor.(b) Connect two 50- resistors in parallel, connect two20- resistors in parallel, and then connect these twocombinations in series with each other.17. 0.846 A downwards in the 8.00- resistor; 0.462 Adownwards in the middle branch; 1.31 A upwards in theright-hand branch

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.2119. (a) 3.00 mA (b) 19.0 V (c) 4.50 V21. 10.7 V23. (a) 0.385 mA, 3.08 mA, 2.69 mA(b) 69.2 V, with c at the higher potential25. I 1 3.5 A, I 2 2.5 A, I 3 1.0 A27. I 30 0.353 A, I 5 0.118 A, I 20 0.471 A29. V 2 3.05 V, V 3 4.57 V, V 4 7.38 V, V 5 1.62 V31. (a) 12 s (b) 1.2 10 4 C33. 1.3 10 4 C35. 0.982 s37. (a) heater, 10.8 A; toaster, 8.33 A; grill, 12.5 A(b) I total 31.6 A, so a 30-A breaker is insufficient.39. (a) 6.25 A (b) 750 W41. (a) 1.2 10 9 C, 7.3 10 9 K ions. Not large, only1e/290 A 2(b) 1.7 10 9 C, 1.0 10 10 Na ions (c) 0.83 A(d) 7.5 10 12 J43. 11 nW45. 7.5 47. (a) 15 (b) I 1 1.0 A, I 2 I 3 0.50 A, I 4 0.30 A, andI 5 0.20 A(c) (V ) ac 6.0 V, (V ) ce 1.2 V, (V ) ed (V ) fd 1.8 V, (V ) cd 3.0 V, (V ) db 6.0 V(d) ac 6.0 W, ce 0.60 W, ed 0.54 W, fd 0.36 W, cd 1.5 W, db 6.0 W49. (a) 12.4 V (b) 9.65 V51. I 1 0, I 2 I 3 0.50 A,53. 112 V, 0.200 55. (a) R x R 2 1 4 R 1(b) R x 2.8 (inadequate grounding)59. (144 V2 )R(R 10.0 ) 23.6 WP load61. (a) 5.68 V (b) 0.227 A63. 0.395 A; 1.50 VChapter 19QUICK QUIZZES1. (b)2. (c)3. (c)4. (a)5. (b)10 R loadCONCEPTUAL QUESTIONS1. The set should be oriented such that the beam is movingeither toward the east or toward the west.3. The proton moves in a circular path upwards on the page.After completing half a circle, it exits the field and movesin a straight-line path back in the direction from whenceit came. An electron will behave similarly, but the directionof traversal of the circle is downward, and the radiusof the circular path is smaller.5. The magnetic force on a moving charged particle is alwaysperpendicular to the particle’s direction of motion. There isno magnetic force on the charge when it moves parallel tothe direction of the magnetic field. However, the force on acharged particle moving in an electric field is never zeroand is always parallel to the direction of the field. Therefore,by projecting the charged particle in different directions,it is possible to determine the nature of the field.7. The magnetic field produces a magnetic force on theelectrons moving toward the screen that produce theimage. This magnetic force deflects the electrons toregions on the screen other than the ones to which theyare supposed to go. The result is a distorted image.9. Such levitation could never occur. At the North Pole,where Earth’s magnetic field is directed downward,toward the equivalent of a buried south pole, a coffinwould be repelled if its south magnetic pole were directeddownward. However, equilibrium would be only transitory,as any slight disturbance would upset the balancebetween the magnetic force and the gravitational force.11. If you were moving along with the electrons, you wouldmeasure a zero current for the electrons, so they wouldnot produce a magnetic field according to your observations.However, the fixed positive charges in the metalwould now be moving backwards relative to you, creatinga current equivalent to the forward motion of the electronswhen you were stationary. Thus, you would measurethe same magnetic field as when you were stationary, butit would be due to the positive charges presumed to bemoving from your point of view.13. A compass does not detect currents in wires near lightswitches, for two reasons. The first is that, because the cableto the light switch contains two wires, one carryingcurrent to the switch and the other carrying it away fromthe switch, the net magnetic field would be very small andwould fall off rapidly with increasing distance. The secondreason is that the current is alternating at 60 Hz. As a result,the magnetic field is oscillating at 60 Hz also. Thisfrequency would be too fast for the compass to follow, sothe effect on the compass reading would average to zero.15. The levitating wire is stable with respect to vertical motion:If it is displaced upward, the repulsive force weakens, andthe wire drops back down. By contrast, if it drops lower,the repulsive force increases, and it moves back up. Thewire is not stable, however, with respect to lateral movement:If it moves away from the vertical position directlyover the lower wire, the repulsive force will have a sidewayscomponent that will push the wire away.In the case of the attracting wires, the hanging wire isnot stable with respect to vertical movement. If it rises, theattractive force increases, and the wire moves even closerto the upper wire. If the hanging wire falls, the attractiveforce weakens, and the wire falls farther. If the wire movesto the right, it moves farther from the upper wire and theattractive force decreases. Although there is a restoringforce component pulling it back to the left, the verticalforce component is not strong enough to hold the wireup, and it falls.17. Each coil of the Slinky ® will become a magnet, because acoil acts as a current loop. The sense of rotation of thecurrent is the same in all coils, so each coil becomes amagnet with the same orientation of poles. Thus, all ofthe coils attract, and the Slinky ® will compress.

A.22 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems19. There is no net force on the wires, but there is a torque.To understand this distinction, imagine a fixed verticalwire and a free horizontal wire (see the figure below).The vertical wire carries an upward current and creates amagnetic field that circles the vertical wire, itself. To theright, the magnetic field of the vertical wire points intothe page, while on the left side it points out of the page,as indicated. Each segment of the horizontal wire (oflength ) carries current that interacts with the magneticfield according to the equation F BI sin . Apply theright-hand rule on the right side: point the fingers of yourright hand in the direction of the horizontal current andcurl them into the page in the direction of the magneticfield. Your thumb points downward, the direction of theforce on the right side of the wire. Repeating the processon the left side gives a force upward on the left side of thewire. The two forces are equal in magnitude and oppositein direction, so the net force is zero, but they create a nettorque around the point where the wires cross.39. 20.0 T toward bottom of page41. 0.167 T out of the page43. (a) 4.00 m (b) 7.50 nT (c) 1.26 m (d) zero45. 4.5 mm47. 31.8 mA49. 2.26 10 4 N away from the center, zero torque51. 1.7 N m53. (a) 0.500 T out of the page (b) 3.89 T parallel toxy-plane and at 59.0° clockwise from x-direction55. 2.13 cm57. (a) 1.33 m/s (b) the sign of the emf is independent ofthe charge59. 1.41 10 6 N61. 13.0 T toward the bottom of the page63. 53 T toward the bottom of the page, 20 T toward thebottom of the page, and 065. (a) 8.00 10 21 kg m/s (b) 8.90°67. 1.29 kW69. (a) 12.0 cm to the left of wire 1 (b) 2.40 A, downwardFBIBIChapter 20QUICK QUIZZES1. b, c, a2. (a)3. (b)4. (c)5. (b)21. (a) The field is into the page. (b) The beam would deflectupwards.PROBLEMS1. (a) horizontal and due east (b) horizontal and 30° Nof E (c) horizontal and due east (d) zero force3. (a) into the page (b) toward the right (c) toward thebottom of the page5. F g 8.93 10 30 N (downward),F e 1.60 10 17 N (upward),F m 4.80 10 17 N (downward)7. 2.83 10 7 m/s west9. 0.021 T in the y-direction11. 8.0 10 3 T in the z-direction13. (a) into the page (b) toward the right (c) toward thebottom of the page15. 7.50 N17. 0.131 T (downward)19. 0.20 T directed out of the page21. ab: 0, bc: 0.040 0 N in x-direction, cd: 0.040 0 N in the z-direction da: 0.056 6 N parallel to the xz-plane and at45° to both the x- and the z-directions23. 9.05 10 4 Nm, tending to make the left-hand side of theloop move toward you and the right-hand side move away.25. (a) 3.97° (b) 3.39 10 3 Nm27. 6.56 10 2 T31. 1.77 cm33. r 3R/435. 20.0 T37. 2.4 mmFCONCEPTUAL QUESTIONS1. According to Faraday’s law, an emf is induced in a wireloop if the magnetic flux through the loop changes withtime. In this situation, an emf can be induced either byrotating the loop around an arbitrary axis or by changingthe shape of the loop.3. As the spacecraft moves through space, it is apparentlymoving from a region of one magnetic field strength to aregion of a different magnetic field strength. The changingmagnetic field through the coil induces an emf and acorresponding current in the coil.5. If the bar were moving to the left, the magnetic force onthe negative charges in the bar would be upward, causingan accumulation of negative charge on the top and positivecharges at the bottom. Hence, the electric field in thebar would be upward, as well.7. If, for any reason, the magnetic field should change rapidly,a large emf could be induced in the bracelet. If thebracelet were not a continuous band, this emf wouldcause high-voltage arcs to occur at any gap in the band. Ifthe bracelet were a continuous band, the induced emfwould produce a large induced current and result inresistance heating of the bracelet.11. As the aluminum plate moves into the field, eddy currentsare induced in the metal by the changing magneticfield at the plate. The magnetic field of the electromagnetinteracts with this current, producing a retardingforce on the plate that slows it down. In a similar fashion,as the plate leaves the magnetic field, a current is induced,and once again there is an upward force to slowthe plate.13. The energy stored in an inductor carrying a current Iis equal to PE L (1/2)LI 2 . Therefore, doubling the currentwill quadruple the energy stored in the inductor.

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.2315. If an external battery is acting to increase the currentin the inductor, an emf is induced in a direction tooppose the increase of current. Likewise, if we attempt toreduce the current in the inductor, the emf that is set uptends to support the current. Thus, the induced emfalways acts to oppose the change occurring in the circuit,or it acts in the “back” direction to the change.17. (a) clockwise (b) The net force exerted on the bar must bezero because it moves at constant speed. The componentof the gravitational force down the incline is balanced by acomponent of the magnetic force up the incline.19. from left to rightCONCEPTUAL QUESTIONS1. For best reception, the length of the antenna should be parallelto the orientation of the oscillating electric field. Be-PROBLEMS1. 5.9 10 2 T m 23. 7.71 10 1 T m 25. (a) B,net 0 (b) 07. (a) 3.1 10 3 T m 2 (b) B,net 09. 0.18 T11. 94 mV13. 2.7 T/s15. (a) 4.0 10 6 T m 2 (b) 18 V17. 10.2 V19. 0.763 V21. (a) toward the east (b) 4.58 10 4 V23. (a) from left to right (b) from right to left25. (a) F N 2 B 2 w 2 v/R to the left (b) 0(c) F N 2 B 2 w 2 v/R to the left27. into the page29. (a) from right to left (b) from right to left (c) from leftto right (d) from left to right31. 1.9 10 11 V33. (a) 18.1 V (b) 035. (a) 60 V (b) 57 V (c) 0.13 s37. 20 mV39. (a) 2.0 mH (b) 38 A/s43. 12 mH45. 1.92 47. 0.140 J49. (a) 18 J (b) 7.2 J51. negative (V a V b )53. (a) 20.0 ms (b) 37.9 V (c) 1.52 mV (d) 51.8 mA55. 1.20 C57. (a) 0.500 A (b) 2.00 W (c) 2.00 W59. 115 kV61. (a) 0.157 mV (end B is positive) (b) 5.89 mV (end A ispositive)63. (a) 9.00 A (b) 10.8 N (c) b is at the higher potential(d) No65. v t mgRB 2 2Chapter 21ANSWERS TO QUICK QUIZZES1. (c)2. (b)3. (b)4. (b), (c)5. (b), (d)cause of atmospheric variations and reflections of the wavebefore it arrives at your location, the orientation of this fieldmay be in different directions for different stations.3. The primary coil of the transformer is an inductor. Whenan AC voltage is applied, the back emf due to the inductancewill limit the current in the coil. If DC voltage is applied,there is no back emf, and the current can rise to ahigher value. It is possible that this increased current willdeliver so much energy to the resistance in the coil that itstemperature rises to the point at which insulation on thewire can burn.5. An antenna that is a conducting line responds to the electricfield of the electromagnetic wave—the oscillatingelectric field causes an electric force on electrons in thewire along its length. The movement of electrons alongthe wire is detected as a current by the radio and is amplified.Thus, a line antenna must have the same orientationas the broadcast antenna. A loop antenna responds to themagnetic field in the radio wave. The varying magneticfield induces a varying current in the loop (by Faraday’slaw), and this signal is amplified. The loop should be inthe vertical plane containing the line of sight to thebroadcast antenna, so the magnetic field lines go throughthe area of the loop.7. The flashing of the light according to Morse code is adrastic amplitude modulation—the amplitude is changingfrom a maximum to zero. In this sense, it is similar tothe on-and-off binary code used in computers and compactdisks. The carrier frequency is that of the light, onthe order of 10 14 HZ. The frequency of the signaldepends on the skill of the signal operator, but it is on theorder of a single hertz, as the light is flashed on and off.The broadcasting antenna for this modulated signal is thefilament of the lightbulb in the signal source. The receivingantenna is the eye.9. The sail should be as reflective as possible, so that themaximum momentum is transferred to the sail from thereflection of sunlight.11. Suppose the extraterrestrial looks around your kitchen.Lightbulbs and the toaster glow brightly in the infrared.Somewhat fainter are the back of the refrigerator and theback of the television set, while the television screen isdark. The pipes under the sink show the same weak glowas the walls, until you turn on the faucets. Then the pipeon the right gets darker and that on the left develops agleam that quickly runs up along its length. The food onthe plates shines, as does human skin, the same color forall races. Clothing is dark as a rule, but your seat and thechair seat glow alike after you stand up. Your face appearslit from within, like a jack-o’-lantern; your nostrils and theopenings of your ear canals are bright; brighter still arethe pupils of your eyes.13. Radio waves move at the speed of light. They can travelaround the curved surface of the Earth, bouncing betweenthe ground and the ionosphere, which has an altitude thatis small compared with the radius of the Earth. Thedistance across the lower 48 states is approximately5 000 km, requiring a travel time that is equal to (5 10 6 m)/(3 10 8 m/s) 10 2 s. Likewise, radio wavestake only 0.07 s to travel halfway around the Earth. Inother words, a speech can be heard on the other side ofthe world (in the form of radio waves) before it is heard atthe back of the room (in the form of sound waves).

A.24 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems15. No. The wire will emit electromagnetic waves only if thecurrent varies in time. The radiation is the result of acceleratingcharges, which can occur only when the current isnot constant.17. The resonance frequency is determined by the inductanceand the capacitance in the circuit. If both L and C are doubled,the resonance frequency is reduced by a factor of two.19. It is far more economical to transmit power at a highvoltage than at a low voltage because the I 2 R loss on thetransmission line is significantly lower at high voltage.Transmitting power at high voltage permits the use ofstep-down transformers to make “low” voltages and highcurrents available to the end user.21. No. A voltage is induced in the secondary coil only if theflux through the core changes with time.PROBLEMS1. (a) 141 V (b) 20.0 A (c) 28.3 A (d) 2.00 kW3. 70.7 V, 2.95 A5. 6.76 W9. 4.0 10 2 Hz11. 17 F15. 3.14 A17. 0.450 T m 219. (a) 0.361 A (b) 18.1 V (c) 23.9 V (d) 53.0°21. (a) 1.4 k (b) 0.10 A (c) 51° (d) voltage leads current23. (a) 89.6 V (b) 108 V25. 1.88 V27. (a) 103 V (b) 150 V (c) 127 V (d) 23.6 V29. (a) 208 (b) 40.0 (c) 0.541 H31. (a) 1.8 10 2 (b) 0.71 H33. 2.29 H35. C min 4.9 nF, C max 51 nF37. 0.242 J39. 0.18% is lost41. (a) 1.1 10 3 kW (b) 3.1 10 2 A (c) 8.3 10 3 A43. 1 000 km; there will always be better use for tax money.45. f red 4.55 10 14 Hz, f IR 3.19 10 14 Hz,E max,f /E max,i 0.5747. 2.94 10 8 m/s49. E max 1.01 10 3 V/m, B max 3.35 10 6 T51. (a) 188 m to 556 m (b) 2.78 m to 3.4 m53. 5.2 10 13 Hz, 5.8 m55. 4.299 999 84 10 14 Hz; 1.6 10 7 Hz(the frequency decreases)57. 99.6 mH59. 1.7 cents61. (a) resistor and inductor (b) R 10 , L 30 mH63. (a) 6.7 10 16 T (b) 5.3 10 17 W/m 2(c) 1.7 10 14 W65. (a) 0.536 N (b) 8.93 10 5 m/s 2 (c) 33.9 days67. 4.47 10 9 JChapter 22QUICK QUIZZES1. (a)2. Beams 2 and 4 are reflected; beams 3 and 5 are refracted.3. (b)4. (c)CONCEPTUAL QUESTIONS1. Sound radiated upward at an acute angle with the horizontalis bent back toward Earth by refraction. This meansthat the sound can reach the listener by this path as wellas by a direct path. Thus, the sound is louder.3. The color will not change, for two reasons. First, despitethe popular statement that color depends on wavelength,it actually depends on the frequency of the light, whichdoes not change under water. Second, when the light entersthe eye, it travels through the fluid within. Thus, evenif color did depend on wavelength, the important wavelengthis that of the light in the ocular fluid, which doesnot depend on the medium through which the light traveledto reach the eye.5. (a) Away from the normal (b) increases (c) remains thesame7. No, the information in the catalog is incorrect. The indexof refraction is given by n c/v, where c is the speed oflight in a vacuum and v is the speed of light in the material.Because light travels faster in a vacuum than in anyother material, it is impossible for the index of refractionof any material to have a value less than 1.9. There is no dependence of the angle of reflection on wavelength,because the light does not enter deeply into thematerial during reflection—it reflects from the surface.11. On the one hand, a ball covered with mirrors sparkles byreflecting light from its surface. On the other hand, afaceted diamond lets in light at the top, reflects it by totalinternal reflection in the bottom half, and sends the lightout through the top again. Because of its high index ofrefraction, the critical angle for diamond in air for totalinternal reflection, namely c sin 1 (n air /n diamond ), issmall. Thus, light rays enter through a large area and exitthrough a very small area with a much higher intensity.When a diamond is immersed in carbon disulfide, thecritical angle is increased to c sin 1 (n carbon disulfide /n diamond ). As a result, the light is emitted from the diamondover a larger area and appears less intense.13. The index of refraction of water is 1.333, quite differentfrom that of air, which has an index of refraction of about 1.The boundary between the air and water is therefore easy todetect, because of the differing diffraction effects above andbelow the boundary. (Try looking at a glass half full ofwater.) The index of refraction of liquid helium, however,happens to be much closer to that of air. Consequently,the defractive differences above and below the helium-airboundary are harder to see.15. The diamond acts like a prism, dispersing the light into itsspectral components. Different colors are observed as aconsequence of the manner in which the index of refractionvaries with the wavelength.17. Light travels through a vacuum at a speed of 3 10 8 m/s.Thus, an image we see from a distant star or galaxy musthave been generated some time ago. For example, thestar Altair is 16 lightyears away; if we look at an image ofAltair today, we know only what Altair looked like 16 yearsago. This may not initially seem significant; however,astronomers who look at other galaxies can get an idea ofwhat galaxies looked like when they were much younger.Thus, it does make sense to speak of “looking backward intime.”PROBLEMS1. 3.00 10 8 m/s3. 114 rad/s for a maximum intensity of returning light5. (b) 3.000 10 8 m/s

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.257. 19.5° above the horizontal9. (a) 1.52 (b) 417 nm (c) 4.74 10 14 Hz(d) 1.98 10 8 m/s11. (a) 584 nm (b) 1.1213. 111°15. (a) 1.559 10 8 m/s (b) 329.1 nm (c) 4.738 10 14 Hz17. five times from the right-hand mirror and six times fromthe left19. 0.388 cm21. 30.4°, 22.3°23. 6.39 ns25. tan 1 (n g )27. 3.39 m29. red 48.22°, blue 47.79°31. (a) 1i 30°, 1r 19°, 2i 41°, 2r 77°(b) First surface: reflection 30°;second surface: reflection 41°33. (a) 31.3° (b) 44.2° (c) 49.8°35. (a) 33.4° (b) 53.4°37. (a) 40.8° (b) 60.6°39. 1.000 0841. (a) 10.7° (b) air (c) Sound falling on the wall frommost directions is 100% reflected.43. 27.5°45. 22.0°47. (a) 53.1° (b) 38.7°49. (a) 38.5° (b) 1.4453. 24.7°55. 1.9359.61. (a) 1.20 (b) 3.40 ns sin 1 √n 2 1 sin cosChapter 23QUICK QUIZZES1. At C.2. (c)3. (a) False (b) False (c) True4. (b)5. An infinite number6. (a) False (b) True (c) FalseCONCEPTUAL QUESTIONS1. You will not be able to focus your eyes on both the pictureand your image at the same time. To focus on the picture,you must adjust your eyes so that an object several centimetersaway (the picture) is in focus. Thus, you are focusingon the mirror surface. But, your image in the mirroris as far behind the mirror as you are in front of it.Thus, you must focus your eyes beyond the mirror, twiceas far away as the picture to bring the image into focus.3. A single flat mirror forms a virtual image of an object dueto two factors. First, the light rays from the object are necessarilydiverging from the object, and second, the lack ofcurvature of the flat mirror cannot convert diverging raysto converging rays. If another optical element is first usedto cause light rays to converge, then the flat mirror can beplaced in the region in which the converging rays arepresent, and it will change the direction of the rays so thatthe real image is formed at a different location. For example,if a real image is formed by a convex lens, and the flatmirror is placed between the lens and the image position,the image formed by the mirror will be real.5. The ultrasonic range finder sends out a sound wave andmeasures the time for the echo to return. Using this information,the camera calculates the distance to the subjectand sets the camera lens. When the camera is facing amirror, the ultrasonic signal reflects from the mirror surfaceand the camera adjusts its focus so that the mirrorsurface is at the correct focusing distance from the camera.But your image in the mirror is twice this distancefrom the camera, so it is blurry.7. Light rays diverge from the position of a virtual imagejust as they do from an actual object. Thus, a virtual imagecan be as easily photographed as any object can. Ofcourse, the camera would have to be placed near theaxis of the lens or mirror in order to intercept the lightrays.9. We consider the two trees to be two separate objects. Thefar tree is an object that is farther from the lens than thenear tree. Thus, the image of the far tree will be closer tothe lens than the image of the near tree. The screen mustbe moved closer to the lens to put the far tree in focus.11. If a converging lens is placed in a liquid having an index ofrefraction larger than that of the lens material, the directionof refractions at the lens surfaces will be reversed, andthe lens will diverge light. A mirror depends only on reflectionwhich is independent of the surrounding material, soa converging mirror will be converging in any liquid.13. This is a possible scenario. When light crosses a boundarybetween air and ice, it will refract in the same manner asit does when crossing a boundary of the same shape betweenair and glass. Thus, a converging lens may be madefrom ice as well as glass. However, ice is such a strong absorberof infrared radiation that it is unlikely you will beable to start a fire with a small ice lens.15. The focal length for a mirror is determined by the law ofreflection from the mirror surface. The law of reflectionis independent of the material of which the mirror ismade and of the surrounding medium. Thus, the focallength depends only on the radius of curvature and noton the material. The focal length of a lens depends onthe indices of refraction of the lens material andsurrounding medium. Thus, the focal length of a lensdepends on the lens material.17. (a) all signs are positive (b) f and p are positive, q isnegative19. (c) the image becomes fuzzy and disappearsPROBLEMS1. on the order of 10 9 s younger3. 10.0 ft, 30.0 ft, 40.0 ft5. 0.268 m behind the mirror; virtual, upright, anddiminished; M 0.026 87. (a) 13.3 cm in front of mirror, real, inverted, M 0.333(b) 20.0 cm in front of mirror, real, inverted, M 1.00(c) No image is formed. Parallel rays leave the mirror.9. Behind the worshipper, 3.33 m from the deepest point inthe niche.11. 5.00 cm13. 1.0 m15. 8.05 cm17. 20.0 cm19. (a) concave with focal length f 0.83 m(b) Object must be 1.0 m in front of the mirror.21. 38.2 cm below the upper surface of the ice

A.26 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems23. 3.8 mm25. n 2.0027. 20.0 cm29. (a) 40.0 cm beyond the lens, real, inverted, M 1.00(b) No image is formed. Parallel rays leave the lens.(c) 20.0 cm in front of the lens, virtual, upright, M 2.0031. (a) 13.3 cm in front of the lens, virtual, upright, M 1/3(b) 10.0 cm in front of the lens, virtual, upright, M 1/2(c) 6.67 cm in front of the lens, virtual, upright, M 2/333. (a) either 9.63 cm or 3.27 cm (b) 2.10 cm35. (a) 39.0 mm (b) 39.5 mm37. at distance 2 f in front of lens39. 40.0 cm41. 30.0 cm to the left of the second lens, M 3.0043. 7.47 cm in front of the second lens; 1.07 cm; virtual, upright45. from 0.224 m to 18.2 m47. real image, 5.71 cm in front of the mirror49. 38.6°51. 160 cm to the left of the lens, inverted, M 0.80053. q 10.7 cm55. 32.0 cm to the right of the second surface (real image)57. (a) 20.0 cm to the right of the second lens; M 6.00(b) inverted(c) 6.67 cm to the right of the second lens; M 2.00;inverted59. (a) 1.99(b) 10.0 cm to the left of the lens(c) inverted61. (a) 5.45 m to the left of the lens(b) 8.24 m to the left of the lens(c) 17.1 m to the left of the lens(d) by surrounding the lens with a medium having arefractive index greater than that of the lens material.63. (a) 263 cm (b) 79.0 cmChapter 24QUICK QUIZZES1. (c)2. (b)3. (b)4. The compact disc.CONCEPTUAL QUESTIONS1. You will not see an interference pattern from the automobileheadlights, for two reasons. The first is that the headlightsare not coherent sources and are therefore incapableof producing sustained interference. Also, theheadlights are so far apart in comparison to the wavelengthsemitted that, even if they were made into coherentsources, the interference maxima and minima wouldbe too closely spaced to be observable.3. The result of the double slit is to redistribute the energy arrivingat the screen. Although there is no energy at the locationof a dark fringe, there is four times as much energyat the location of a bright fringe as there would be withonly a single narrow slit. The total amount of energy arrivingat the screen is twice as much as with a single slit, as itmust be according to the law of conservation of energy.5. One of the materials has a higher index of refraction thanwater, and the other has a lower index. The material withthe higher index will appear black as it approaches zerothickness. There will be a 180° phase change for the lightreflected from the upper surface, but no such phasechange for the light reflected from the lower surface,because the index of refraction for water on the otherside is lower than that of the film. Thus, the two reflectionswill be out of phase and will interfere destructively.The material with index of refraction lower than waterwill have a phase change for the light reflected from boththe upper and the lower surface, so that the reflectionsfrom the zero-thickness film will be back in phase and thefilm will appear bright.7. For incidence normal to the film, the extra path lengthfollowed by the reflected ray is twice the thickness ofthe film. For destructive interference, this must be a distanceof half a wavelength of the light in the material ofthe film. For a film in air, no 180° phase change will occurin these reflections, so the thickness of the film must beone-quarter wavelength, which is the same as the conditionfor constructive interference of reflected light. Thismeans that the transmitted light is a minimum when thereflected light is a maximum, and vice versa.9. Since the light reflecting at the lower surface of the filmundergoes a 180° phase change, while light reflectingfrom the upper surface of the film does not undergo sucha change, the central spot (where the film has near zerothickness) will be dark. If the observed rings are not circular,the curved surface of the lens does not have a truespherical shape.11. For regional communication at the Earth’s surface, radiowaves are typically broadcast from currents oscillating intall vertical towers. These waves have vertical planes ofpolarization. Light originates from the vibrations of atomsor electronic transitions within atoms, which representoscillations in all possible directions. Thus, light generallyis not polarized.13. Yes. In order to do this, first measure the radar reflectivityof the metal of your airplane. Then choose a light,durable material that has approximately half the radarreflectivity of the metal in your plane. Measure its indexof refraction, and place onto the metal a coating equal inthickness to one-quarter of 3 cm, divided by that index.Sell the plane quick, and then you can sell the supposedenemy new radars operating at 1.5 cm, which the coatedmetal will reflect with extra-high efficiency.15. If you wish to perform an interference experiment, youneed monochromatic coherent light. To obtain it, youmust first pass light from an ordinary source through aprism or diffraction grating to disperse different colorsinto different directions. Using a single narrow slit, selecta single color and make that light diffract to cover bothslits for a Young’s experiment. The procedure is muchsimpler with a laser because its output is already monochromaticand coherent.17. Strictly speaking, the ribs do act as a diffraction grating,but the separation distance of the ribs is so much largerthan the wavelength of the x-rays that there are no observableeffects.19. As the edge of the Moon cuts across the light from thestar, edge diffraction effects occur. Thus, as the edge ofthe Moon moves relative to the star, the observed lightfrom the star proceeds through a series of maxima andminima.21. Larger. From Brewster’s law, n tan p , we see that theangle increases as n increases.

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.27PROBLEMS1. 1.58 cm3. (a) 2.6 mm (b) 2.62 mm5. (a) 36.2° (b) 5.08 cm (c) 5.08 10 14 Hz7. (a) 55.7 m (b) 124 m9. 75.0 m11. 11.3 m13. 148 m15. 91.9 nm17. 550 nm19. 0.500 cm21. (a) 238 nm (b) will increase (c) 328 nm23. 4.35 m25. 4.75 m27. No, the wavelengths intensified are 276 nm, 138 nm,92.0 nm, . . .29. 4.22 mm31. (a) 1.1 m (b) 1.7 mm33. 1.20 mm, 1.20 mm35. (a) 479 nm, 647 nm, 698 nm (b) 20.5°, 28.3°, 30.7°37. 5.91° in first order; 13.2° in second order; and26.5° in third order39. 44.5 cm41. 9.13 cm43. (a) 25.6° (b) 19.0°45. (a) 1.11 (b) 42.0°47. (a) 56.7° (b) 48.8°49. 31.2°53. 6.89 units55. (a) 413.7 nm, 409.7 nm (b) 8.6°57. 0.156 mm59. 2.50 mm61. Any positive integral multiple of 210 nm63. (a) 16.6 m (b) 8.28 m65. 127 m67. 0.350 mm69. 115 nmChapter 25QUICK QUIZZES1. (c)2. (a)CONCEPTUAL QUESTIONS1. The observer is not using the lens as a simple magnifier.For a lens to be used as a simple magnifier, the object distancemust be less than the focal length of the lens. Also,a simple magnifier produces a virtual image at the normalnear point of the eye, or at an image distance of aboutq 25 cm. With a large object distance and a relativelyshort image distance, the magnitude of the magnificationby the lens would be considerably less than one. Mostlikely, the lens in this example is part of a lens combinationbeing used as a telescope.3. The image formed on the retina by the lens and cornea isalready inverted.5. There will be an effect on the interference pattern—itwill be distorted. The high temperature of the flame willchange the index of refraction of air for the arm of theinterferometer in which the match is held. As the indexof refraction varies randomly, the wavelength of the lightin that region will also vary randomly. As a result, theeffective difference in length between the two arms willfluctuate, resulting in a wildly varying interferencepattern.7. Large lenses are difficult to manufacture and machinewith accuracy. Also, their large weight leads to sagging,which produces a distorted image. In reflecting telescopes,light does not pass through glass; hence, problemsassociated with chromatic aberrations are eliminated.Large-diameter reflecting telescopes are also technicallyeasier to construct. Some designs use a rotating pool ofmercury as the reflecting surface.9. In order for someone to see an object through a microscope,the wavelength of the light in the microscope mustbe smaller than the size of the object. An atom is muchsmaller than the wavelength of light in the visible spectrum,so an atom can never be seen with the use of visiblelight.11. farsighted; convergingPROBLEMS1. 30.0 cm beyond the lens, M 1/53. 177 m5. f/1.47. f/8.09. 40.0 cm11. 23.2 cm13. (a) 2.00 diopters (b) 17.6 cm15. 17.0 diopters17. (a) 5.8 cm (b) m 4.319. (a) 4.07 cm (b) m 7.1421. (a) M 1.22 (b) / 0 6.0823. 2.1 cm25. m 11527. f o 90 cm, f e 2.0 cm29. (b) fh/p (c) 1.07 mm31. (a) m 1.50 (b) m 1.9033. 492 km35. 0.40 rad37. 9.1 10 7 km39. 9.8 km41. No. A resolving power of 2.0 10 5 is needed, and thatavailable is only 1.8 10 5 .43. 50.4 m45. 4047. 98 fringe shifts49. (a) 2.67 diopters (b) 0.16 diopter too low51. (a) 44.6 diopters (b) 3.03 diopters53. (a) 1.0 10 3 lines (b) 3.3 10 2 lines55. m 10.757. (a) m 4.0 (b) m 3.0Chapter 26QUICK QUIZZES1. (a)2. No. From your perspective you’re at rest with respect tothe cabin, so you will measure yourself as having yournormal length, and will require a normal-sized cabin.3. (a), (e); (a), (e)4. (a) False (b) False (c) True (d) False5. (a)

A.28 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and ProblemsCONCEPTUAL QUESTIONS1. An ellipsoid. The dimension in the direction of motionwould be measured to be less than D.3. This scenario is not possible with light. Light waves are describedby the principles of special relativity. As you detectthe light wave ahead of you and moving away from you(which would be a pretty good trick—think about it!), itsvelocity relative to you is c. Thus, you will not be able tocatch up to the light wave.5. No. The principle of relativity implies that nothing cantravel faster than the speed of light in a vacuum, which is3.00 10 8 m/s.7. The light from the quasar moves at 3.00 10 8 m/s. Thespeed of light is independent of the motion of the sourceor the observer.9. For a wonderful fictional exploration of this question, geta “Mr. Tompkins” book by George Gamow. All of the relativityeffects would be obvious in our lives. Time dilationand length contraction would both occur. Driving homein a hurry, you would push on the gas pedal not to increaseyour speed very much, but to make the blocksshorter. Big Doppler shifts in wave frequencies wouldmake red lights look green as you approached and makecar horns and radios useless. High-speed transportationwould be both very expensive, requiring huge fuel purchases,as well as dangerous, since a speeding car couldknock down a building. When you got home, hungry forlunch, you would find that you had missed dinner; therewould be a five-day delay in transit when you watch a liveTV program originating in Australia. Finally, we would notbe able to see the Milky Way, since the fireball of the BigBang would surround us at the distance of Rigel orDeneb.11. A photon transports energy. The relativistic equivalenceof mass and energy means that is enough to give itmomentum.13. Your assignment: measure the length of a rod as it slidespast you. Mark the position of its front end on the floorand have an assistant mark the position of the back end.Then measure the distance between the two marks. Thisdistance will represent the length of the rod only if thetwo marks were made simultaneously in your frame ofreference.PROBLEMS1. (a) t OB 1.67 10 3 s, t OA 2.04 10 3 s(b) t BO 2.50 10 3 s, t AO 2.04 10 3 s(c) t 90 s3. 5.0 s5. (a) 20 m (b) 19 m (c) 0.31c7. (a) 1.3 10 7 s (b) 38 m (c) 7.6 m9. (a) 2.2 s (b) 0.65 km11. 0.950c13. Yes, with 19 m to spare15. (a) 39.2 s (b) Accurate to one digit17. 3.3 10 5 m/s19. 0.285c21. 0.54c to the right23. 0.357c25. 0.998c toward the right27. (a) 54 min (b) 52 min29. c(√3/2)31. 0.786c33. 18.4 g/cm 335. 1.98 MeV37. 2.27 10 23 Hz, 1.32 fm for each photon39. (a) 3.10 10 5 m/s (b) 0.758c41. 1.42 MeV/c43. (a) 0.80c (b) 7.5 10 3 s (c) 1.4 10 12 m, 0.38c45. 0.37c in the x-direction47. (a) v/c 1 1.12 10 10 (b) 6.00 10 27 J(c) $2.17 10 2049. 0.80c51. (a) 0.946c (b) 0.160 ly (c) 0.114 yr (d) 7.50 10 22 J53. (a) 7.0 s (c) 1.1 10 4 muons59. 5.45 yr; Goslo is older.Chapter 27QUICK QUIZZES1. (b)2. (c)3. (c)4. (b)CONCEPTUAL QUESTIONS1. The shape of an object is normally determined by observingthe light reflecting from its surface. In a kiln, the objectwill be very hot and will be glowing red. The emittedradiation is far stronger than the reflected radiation, andthe thermal radiation emitted is only slightly dependenton the material from which the object is made. Unlike reflectedlight, the emitted light comes from all surfaceswith equal intensity, so contrast is lost and the shape ofthe object is harder to discern.3. The “blackness” of a blackbody refers to its ideal propertyof absorbing all radiation incident on it. If an observedroom temperature object in everyday life absorbs all radiation,we describe it as (visibly) black. The black appearance,however, is due to the fact that our eyes are sensitiveonly to visible light. If we could detect infrared light withour eyes, we would see the object emitting radiation. Ifthe temperature of the blackbody is raised, Wien’s lawtells us that the emitted radiation will move into the visiblerange of the spectrum. Thus, the blackbody could appearas red, white, or blue, depending on its temperature.5. All objects do radiate energy, but at room temperature thisenergy is primarily in the infrared region of the electromagneticspectrum, which our eyes cannot detect. (Pit vipershave sensory organs that are sensitive to infrared radiation;thus, they can seek out their warm-blooded prey in what wewould consider absolute darkness.7. Most metals have cutoff frequencies corresponding tophotons in or near the visible range of the electromagneticspectrum. AM radio wave photons have far too littleenergy to eject electrons from the metal.9. We can picture higher frequency light as a stream of photonsof higher energy. In a collision, one photon can giveall of its energy to a single electron. The kinetic energy ofsuch an electron is measured by the stopping potential.The reverse voltage (stopping voltage) required to stopthe current is proportional to the frequency of the incominglight. More intense light consists of more photonsstriking a unit area each second, but atoms are sosmall that one emitted electron never gets a “kick” frommore than one photon. Increasing the intensity of the

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.29light will generally increase the size of the current, butwill not change the energy of the individual electronsthat are ejected. Thus, the stopping potential remainsconstant.11. Wave theory predicts that the photoelectric effect shouldoccur at any frequency, provided that the light intensity ishigh enough. However, as seen in photoelectric experiments,the light must have sufficiently high frequency forthe effect to occur.13. (a) Electrons are emitted only if the photon frequency isgreater than the cutoff frequency.15. No. Suppose that the incident light frequency at whichyou first observed the photoelectric effect is above the cutofffrequency of the first metal, but less than the cutofffrequency of the second metal. In that case, the photoelectriceffect would not be observed at all in the secondmetal.17. The frequency of the scattered photon must decrease,because some of its energy is transferred to the electron.PROBLEMS1. (a) 3 000 K (b) 20 000 K3. 500 nm5. (a) 2.49 10 5 eV (b) 2.49 eV (c) 249 eV7. 2.27 10 30 photons/s9. (a) 2.3 10 31 (b) E/E 4.3 10 3211. (a) 2.24 eV (b) 555 nm (c) 5.41 10 14 Hz13. 234 nm15. 148 days, incompatible with observation17. 4.8 10 14 Hz, 2.0 eV19. 1.2 10 2 V and 1.2 10 7 V, respectively21. 41.4 kV23. 0.078 nm25. 0.281 nm27. 1.78 eV, 9.47 10 28 kg m/s29. 70°31. 1.18 10 23 kg m/s, 478 eV33. (a) 1.2 eV (b) 6.5 10 5 m/s35. (a) 1.46 km/s (b) 7.28 10 11 m37. (a) 10 2 MeV (b) No. With kinetic energy much largerthan the magnitude of the negative potential energy, theelectron would immediately escape.39. 3.58 10 13 m41. (a) 15 keV (b) 1.2 10 2 keV43. 10 6 m/s45. 116 m/s47. 5 200 K; clearly, a firefly is not at that temperature, sothis cannot be blackbody radiation.49. 18.2°51. 1.36 eV53. 2.00 eV55. (a) 0.022 0c (b) 0.999 2c57. (b) 3.72 km/s59. (b) 5.19 10 16 m61. (a) 0.263 kg (b) 1.81 W(c) 0.015 3°C/s 0.919°C/min (d) 9.89 m(e) 2.01 10 20 J (f) 8.98 10 19 photon/sChapter 28QUICK QUIZZES1. (b)2. (a)3. (a) 5 (b) 9 (c) 254. (d)CONCEPTUAL QUESTIONS1. If the energy of the hydrogen atom were proportional to n(or any power of n), then the energy would become infiniteas n grew to infinity. But the energy of the atom is inverselyproportional to n 2 . Thus, as n grows to infinity, the energyof the atom approaches a value that is above the groundstate by a finite amount, namely, the ionization energy13.6 eV. As the electron falls from one bound state to another,its energy loss is always less than the ionization energy.The energy and frequency of any emitted photon are finite.3. The characteristic x-rays originate from transitions withinthe atoms of the target, such as an electron from the Lshell making a transition to a vacancy in the K shell. The vacancyis caused when an accelerated electron in the x-raytube supplies energy to the K shell electron to eject it fromthe atom. If the energy of the bombarding electrons wereto be increased, the K shell electron will be ejected fromthe atom with more remaining kinetic energy. But the energydifference between the K and L shell has not changed,so the emitted x-ray has exactly the same wavelength.5. A continuous spectrum without characteristic x-rays is possible.At a low accelerating potential difference for theelectron, the electron may not have enough energy toeject an electron from a target atom. As a result, there willbe no characteristic x-rays. The change in speed of theelectron as it enters the target will result in the continuousspectrum.7. The hologram is an interference pattern between lightscattered from the object and the reference beam. If anythingmoves by a distance comparable to the wavelengthof the light (or more), the pattern will wash out. Theeffect is just like making the slits vibrate in Young’s experiment,to make the interference fringes vibrate wildly sothat a photograph of the screen displays only the averageintensity everywhere.9. If the Pauli exclusion principle were not valid, theelements and their chemical behavior would be grosslydifferent, because every electron would end up in the lowestenergy level of the atom. All matter would thereforebe nearly alike in its chemistry and composition, since theshell structures of each element would be identical. Mostmaterials would have a much higher density, and the spectraof atoms and molecules would be very simple, resultingin the existence of less color in the world.11. The three elements have similar electronic configurations,with filled inner shells plus a single electron in an sorbital. Since atoms typically interact through their unfilledouter shells, and since the outer shells of theseatoms are similar, the chemical interactions of the threeatoms are also similar.13. Each of the eight electrons must have at least one quantumnumber different from each of the others. They candiffer (in m s ) by being spin-up or spin-down. They can differ(in ) in angular momentum and in the general shapeof the wave function. Those electrons with 1 can differ(in m ) in orientation of angular momentum.15. Stimulated emission is the reason laser light is coherentand tends to travel in a well-defined parallel beam. Whena photon passing by an excited atom stimulates that atomto emit a photon, the emitted photon is in phase with the

A.30 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and ProblemsCONCEPTUAL QUESTIONS1. Isotopes of a given element correspond to nuclei with differentnumbers of neutrons. This will result in a variety ofdifferent physical properties for the nuclei, including theobvious one of mass. The chemical behavior, however, isgoverned by the element’s electrons. All isotopes of agiven element have the same number of electrons and,therefore, the same chemical behavior.3. An alpha particle contains two protons and two neutrons.Because a hydrogen nucleus contains only one proton, itcannot emit an alpha particle.5. In alpha decay, there are only two final particles: thealpha particle and the daughter nucleus. There are alsotwo conservation principles: of energy and of momentum.As a result, the alpha particle must be ejected with a discreteenergy to satisfy both conservation principles. However,beta decay is a three-particle decay: the beta particle,the neutrino (or antineutron), and the daughter nucleus.As a result, the energy and momentum can be shared in avariety of ways among the three particles while still satisfyingthe two conservation principles. This allows a continuousrange of energies for the beta particle.7. The larger rest energy of the neutron means that a freeproton in space will not spontaneously decay into a neutronand a positron. When the proton is in the nucleus,however, the important question is that of the total restenergy of the nucleus. If it is energetically favorable for thenucleus to have one less proton and one more neutron,then the decay process will occur to achieve this lower energy.9. Carbon dating cannot generally be used to estimate theage of a stone, because the stone was not alive to take upcarbon from the environment. Only the ages of artifactsthat were once alive can be estimated with carbon dating.11. The protons, although held together by the nuclear force,are repelled by the electrostatic force. If enoughprotons were placed together in a nucleus, the electrostaticforce would overcome the nuclear force, which isbased on the number of particles, and cause the nucleusto fission.The addition of neutrons prevents such fission. Theneutron does not increase the electrical force, being electricallyneutral, but does contribute to the nuclear force.13. The photon and the neutrino are similar in that both particleshave zero charge and very little mass. (The photonhas zero mass, but recent evidence suggests that certainkinds of neutrinos have a very small mass.) Both musttravel at the speed of light and are capable of transferringboth energy and momentum. They differ in that the phooriginalphoton and travels in the same direction. As thisprocess is repeated many times, an intense, parallel beamof coherent light is produced. Without stimulated emission,the excited atoms would return to the ground stateby emitting photons at random times and in random directions.The resulting light would not have the usefulproperties of laser light.17. The atom is a bound system. The atomic electron doesnot have enough kinetic energy to escape from its electricalattraction to the nucleus. The electrical potential energyof the atom is negative and is greater than the kineticenergy, so the total energy of the atom is negative.19. (a) The wavelength of photon A is greater. (b) Theenergy of photon B is greater.PROBLEMS1. 656 nm, 486 nm, and 434 nm3. (a) 2.3 10 8 N (b) 14 eV5. (a) 1.6 10 6 m/s (b) No, v/c 5.3 10 3 1(c) 0.46 nm (d) Yes. The wavelength is roughly the samesize as the atom.7. (a) 0.212 nm (b) 9.95 10 25 kg m/s(c) 2.11 10 34 J s(d) 3.40 eV (e) 6.80 eV (f ) 3.40 eV11. E 1.51 eV (n 3) to E 3.40 eV (n 2)13. (a) 0.967 eV (b) 0.266 eV15. (a) 122 nm, 91.1 nm (b) 1.87 10 3 nm, 820 nm17. 97.2 nm19. (a) 488 nm (b) 0.814 m/s21. (d) n 2.53 10 74 (e) No. At such large quantumnumbers, the allowed energies are essentiallycontinuous.23. (a) 2.47 10 14 Hz, f orb 8.23 10 14 Hz(b) 6.59 10 3 Hz, f orb 6.59 10 3 Hz. For large n,classical theory and quantum theory approach each otherin their results.25. 4.42 10 4 m/s27. (a) 122 eV (b) 1.76 10 11 m29. (a) 0.026 5 nm (b) 0.017 6 nm (c) 0.013 2 nm31. 1.33 nm33. n 3, 1, m 1, m s 1/2; n 3, 1, m 0,m s 1/2; n 3, 1, m 1, m s 1/235. Fifteen possible states, as summarized in the followingtable:n 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2m 2 2 2 1 1 1 0 0 0 1 1 1 2 2 2m s 1 0 1 1 0 1 1 0 1 1 0 1 1 0 137. (a) 30 possible states (b) 3639. (a) n 4 and 2 (b) m (0, 1, 2), m s 1/2(c) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 2 5s 2 [Kr] 4d 2 5s 241. 0.160 nm43. L shell: 11.7 keV; M shell: 10.0 keV; N shell: 2.30 keV45. (a) 10.2 eV (b) 7.88 10 4 K47. (a) 8.18 eV, 2.04 eV, 0.904 eV, 0.510 eV, 0.325 eV(b) 1.09 10 3 nm and 609 nm49. The four lowest energies are 10.39 eV, 5.502 eV, 3.687 eV, and 2.567 eV (b) The wavelengths of theemission lines are 158.5 nm, 185.0 nm, 253.7 nm,422.5 nm, 683.2 nm, and 1 107 nm(c) 1.31 10 6 m/s51. (a) 4.24 10 15 W/m 2 (b) 1.20 10 12 J55. (a) E n ( 1.49 10 4 eV)/n 2 (b) n 4 : n 157. (a) 9.03 10 22 m/s 2 (b) 4.63 10 8 W(c) 10 11 sChapter 29QUICK QUIZZES1. (c)2. (c)3. (a)4. (a) and (b)5. (b)

Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.31ton has spin (intrinsic angular momentum) and is involvedin electromagnetic interactions, while the neutrinohas spin /2, and is closely related to beta decays.15. Since the two samples are of the same radioactive nuclide,they have the same half-life; the 2:1 difference in activityis due to a 2:1 difference in the mass of each sample. After5 half lives, each will have decreased in mass by a power of2 5 32. However, since this simply means that the massof each is 32 times smaller, the ratio of the masses will stillbe (2/32) : (1/32), or 2:1. Therefore, the ratio of theiractivities will always be 2:1.PROBLEMS1. A 2, r 1.5 fm; A 60, r 4.7 fm; A 197, r 7.0 fm;A 239, r 7.4 fm3. 1.8 10 2 m5. (a) 27.6 N (b) 4.16 10 27 m/s 2 (c) 1.73 MeV7. (a) 1.9 10 7 m/s (b) 7.1 MeV931979. 8.66 MeV/nucleon for 41 Nb , 7.92 MeV/nucleon for 79 Au11. 3.54 MeV2313. 0.210 MeV/nucleon greater for 11 Na , attributable to lessproton repulsion15. 0.46 Ci17. (a) 9.98 10 7 s 1 (b) 1.9 10 10 nuclei19. 1.0 h21. 4.31 10 3 yr23. (a) 5.58 10 2 h 1 , 12.4 h (b) 2.39 10 13 nuclei(c) 1.9 mCi208 95 14425. , 37 Rb81 Tl , 60 Nd40 94 427. , ,29. e 42 Mo56decay, 27 Co2 He20 Ca: 5626 Fe e 31. (a) cannot occur spontaneously(b) can occur spontaneously33. 18.6 keV35. 4.22 10 3 yr3037. (a) 15 P (b) 2.64 MeV21 14439. (a) 10 Ne (b) 54 Xe (c) X e , X 13 1041. (a) 6 C (b) 5 B19719843. (a) 79Au n : 80 Hg e (b) 7.88 MeV145. (a) 0 n (b) Fluoride mass 18.000 953 u47. 18.8 J49. 24 d51. (a) 8.97 10 11 electrons (b) 0.100 J (c) 100 rad53. 46.5 d55. Q 3.27 MeV 0, no threshold energy required57. (a) 2.52 10 24 (b) 2.29 10 12 Bq (c) 1.07 10 6 yr59. (a) 4.0 10 9 yr (b) It could be no older. The rock couldbe younger if some 87 Sr were initially present.61. 54 Ci63. 2.3 10 2 yr65. 4.4 10 8 kg/hChapter 30QUICK QUIZZES1. (c)2. (a)3. (b)4. (d)CONCEPTUAL QUESTIONS1. The experiment described is a nice analogy to the Rutherfordscattering experiment. In the Rutherford experiment,alpha particles were scattered from atoms and the scatteringwas consistent with a small structure in the atom containingthe positive charge.3. The largest charge quark is 2e/3, so a combination of onlytwo particles, a quark and an antiquark forming a meson,could not have an electric charge of 2e. Only particlescontaining three quarks, each with a charge of 2e/3, cancombine to produce a total charge of 2e.5. Until about 700 000 years after the Big Bang, the temperatureof the Universe was high enough for any atoms thatformed to be ionized by ambient radiation. Once theaverage radiation energy dropped below the hydrogenionization energy of 13.6 eV, hydrogen atoms could formand remain as neutral atoms for relatively long period oftime.7. In the quark model, all hadrons are composed of smallerunits called quarks. Quarks have a fractional electric1charge and a baryon number of 3. There are six flavors ofquarks: up (u), down (d), strange (s), charmed (c), top(t), and bottom (b). All baryons contain three quarks,and all mesons contain one quark and one antiquark. Section30.12 has a more detailed discussion of the quarkmodel.9. Baryons and mesons are hadrons, interacting primarilythrough the strong force. They are not elementary particles,being composed of either three quarks (baryons) ora quark and an antiquark (mesons). Baryons have a1 3nonzero baryon number with a spin of either2or2.Mesons have a baryon number of zero and a spin of either0 or 1.11. All stable particles other than protons and neutrons havebaryon number zero. Since the baryon number must beconserved, and the final states of the kaon decay containno protons or neutrons, the baryon number of all kaonsmust be zero.13. Yes, but the strong interaction predominates.15. Unless the particles have enough kinetic energy to producea baryon–antibaryon pair, the answer is no. Antibaryonshave a baryon number of 1, baryons have abaryon number of 1, and mesons have a baryon numberof 0. If such an interaction were to occur and produce abaryon, the baryon number would not be conserved.17. Baryons and antibaryons contain three quarks, whilemesons and antimesons contain two quarks. Quarks havea spin of 1/2; thus, three quarks in a baryon can onlycombine to form a net spin that is half-integral. Likewise,two quarks in a meson can only combine to form a netspin of 0 or 1.19. For the first decay, the half-life is characteristic of thestrong interaction, so the 0 must have S 0, and strangenessis conserved. The second decay must occur via theweak interaction.PROBLEMS1. 1.1 10 16 fissions3. 126 MeV5. (a) 16.2 kg (b) 117 g7. 2.9 10 3 km (1 800 miles)9. 1.01 g811. (a) 4Be (b) 12 C (c) 7.27 MeV613. 3.07 10 22 events/yr15. (a) 3.44 10 30 J (b) 1.56 10 8 yr17. (a) 4.53 10 23 Hz (b) 0.622 fm

A.32 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems19. 10 23 s21. 10 18 m23. (a) conservation of electron-lepton number and conservationof muon-lepton number (b) conservation of charge(c) conservation of baryon number (d) conservation ofbaryon number (e) conservation of charge25.27. (a) (b) (c) e (d) e (e) (f) and e29. (a) not allowed; violates conservation of baryon number(b) strong interaction (c) weak interaction (d) weak interaction(e) electromagnetic interaction31. (a) not conserved (b) conserved (c) conserved (d) notconserved (e) not conserved (f) not conserved33. (a) charge, baryon number, L e , L (b) charge, baryonnumber, L e , L , L (c) charge, L e , L , L , strangenessnumber (d) charge, baryon number, L e , L , L , strangenessnumber (e) charge, baryon number, L e , L , L ,strangeness number (f) charge, baryon number, L e , L ,L , strangeness number35. 3.34 10 26 electrons, 9.36 10 26 up quarks, 8.70 10 26down quarks37. (a) (b) (c) K 0 (d) 41. a neutron, udd43. 70.45 MeV45. 18.8 MeV47. (a) electron-lepton and muon-lepton numbers not conserved(b) electron-lepton number not conserved(c) charge not conserved(d) baryon and electron-lepton numbers not conserved(e) strangeness violated by 2 units49. (a) 2 10 24 nuclei (b) 0.6 kg51. (a) 1 baryon before and zero baryons after decay. Baryonnumber is not conserved.(b) 469 MeV, 469 MeV/c(c) 0.999 999 4c53. (b) 12 days55. 26 collisions39.Net QuarksReaction At Quark Level (before and after) p : K 0 0ud uud : ds uds p : K ud uud : us uusK p : us uud :K K 0 us d s sss1 up, 2 down,0 strange3 up, 0 down,0 strange1 up, 1 down,1 strange(d) The mystery particle is a 0 or a 0 .

IndexPage numbers followed by “f”indicate figures; page numbersfollowed by “t” indicate tables; pagenumbers followed by “n” indicatefootnotes.AAberrationschromatic, 778–779, 778flens, 777–779, 778fmirror, 777–779spherical, 758, 758f, 778, 778f, 779Absorption, selective, polarization by,805–807, 805f, 806fAbsorption process, stimulated, 921,921fAbsorption spectrum, 905–906, 905fAcceptor atoms, 927fAccommodation, 821age-related reduction in, 822Action potentials, 612, 614fAir cleaner, electrostatic, 544Airplane, wing of, electrified,motional emf and, 669Alkali metals, 918–919Alpha decay, 949–950, 949fAlpha particles, 904Alternating current (AC) circuits,693–707capacitors in, 696–697, 696f, 719inductors in, 697–698, 697f, 698f,719power in, 702–704, 719resistors in, 693–696, 694f, 695ftransformer for, 705–707, 705f,707fAlternating current generators,673–674, 673f, 674f, 683Ampere, 569, 586definition of, 643Ampère, André-Marie, 641, 641fAmpère’s circuital law, 641, 650applied to solenoid, 648long straight wire and, 641–642,642f, 650Analyzer, 805Anderson, Carl, 985–986Anglecritical, 742, 746of deviation, 736polarizing, 807of refraction, 731, 731ffor glass, 733–734Angular magnificationof lens, 825, 826f, 837of telescope, 829, 838Antenna, electromagnetic waveproduction by, 709–710,710fAntineutrino, 951, 952, 966Antiparticle(s), 865, 985–986of electron, 945, 952Antiquarks, properties of, 994tApertures, single-slit and circular,resolution of, 831–836,831f, 832f, 833f, 838Apnea monitors, Faraday’s law and,666, 667fAppliances, consumer, third wire on,611–612, 611fAqueous humor, 820, 821fAstigmatism, 822Atmospheric electric fields, measuring,512Atmospheric refraction, 768–769,768fAtom(s)acceptor, 927donor, 927excited states of, 893, 921–922,921f, 922f, 932hydrogenBohr’s model of, 906–907, 906f,908fde Broglie waves and, 912–913,913fquantum mechanics and,913–915models ofearly, 903–904, 904fRutherford’s planetary, 904, 904fThomson’s, 904, 904fnucleus of, 904Atomic number of nuclei, 940, 965Atomic physics, 903–938atomic spectra in, 904–906, 905fatomic transitions in, 921–922, 921f,922fBohr theory of hydrogen and,906–912characteristic x-rays in, 920–921,920f, 932de Broglie waves and hydrogenatom and, 912–913, 913felectron clouds in, 916–917, 917fenergy bands in solids in, 924–927,925f, 926f, 927fexclusion principle in, 917–919lasers and, 922–924, 923f, 924f, 932periodic table and, 917–919Atomic physics (Continued)quantum mechanics and hydrogenatom and, 913–915spin magnetic quantum numberand, 915–916Atomic spectra, 904–906, 905fAtomic transitions, 921–922, 921f,922fAurora borealis, origin of colors in,906Automobilein lightning storm, 515Average power, in AC circuit, 703,719Axial myopia, 822Axons, 612–614, 613fBBacteria, magnetic, 627–628Bacterial growth projection, 606Balmer, Johann, 905Balmer series, 905, 905ffor hydrogen, 909–910, 909fBand(s)energy, in solids, 924–927, 925f,926f, 927fof energy levels, 925, 925fBand gap, 925, 925fBardeen, John, 929Baryon number, 989–990Baryons, 988, 988t, 1002quark composition of, 994f,985fBattery(ies)automobile, electric potentialdifference in, 537in direct current circuits, 592–601,593f, 594f, 596fBecquerel, 946Becquerel, Henri, 939Bednorz, J. Georg, 579Beta decay, 950–952, 951fBig Bang theory, 999–1000, 999f,1001fBinding energyof deuteron, 943–944of nucleus, 943–944, 944f,965Black hole, 868Blackbody radiation, 875–877, 875f,876f, 895Bohr, Niels, 907fBohr radius, 907–908I.1

I.2 IndexBohr theory, of hydrogen, 908–912modification of, 910–912, 912f,931Bohr’s correspondence principle,910, 931Bosons, W and Z, 985, 985tBradycardia, definition of, 585–586Bragg, W. H., 884Bragg, W. L., 884Bragg’s law, 884, 896Brattain, Walter, 929Bremsstrahlung, 881Brewster, Sir David, 808Brewster’s angle, 808Brewster’s law, 808, 812Bright fringes, 788–789, 788f, 811in single-slit diffraction, 799, 799fBrushes, in electric motors, 636f,637Bubble chamber, as track detector,965, 965fCCamera, 819–820, 820fCamera flash attachments, capacitorin, 546Capacitance, 545–548of circuit in resonance, 705equivalent, 553–554of parallel-plate capacitor, 545Capacitive reactance, 696–697, 719Capacitor(s), 545–560, 561in AC circuits, 696–697, 696f, 719applications of, 555, 558–559charged, energy stored in,554–556, 554f, 561charging/discharging of, in RCcircuit, 607–609combinations of, 548–554parallel, 548–550, 549f, 561series, 550–554, 551f, 561commercial designs for, 558, 558fdefinition of, 545with dielectrics, 556–560, 557f,558f, 560f, 561paper-filled, 559parallel-plate, 546–548problem-solving strategy for, 552Carbon dating, 948–951Cardiac pacemaker, 584–585Cardioverter defibrillators,implanted, 585–586, 585f,586tCarlson, Chester, 544Case ground, 611CAT scans, 960–962, 961f, 962fCatapult, space, motional emf and,668, 668fCharacteristic x-rays, 920–921, 920f,932Charge, conservation of, junctionrule and, 601, 615Charge carriers, 569in conductor, 570, 570fin current, 570–571, 570f, 586Chargingby conduction, 499, 499fby induction, 500, 500fCharm, 994–995, 1002Chromatic aberration, 778–779, 778fChromodynamics, quantum, 9996,1002Ciliary muscle, in accommodation,821Circuit(s)alternating current, 693–707 (Seealso Alternating current(AC) circuits)current measurements in, 572–573direct current, 592–623complex, Kirchhoff’s rules and,601–605, 601f, 603f, 604fhousehold, 609–610, 610fintegrated, in semiconductordevices, 930–931, 931fRC, 605–609, 606f, 607f, 615in resonance, capacitance of, 705resonance frequency of, 709RL, 680–682, 683–684RLC series, 699–702, 719resonance in, 704–705, 704fsymbols for, 548, 548fvoltage measurements in, 572–573Circuit breakers, 598in household circuits, 610, 610fCircuit diagram, 572–573, 572fCircuit elements, symbols for, 548,548fCircular aperture, limiting angle for,833, 838Cloud charger, as track detector, 965Coherent light sources, interferenceand, 786–787Collector of pnp transistor, 929, 930fColliders, 998Color charge, 9996Color force, 996, 1002between quarks, 996–997, 1002Compact disks (CDs)tracking information on,diffraction grating in, 803,803fusing interference to read, 796–797,796fCompound microscope, 827–829,827f, 837–838Compton, Arthur Holly, 885fCompton effect, 885–887, 885f, 896Compton shift, 885, 896Compton wavelength, 885, 886Computed axial tomography (CATscans), 960–962, 961f,962fComputer keyboard, capacitor in,546, 547fConcave mirrors, 757–759, 757f,758f, 759f, 761fimages formed by, 762–763Conductioncharging by, 499, 499fof electrical signals by neurons,612–614Conduction band, 925, 926fConductor(s), 499–500, 523charge carriers in, 570, 570fcharged, potentials and,541–542definition of, 499in electrostatic equilibrium,512–515, 523energy bands of, 925, 926ffield lines and, 514isolated, properties of, 512–513parallel, magnetic force between,643–644, 643fsuper, 579–580, 580fcritical temperatures for, 579tCones, 820–821Conservationof charge, junction rule and, 601,615of energyLenz’s law and, 671loop rule and, 602, 615Conservation laws, 989–991Constructive interference, 788–789,788f, 811in thin films, 792Converging lenses, 769, 770images formed by, 773–774,774fConvex mirrors, 759–765, 759f, 760f,761f, 762fimages formed by, 764Cornea, 820, 821fCosmic connection, 999–1001Coulomb, 499, 501definition of, 643Coulomb, Charles, 500–505, 501fCoulomb constant, 501, 523Coulomb’s law, 500–505, 523Critical angle, 742, 746Critical temperature, 579for superconductors, 579tCrystalline lens, 820, 821, 821fCrystalsdiffraction of x-rays by, 883–885,883f, 884f, 895–896liquid, 810–811, 810fCurie, 946Curie, Marie, 945, 945fCurie, Pierre, 945

Current, 568–573, 586direction of, 569, 569fdrift speed and, 570–571, 586induced, magnetic induction and,663in lightbulb, 569measurement of, in circuits,572–573, 572fmicroscopic view of, 570–572,586notations for, 695tfor resistors in series, 593rms, 694–696, 718in superconductors, 579Current loop(s)in electric motors, 636–637,636fmagnetic field of, 644–645,644ftorque on, 634–636, 634f, 635f,650Cutoff frequency of light, 878Cutoff wavelength, 879Cyclotron equation, 637, 638–639DDark fringes, 788–789, 788f, 811in single-slit diffraction, 799Datingcarbon, 952–955radioactive, 953–954Daughter nucleus, 949Davisson, C. J., 888Davisson-Germer experiment,888–889De Broglie, Louis, 887fDe Broglie waves, hydrogen atomand, 912–913, 913fDe Broglie’s wavelength, 888,896Decayalpha, 949–950, 949fbeta, 950–952, 951fexponential, 946, 946fgamma, 952neutron, 990processes of, 948–955, 949f, 951f,953f, 955fspontaneous, 949Decay constant, of radioactivematerial, 945–946,966Decay rate, 945, 966Defibrillator(s)capacitor in, 555–556implanted cardioverter, 585–586,585f, 586tDendrites, 612, 613fDeoxyribonucleic acid (DNA),double-helix structure of,884, 884fDepolarization wave, 583, 584fDepth of field, of camera, 820Destructive interference, 788f, 789,811in single-slit diffraction, 799in thin films, 792–793Deuterium-deuterium reaction,982Deuteron, binding energy of,943–944Deviation, angle of, 736Dielectric constant, 556, 557t, 561Dielectric strength, 557, 557tDielectricsatomic description of, 559–560,560fcapacitors with, 556–560, 557f,558f, 560f, 561Diffraction, 727, 797–804single-slit, 798–800, 811of x-rays by crystals, 883–885, 883f,884f, 895–896Diffraction grating, 800–804, 801f,802f, 803f, 811–812in CD tracking, 803, 803fprism vs., 802resolving power of, 835–836, 838Diffraction pattern, Fraunhofer, 798,798fDiffuse reflection, 728, 728fDigital video disks (DVDs), usinginterference to read,796–797Diode, 574Diopters, 822–823, 837Dip angle, 627Dirac, Paul Adrien Maurice, 985,985fDirect current circuitscomplex, Kirchhoff’s rules and,601–605, 601f, 603f, 604fDirect current generators, 674–675,675fDispersionof light, into spectrum, 738–739,738f, 739fprisms and, 736–738, 737f, 746Diverging lenses, 769, 770, 774–776,775fDNA (deoxyribonucleic acid),double-helix structure of,884, 884fDomains, magnetic, 648–649, 648f,649fDonor atom, 927Doping, in semiconductors, 927Doppler effect, for electromagneticwaves, 718Drift speed, 570–571, 586EIndex I.3Earth, magnetic field of, 626–628deflection of lightning strike by,633Eightfold way, 992–983, 983fEinstein, Albert, 843f, 849fon light quanta, 727mass-energy equivalence equationof, 861on photoelectric effect, 879special theory of relativity of, 844,849–858, 851f, 852f, 854f,855f, 856f, 868theory of gravitation of, 867–868EKGs (electrocardiograms),583–584, 584fElectric charge(s)conservation of, 499negative versus positive, 498, 498fproperties of, 497–499, 523Electric current, 568–573. See alsoCurrentElectric dipole, 510, 511fElectric field(s), 505–509, 523atmospheric, measuring, 512in atom smashers, 537–538of charged thin spherical shell,520–521, 520fof nonconducting plane sheet ofcharge, 521–522Electric field lines, 510–512, 510f,511f, 523conductors and, 514Electric flux, 517–519, 517f, 523Electric force(s)electric fields and, 506gravitational force and, 502properties of, 500on proton, 507Electric motors, 636–637, 636f,637fElectric potential, 531–542, 561created by point charge, 538–541,539ffinding, 540–541, 540fproblem-solving strategy for, 540Electric potential difference, 534,561Electric potential energychange in, 532work and, 531–535Electrical activityof heart, 583–586, 583f, 584fElectrical charges, quantized, 499Electrical energy, power and,580–583Electrical potential energy, 538–541,561Electrical resistivity, 587Electrical safety, 611–612

I.4 IndexElectrical signals, conduction of, byneurons, 612–614Electrical storms, driver safetyduring, 515Electrified airplane wing, motionalemf and, 669Electrocardiograms (EKGs),583–584, 584fElectromagnet, 646Electromagnetic force, 984, 1002Electromagnetic induction, 555Faraday’s law of, 663–667, 683Electromagnetic pump, medical usesof, 633, 633fElectromagnetic radiation, light and,887Electromagnetic waves, 693, 708,719–720Doppler effect for, 718intensity of, 712production of, by antenna,709–710, 710fproperties of, 710–715spectrum of, 715–717, 720Electron(s), 498antiparticle of, 945binding, nucleons and, 944charge and mass of, 501tin copper wire, drift speed of, 571locating, position-momentumuncertainty principle in,893mass number of, 951number of, in filled subshells andshells, 918trelativistic momentum of, 858, 868wavelength of, 889Electron clouds, 916–917, 917fElectron-lepton number, law ofconservation of, 990Electron microscope, 889–890,890fElectron volt, 542, 561Electronic configuration, forelements, 918–919, 919tElectrostatic air cleaner, 544Electrostatic equilibrium, conductorsin, 512–515, 523Electrostatic precipitator, 543–544,543fElectroweak force, 984, 1002Electroweak theory, 997–999, 998fElements, electronic configurationfor, 918–919, 919tEmfback, motors and, 676–677, 676finduced, 660–663, 661falternating current generatorand, 675–676magnetic flux and, 661–663,661f, 662fmagnetic induction and, 663Emf (Continued)motional, 667–670, 667f, 668f, 683self-induced, 677–680, 677f, 678f,684sources of, 592–593, 614Emissionspontaneous, 922, 922f, 932stimulated, 922, 922f, 932Emission spectrum, 905, 905fEmitter, of pnp transistor, 929, 930fEndothermic reactions, 958, 966Energybindingof deutron, 943–945of nucleus, 943–944, 944f, 965conservation ofjunction rule and, 601, 615Lenz’s law and, 671electrical, power and, 580–583electrical potential, 538–541, 561half-life and, 949of hydrogen atom, 907ionization, 908kinetic, 860, 869conversion of mass to, inuranium fission, 864–865mass and, equivalence of, 861of photon, 887potential (See Potential energy)relativistic momentum and,862–865rest, 861, 869solar, 713–714stored in charged capacitor,554–556, 554f, 561stored in magnetic field, 682–683threshold, 958, 966total, 861, 869of x-ray, estimating, 920Energy bands in solids, 924–927,925f, 926f, 927fEquationlens maker’s, 771mass-energy equivalence, 861mirror, 758–759, 779photoelectric effect, 879thin-lens, 79, 770–771, 779Equilibrium, electrostatic,conductors in, 512–515,523Equipotential surfaces, 542–543,561Equipotentials, 542–543, 543fEquivalence, principle of, 866Equivalent capacitance, 553–554Equivalent resistanceof parallel combination of resistors,596f, 597, 615of series combination of resistors,594, 600–601, 615Ether, luminiferous, 846Event horizon, 868Excited states, of atoms, 921–922,921f, 922f, 932Exclusion principle, 912, 917–919,932quark model and, 996Exothermic reactions, 958, 966Exponential decay, 946, 946fEye(s), 820–825, 821f, 822f, 823fcat’s, resolution in, 833conditions of, 821–823evolution of, sun and, 717Ff -number, of camera lens, 820, 837Far point, 821Faraday, Michael, 505, 661fexperiment of, on currentproduction by changingmagnetic field, 661, 661fice-pail experiment of, 514,514flaw of induction of, 663–667,683applications of, 665–666, 665f,666f, 667fmotional emf and, 667–670,667f, 668f, 683Farads, 545Farsightedness, 821–822, 822f,823–824, 837Femtometer, 941Fermi, 941Fermi, Enrico, 951, 952fFerromagnetic materials, 649Feynman, Richard P., 987, 987fFeynman diagram, 987, 987fFiber optics, 744–745Fibrillation, 584Field(s)electric, 505–509, 506f, 523 (Seealso Electric field(s))magnetic, 628–631, 649Earth’s, 626–628deflection of lightning strikeby, 633Lenz’s law and, 671Field forces, Coulomb force as, 502Filmssoap, interference in, 794thin, interference in, 792–796,792f, 793f, 794f, 795f,811wedge-shaped, interference in,795–796, 795fFingerprints, magnetic field patternsand, 626Fireball, primordial, observation ofradiation from, 1000–1001

Index I.5Fissionnuclear, 973–976, 974f, 1002uranium, 973–975conversion of mass to kineticenergy in, 864–865Fission fragments, 974Focal lengthfor concave mirror, 759, 759ffor lens, 769, 769fFocal pointfor concave mirror, 758–759,759ffor lens, 769, 770fForce(s)electricelectric fields and, 506gravitational force and, 502on proton, 507field, Coulomb force as, 502gravitational, electric force and,502in nature, fundamental, 984–985,1002Franklin, Benjamin, 498felectricity and, 498Fraunhofer diffraction pattern, 798,798fFresnel bright spot, 798Fringe shift, 837Fringes, interference, 787f, 788–789,788f, 811diffraction and, 798Newton’s rings as, 793–794, 793fin single-slit diffraction, 799,799fFrisch, Otto, 973Full-wave rectifier, 929, 929fFuses, 598in household circuits, 610Fusion, nuclear, 980–984, 1002in sun, 980–981, 980fGGalilean relativity, 844–845, 845fGamma decay, 952Gamma rays, 716f, 717, 952, 966Gamow, George, 1000fGas(es)identification of, usingspectrometer, 737noble, 918radon, activity of, 948Gauss, 629, 649Gauss, Karl Friedrich, 517Gauss’s law, 519–523, 523Geiger, Hans, 904, 939Geiger counter, 963–964, 963fGell-Mann, Murray, 993, 983fGenerator(s), 673–677, 683alternating current, 673–674, 673f,674f, 683induced emf in, 675–676direct current, 674–675, 675fVan de Graaff, 516–517, 516fGenetic damage, 959Geometric optics, ray approximationin, 728, 728fGermer, L. H., 888Glaser, D., 965Glashow, Sheldon, 997–998Glass, angle of refraction for, 733–734Gluons, 985, 985tin force between quarks, 996Goeppert-Mayer, Maria, 942fGoudsmit, Samuel, 912, 915Grand unification theory, 999Grating, diffraction. See DiffractiongratingGravitational force, 984–985, 1002electric force and, 502Gravitational potential energy,electric potential energyand, 533Gravitational property, of mass,865–866, 866fGravitons, 985, 985tGrimaldi, Francesco, 727Ground-fault interrupters (GFIs), 612Faraday’s law and, 665–666, 665f,666fGround state, 908Grounding, 500Guitar, electric, sound production by,Faraday’s law and, 666, 666fHHadrons, 988, 988t, 1002quark composition of, 994tHahn, Otto, 973Hale telescope, 830, 830fHalf-lifeenergy and, 949of radioactive substance, 946–948,966Half-wave rectifier, 929Halogens, 919Heart, electrical activity of, 583–586,583f, 584fHeisenberg, Werner, 891–892,891fHeliumdiscovery of, 906electronic arrangement in, 918singly ionized, 911Henry, 678Henry, Joseph, 678fHertz, Heinrich Rudolf, 708fon Maxwell’s predictions, 708–709Higgs boson, 998Holes, for valence band, 927, 927fHolography, 924, 924fHubble telescope, 831Huygens, Christian, 726, 727fHuygens’s principle, 739–742, 740f,741f, 746applied to reflection andrefraction, 740–742, 741fHydrogenBalmer series for, 909–910, 909fBohr theory of, 906–912modification of, 910–912, 912f,931electronic arrangement in, 917–918Hydrogen atomde Broglie waves and, 912–913,913fquantum mechanics and, 913–915,914tHyperopia, 821–822, 822f, 823–824,837IImage(s)formed by concave mirrors,762–763formed by converging lens,773–774, 774fformed by convex mirrors, 764formed by refraction, 765–768,779just resolved, 832, 832f, 838real, 754, 779virtual, 754, 779Image distance, 754, 779Image point, in concave mirror, 757,757f, 758Impedance, of RLC circuit, 700, 701t,719Implanted cardioverter defibrillators(ICDs), 585–586, 585f, 586tIncoherent light sources,interference and, 787Index of refraction, 732, 732t, 736f,745Induced emf, 660–663, 661falternating current generator and,675–676magnetic flux and, 661–663, 661f,662fmagnetic induction and, 663Induced voltages, 660–692Inductance, 678, 684calculating, 678, 679–680of solenoid, 679–680

I.6 IndexInductioncharging by, 500, 500felectromagnetic, 555Faraday’s law of, 663–667, 683Inductive reactance, 697, 719Inductor(s)in AC circuits, 697–698, 697f, 698f,719in RL circuits, 680–682Inertial property of mass, 865–866Infrared waves, 716–717, 716f, 717fInstruments, optical, 819–842. Seealso Optical instrumentsInsulator(s), 499–500definition of, 499, 523energy bands of, 925–926, 926fIntegrated circuit, in semiconductordevices, 930–9331, 931fIntensity, of electromagnetic wave, 712Interferencein light wavesconditions for, 786–787, 811constructive, 788–789, 788f, 811destructive, 788f, 789, 811Young’s double-slit, 787–791,787f, 788f, 789f, 811to read CDs and DVDs, 796–797,796ftelevision signal, 790in thin films, 792–796, 792f, 793f,794f, 795f, 811Newton’s rings and, 793–796,793fInterferometer, Michelson, 836–837,836fIonization energy, 908Iris, 820, 821, 821fIrradiation, of food and medicalequipment, 960Isotopes, of elements, 940, 965JJunction rule, 601, 601f, 602Junction transistor, in semiconductordevices, 929–930, 930fJust resolved images, 832, 832f, 838KKeyboard, computer, capacitor in,546, 547fKilby, Jack, 930Kilowatt-hour, 581, 587Kinetic energy, 860, 869conversion of mass to, in uraniumfission, 864–865Kirchhoff, Gustav, 602fKirchhoff’s rulesapplications of, 603–605, 603f,604fcomplex DC circuits and, 601–605,601f, 603f, 604f, 615problem-solving strategy for, 602LLand, E. H., 805Laser printers, 544, 545fLasers, 922–924, 923f, 924f, 932Laue pattern, 883Law(s)Bragg’s, 884, 896Brewster’s, 808, 812of conservation, 989–991of strangeness, 991Malus’s, 806of refraction, 732–736, 745Snell’s, 733, 745Wien’s displacement, 875, 895Lawson’s criterion, 982, 1002Length, proper, 856Length contraction, 856–858, 856f,868Lens(es)aberrations of, 777–779, 778fangular magnification of, 825, 826f,837camera, f-number of, 820, 837converging, 769, 770images formed by, 773–774, 774fcrystalline, 820, 821, 821fdiverging, 769, 770, 774–776, 775fin diving masks, vision and, 773power of, 823prescribing, 823–825thin, 769–777, 769f, 770f, 771f,772f, 774f, 775f, 776fcombinations of, 776–777, 776fray diagrams for, 771–776, 772f,774f, 775f, 776fsign conventions for, 771tLens maker’s equation, 771Lenz’s law, 664, 670–673, 683applications of, 672–673, 672fLepton number, 990–991Leptons, 988t, 989, 1002Light, 726–753dispersion of, into spectrum,738–739, 738f, 739fdual nature of, 887–890electromagnetic radiation and, 887as electromagnetic wave, 711–712laser, refraction of, in DVD,735–736nature of, 726–727, 745Light (Continued)passing through slab, 734–735photon theory of, 879reflection of, 727, 745refraction of, 727, 730–731, 731f,7452speed of, 845–846, 845f, 846fether wind theory of, 846, 846f,847fin fused quartz, 734Michelson-Morley experimenton, 846–849, 847f, 849fultraviolet, 716f, 717visible, 716f, 717wave theory of, 726–727Light sourcescoherent, interference and,786–787incoherent, interference and, 787measuring wavelength of, 790–791Light waveslinearly polarized, 805, 805fpolarization of, 804–811, 812unpolarized, 805, 805fLightbulb(s)aging, dimming of, 576brightness of, in various circuits,598–599Christmas, in series, 594–595, 594fcost of operating, 582current in, 569failure of, timing of, 581three-way, 599Lightning rods, as conductors, 514Lightning storm, driver safety during,515Lightning strike, deflection of, byEarth’s magnetic field, 633Liquid crystals, 810–811, 810fLithium, electronic arrangement in,918Lloyd’s mirror, 791, 791fLoad resistance, 593Loop rule, 601, 603f, 604fLuminiferous ether, 846MMagnet(s), 624–626permanent, 649superconducting, 579–580, 580fMagnetic bacteria, 627–628Magnetic declination, 627, 627fMagnetic domains, 648–649, 648f,649fMagnetic field(s), 628–631, 649of current loop, 644–645, 644ftorque on, 634–636, 634f, 635f,650

Index I.7Magnetic field(s) (Continued)Earth’s, 626–628deflection of lightning strike by,633energy stored in, 682–683Lenz’s law and, 671of long, straight wire, 640–643,640f, 641f, 642f, 650motion of charged particles in,637–640, 637f, 638f, 639f,650of solenoid, 646–648, 646fMagnetic field confinement,976–978, 977fMagnetic field lines, 625–626, 626fMagnetic fluxchange in, Faraday’s law ofinduction and, 663induced emf and, 661–663, 661f,662fMagnetic forceon current-carrying conductor,631–634, 631f, 632f, 633f,649between parallel conductors,643–644, 643fMagnetic materials, soft versus hard,625Magnetic resonance imaging (MRI),962–963, 963fMagnetism, 624–659Magnificationangularof lens, 825, 826f, 837of telescope, 829, 838of compound microscope,827–829, 827f, 837–838lateral, 755of mirror, 755, 779of refracting surface, 765, 779for thin lens, 770, 779Magnifier, simple, 825–827, 825f, 826fMalus’s law, 806Marsden, Ernest, 908, 939Massconversion of, to kinetic energy inuranium fission,864–865energy and, equivalence of, 861gravitational attraction of, for othermasses, 865–866, 866finertial property of, 865–866of nucleus, 940, 940tMass numberof electron, 951of nuclei, 940, 965Mass spectrometer, 638–639Materialselectrically charged, 498ferromagnetic, 649nonohmic, 574, 574fohmic, 574, 574fMaterials (Continued)optical activity of, 808–809, 809fresistivity of, 575–577, 576tMatter, radiation damage in, 959–960Maximain interference pattern of diffractiongrating, 801–802, 801fsecondary, 797Maximum angular magnification, oflens, 826–827, 837Maxwell, James Clerk, 707, 708fon electromagnetism, 707–708Hertz on, 708–709on light, 727Medical applications, of radiation,959–963, 961f, 962f, 963fMeitner, Lise, 973Mendeleev, Dmitri, 918Mesons, 986–988, 1002quark composition of, 994t, 995fMetal detectors, in airports as seriesresonance circuits, 704Meteoroids, light streaks of, 922Metric, 867Michelson, Albert A., 836experiment of, on speed of light,846–849, 847f, 849fMichelson interferometer, 836–837,836fMichelson–Morley experiment, onspeed of light, 846–849,847f, 849fMicroscopecompound, 827–829, 827f,837–838electron, 889–890, 890flimiting resolution of, 833–834scanning tunneling, 894–895,894f, 895fMicrowaves, 716, 716fpolarizing, 806Millikan, Robert Andrews, 499oil-drop experiment of, 515–516,515fMinima, 797–798Mirage, 768f, 769Mirror(s)concave, 757–759, 757f, 758f, 759f,761fconvex, 759–765, 759f, 760f, 761f,762fdiverging, 759flat, 754–757, 755f, 756f, 757f, 779Lloyd’s, 791, 791fray diagrams for, 760–762, 761frearview, day and night settings for,756–757, 756fsign conventions for, 759–760,760tspherical, images formed by,757–759Mirror equation, 758–759, 779Moderator, in nuclear reactor, 972Momentumof photon, 887–888relativistic, 858, 868energy and, 862–865Monitor, apnea, Faraday’s law and,666, 667fMorley, Edward W., experiment of,on speed of light, 846–849,847f, 849fMoseley, Henry G. J., 920Motion of proton between twocharged plates, 537Motional emf, 667–670, 667f, 668fMotorsback emf and, 676–677, 676felectric, 636–637, 636f, 637fMRI (magnetic resonance imaging),962–963, 963fMüller, K. Alex, 579Multimeter, digital, 572f, 573Muons, 853–854, 854f, 981Muscles, ciliary, in accommodation,821Myopia, 822, 822f, 824–825, 837Nn-type semiconductors, 927Nature, fundamental forces in,984–985, 1002Near point, 821Nearsightedness, 822, 823f, 824–825,837Ne’eman, Yuval, 993Neurons, conduction of electricalsignals by, 612–614Neutrino, 951, 952, 966Neutroncharge and mass of, 501tdiscovery of, 956Neutron capture, in nuclear reactors,998Neutron decay, 990Neutron energies, regulation of, innuclear reactors, 977–978Neutron leakage, in nuclear reactors,977Neutron number, of nuclei, 940Newton, Isaac, on time, 850Newtonian focus, 830Newton’s law of motionfirst, Galilean relativity and, 844second, motion of charged particlein magnetic field and, 637,650third, magnetic force between twoparallel conductors and,643

I.8 IndexNewton’s rings, 793–796, 793fNoble gases, 918Nonohmic material, 574, 574fNorth poleEarth’s geographic, 626of magnet, 624Noyce, Robert, 930npn transistor, in semiconductordevices, 929Nuclear fission, 973–976, 974f, 1002Nuclear force, 942Nuclear fusion, 980–984, 1002in sun, 980–981, 980fNuclear magnetic resonance, 962Nuclear physics, 939–972decay processes in, 948–955, 949f,951f, 953f, 955fnatural radioactivity in, 955, 955f,955tnuclear reactions in, 955–958, 966properties of nuclei in, 940–942,940t, 941f, 943fradioactivity in, 945–948, 945f,946f, 966Nuclear reactions, 955–958, 966Nuclear reactors, 976–980, 976f,977f, 978f, 1002safety of, 979–980Nucleus(i)of atom, 904atomic number of, 940, 965binding energy of, 943–944, 944f,965charge of, 940daughter, 949mass number of, 940, 965mass of, 940, 940tneutron number of, 940parent, 949properties of, 940–942, 940t, 941f,943fsize of, 941–942, 941fstability of, 942, 943fNumbers, right-hand rule, 640OObject distance, 754, 779Occhialini, Guiseppe P. S., 987Occupational radiation, exposurelimits for, 960Oersted, Hans Christian, 640, 640fOhm, Georg Simon, 574, 574fOhm-meter, 575, 587Ohmic material, 574, 574fOhms, 574, 586Ohm’s law, 574, 586Ohm’s law, for AC circuit, 697, 718Omega minus particle, 993Onnes, H. Kamerlingh, 579Open-circuit voltage, 593Optical activity, 808–809, 809fOptical instruments, 819–842camera as, 819–820, 820fcompound microscope as,827–829, 827f, 837–838Michelson interferometer as,836–837, 836ftelescope as, 829–831, 829f, 830f,831f, 838Opticsfiber, 744–745geometric, ray approximation in,728, 728fwave, 786–818Orbital magnetic quantum number,912, 932for hydrogen atom, 914, 914tOrbital quantum number, 912, 932for hydrogen atom, 914, 914tOrder numberof diffraction pattern, 801–802of fringe, 789, 811Oscillator, macroscopic, quantized, 877Pp-n junction, in semiconductordevices, 928–929, 928f,929fp-type semiconductors, 927Pacemakers, cardiac, 584–585Pair annihilation, 865, 865f, 869Pair production, 865, 865f, 869Parallel combination of capacitors,548–550, 549f, 561Parallel-plate capacitor, 546–548Parent nucleus, 949Particle(s)alpha, 904chargedmotion of, in magnetic field,637–640, 637f, 638f, 639f,650trapping of, by magnetic field, 638classification of, 988t, 998–999,1002elementary, 984motion of, perpendicular touniform magnetic field, 637strange, 991–992wave properties of, 887–888Particle physicsbeginning of, 986–988, 987fproblems and perspectives in, 1001Particle theory of light, 879Path difference, 788, 789fPauli, Wolfgang, 915, 918fPauli exclusion principle, 912,917–919, 932Penning trap, 638Penzias, Arno A., 1000, 1000fPeriodic table, 918–919Periscopes, submarine, internalreflection in, 743Permeability, of free space, 641,650Permittivity of free space, 546, 561PET (positron emissiontomography), 986Phase angle , for RLC circuit,699–700, 700t, 719Phasor diagram, for RLC circuit, 699,699f, 700fPhasors, in RLC circuit, 699Photocells, 880Photodielectric effect, 727Photoelectric effect, 877–878, 878f,895for sodium, 879–880Photoelectrons, 877, 878Photographic emulsion, as trackdetector, 965Photographs, flash, red eyes in, 729Photomultiplier (PM) tube, 964,964fPhoton(s), 277, 985, 985tenergy of, 879momentum of, 887–888virtual, 987, 987fPhoton theory, of light, 879Physicsatomic, 903–938 (See also Atomicphysics)nuclear, 939–972 (See also Nuclearphysics)quantum, 874–902 (See alsoQuantum physics)Pion, 987Planck, Max, 878fPlanck’s constant, 727, 745, 876–877,895Plane wave, 710–711Plasma confinement time, inthermonuclear reactor,982Plasma ion density, in thermonuclearreactor, 982pnp transistor, in semiconductordevices, 929, 930fPoint charge, electric potentialcreated by, 538–541,539fPolarization, 500capacitance and, 559–560, 560fof light waves, 804–811, 812by reflection, 807–808, 808fby scattering, 808, 808fby selective absorption, 805–807,805f, 806f

Polarizer, 805–806, 805f, 806–807Polarizing angle, 808, 812Polaroid, 805PolesEarth’s geographic, 626of magnet, 624Population inversion, 923Positron emission tomography (PET),986Positrons, 945, 952, 985–986, 1002Potential(s)action, 612, 614fcharged conductors and,541–542electric, 531–542, 561potential energy and, 536stopping, 878Potential difference, voltage as, 549Potential energydue to point charges, 538–541electricchange in, 532work and, 531–535gravitational, electric potentialenergy and, 533potential and, 536Powell, Cecil Frank, 987Powerin AC circuit, 702–704, 719average, in AC circuit, 703, 719converted by electric heater,582–583distribution of, to city, transformersin, 706–707, 707felectrical energy and, 580–583of lens, 823, 837level of, control on, in nuclearreactors, 978–979, 978fresolving, of diffraction grating,835–836, 838Power factor, in AC circuit, 703, 719Precipitator, electrostatic, 543–544,543fPresbyopia, 822Primary coil, 660Primordial fireball, observation ofradiation from, 1000–1001Principal axis, of mirror, 757–758,757f, 758fPrincipal quantum number, 912, 932Prism(s)diffraction grating vs., 802dispersion and, 736–738, 737f, 746Prism spectrometer, 736–737, 738fProblem-solving strategyfor capacitors, 552for electric forces and fields, 508for electric potential, 540for Kirchhoff’s rules, 602for resistors, 600for thin-film interference, 794Proper time, 852Proton-proton cycle, in sun, 974–975Proton(s), 498charge and mass of, 501telectric force on, 507motion of, between two chargedplates, 537motion of, in magnetic field,630–631, 631fPumps, electromagnetic, medicaluses of, 633, 633fPupil, 820Purkinje fibers, 583f, 584QQ values, 957–958Quantized charge, 499Quantum chromodynamics, 996,1002Quantum mechanicshydrogen atom and, 913–915, 914ttheory of, 891, 896Quantum number(s), 876for 2 p subshell, 916energy and, 908, 908forbital, 912, 932for hydrogen atom, 914, 914torbital magnetic, 912, 932for hydrogen atom, 914, 914tprincipal, 912, 932spin magnetic, 912, 915–916, 932Quantum physics, 874–902blackbody radiation in, 875–877,875f, 876f, 895Compton effect of, 885–887, 885f,896diffraction of x-rays by crystals in,883–885, 883f, 884f,895–896photoelectric effect and, 877–878,878f, 895photon theory of light and, 879Planck’s hypothesis in, 876–877scanning tunneling microscope in,894–895, 894f, 895funcertainty principle in, 891–893,892f, 896wave function in, 890–891, 896wave properties of particles in,887–888x-rays in, 880–885, 881f, 882f, 883f,884f, 895–896Quarks, 499n, 984, 996–997, 994t,995f, 995t, 996f, 997f, 1002color force between, 996–997, 1002colored, 996–997, 996f, 997f,1002force between, 996properties of, 994tRIndex I.9Radiationblackbody, 875–877, 875f, 876f,895electromagnetic, light and, 887medical applications of, 959–963,961f, 962f, 963fobservation of, from primordialfireball, 1000–1001occupational, exposure limits for,960thermal, 874–876color of stars and, 875from human body, 875–876Radiation damage, in matter,959–960Radiation detectors, 963–965,963–965, 963f, 963f, 964f,965fRadio, tuning of, series resonancecircuit and, 704Radio-wave transmission, 709–710Radio waves, 716, 716f, 717fRadioactive materialdecay constant of, 945–946, 966half-life of, 946–948, 966Radioactive tracing, 960Radioactivity, 945–948, 945f, 946f,966artificial, 955for carbon dating, 952–953natural, 955, 955f, 955tpractical uses of, 952–954Radiumactivity of, 947decaying, 950Radon detecting, radioactivity in, 953Radon gas, activity of, 948Rainbow, 738–739, 739f, 746Ray approximation, in geometricoptics, 728, 728fRay diagramsfor mirrors, 760–762, 761ffor thin lenses, 771–776, 772f,774f, 775f, 776fRayleigh’s criterion, 832, 838Rays, gamma, 716f, 717, 952, 966RBE (relative biologicaleffectiveness), 959, 959tRC circuits, 605–609, 606f, 607f, 615Reactancecapacitive, 696–697, 719inductive, 697, 719Reactorsfusion, 981–982nuclear, 976–980, 976f, 977f, 978f,1002Real image, 754, 779Rectifier, in semiconductor devices,929, 929f

I.10 IndexReflecting telescope, 829–830, 830fReflectionchange of phase due to, 791–792,791fdiffuse, 728, 728fHuygen’s principle applied to,740–742, 741fof light, 727, 745polarization by, 807–808, 808fspecular, 728–729, 728ftotal internal, 742–745, 746Refracting surfacesflat, 766–768, 766f, 767fsign conventions for, 766tRefracting telescope, 829, 830, 830fRefractionangle of, 731, 731ffor glass, 733–734atmospheric, 768–769, 768fHuygens’s principle applied to,740–742, 741fimages formed by, 765–768, 779index of, 732, 732t, 736f, 745of laser light in DVD, 735–736law of, 732–736, 745Snell’s, 733, 745of light, 727, 730–731, 731f, 745Refractive myopia, 822Relative biological effectiveness(RBE), 959, 959tRelativistic energy, 860–861Relativistic momentum, 858, 868energy and, 862–865Relativity, 843–873Einstein’s principle of, 849–850Galilean, principle of, 844–845,845fgeneral, 865–868, 866f, 867fspecial theory of, 844, 849–858,851f, 852f, 854f, 855f, 856f,868consequences of, 850–858length contraction and,856–858, 856f, 868simultaneity and relativity of timeand, 850–851, 851ftime dilation and, 851–855,852f, 854f, 868twin paradox and, 855–856, 855fRem, 959Resistance, 573, 586–587equivalentof parallel combination ofresistors, 596f, 597,600–601, 615of series combination ofresistors, 594, 615load, 593of nichrome wire, 576–577of steam iron, 575temperature variation of, 577–579Resistance thermometer, platinum,578–579Resistivity, 575–577, 576t, 586electrical, 587temperature coefficient of, 578, 587Resistor(s), 574, 575fin AC circuits, 693–696, 694f, 695fin parallel, 596–601, 596f, 597f,598f, 599fin three-way lightbulb, 599power delivered to, 587power dissipated by, 580–581problem-solving strategy for, 600in series, 593–596, 594fResolving power, of diffractiongrating, 835–836, 838Resonancecircuit in, capacitance of, 705in RLC series circuit, 704–705,704fResonance frequency, 704of circuit, 709Resonators, 876Rest energy, 861, 869Retina, 820–821, 821fRetroreflection, 729Richter, Burton, 995Right-hand rulenumber 1, 629, 629f, 649number 2, 640, 640fRL circuits, 680–682, 683–684RLC series circuit, 699–702, 719resonance in, 704–705, 704frms current, 694–696, 718rms voltage, 695, 718Roadway flashers as RC circuits,606–607Rods, 820Roentgen, Wilhelm, 880–881Rubbia, Carlo, 998Rutherford, Ernest, 904, 939, 940f,941, 955–956planetary model of, of atom, 904,904fRydberg constant, 905SSafetyelectrical, 611–612reactor, 979–980Salam, Abdus, 997–998Scalar quantity, electric potential as,536, 539Scanning tunneling microscope,894–895, 894f, 895fScattering, polarization of light wavesby, 808, 808fSchrödinger, Erwin, 890–891, 891fSchwarzschild radius, 868Scintillation counter, 964, 964fSecondary coil, 660Self-induced emf, 677–680, 677f,678f, 684Self-inductance, 677–680, 677f, 678f,684Self-sustained chain reaction, innuclear reactor, 977Semiconductor(s), 499energy bands of, 926–927, 926f,927fSemiconductor devices, 928–929,928f, 929f, 930f, 931fjunction transistor in, 929–930,930fp-n junction in, 928–929, 928f,929fSemiconductor diode detector,964Series combination of capacitors,550–554, 551f, 561Shells, 912, 912tnumber of electrons in, 918tShockley, William, 929SI unit(s)for capacitance, 545for change in electric potentialenergy, 532for charge, 643for current, 569, 586, 643for electric charge, 499, 501for electric potential differencebetween two points, 536for electrical field, 505for emf, 592for inductance, 678for magnetic field, 629for magnetic flux, 661for radioactivity, 946for resistance, 574for resistivity, 587Sign conventions for mirrors,759–760, 760tSimultaneity, relativity of time and,850–851, 851fSingularity, 868Slit, limiting angle for, 832–833,838Smoke detectors, radioactivematerials in, 953, 953fSnell’s law, of refraction, 733, 745Sodium, photoelectric effect for,879–880Solar cells, nonreflective coatings for,794–795, 794fSolar energy, 713–714SolenoidAmpère’s law applied to, 648inductance of, 679–680magnetic field of, 646–648, 646fSolids, energy bands in, 924–927,925f, 926f, 927fSomatic damage, 959Sommerfeld, Arnold, 912Sound waves, diffraction of, 800

South poleEarth’s geographic, 626of magnet, 624Space catapult, motional emf and,668, 668fSpace travellength contraction and, 856–858,856fspecial theory of relativity and,855–856, 855fSpacetime, curvature of, 867Spectral lines, 737Spectrometerdiffraction grating, 802, 802fmass, 638–639prism, 736–737, 738fSpectrum(a), 736, 736fabsorption, 905–906, 905fatomic, 905–906, 905fdispersion of light into, 738–739,738f, 739felectromagnetic, 715–717, 720emission, 905, 905fSpecular reflection, 728–729, 728fSpeed(s)drift, 570–571, 586of lightether wind theory of, 846, 846f,847fMichelson-Morley experiment,846–849, 847f, 849fSpherical aberration, 758, 758f, 778,778f, 779Spin magnetic quantum number,912, 915–916, 932Spontaneous decay, 949Spontaneous emission, 922, 922f, 932Standard Model, 998Stars, colors of, 875Step-down transformer, 706Step-up transformer, 706Stimulated absorption process, 921,921f, 932Stopping potential, 878Storms, electrical, driver safetyduring, 515Strange particles, 991–992Strangeness, 991–992, 1002Strassman, Fritz, 973Strong force, 984, 1002Stud finders, capacitors in, 558Submarine periscopes, internalreflection in, 743Subshells, 912, 912tnumber of electrons in, 918tSun, fusion in, 980–981, 980fSuperconductors, 579–580, 580fcritical temperatures for, 579tSuperposition principle, 503in calculating electric field, 506electric potential for two or morecharges and, 539Symmetry breaking, 998TTachycardia, definition of, 585Tape recorders, Lens’s law and,672–673, 672fTelecommunications, fiber optics in,745Telescope, 829–831, 829f, 830f, 831f,838resolution of, 834Television signal interference, 790Temperature, critical, 579for superconductors, 579tTemperature coefficientsof resistivity, 578, 587for various materials, 576tTemperature variationof resistance, 577–579Tesla, 629, 649Tesla, Nikola, 701fThermal radiation, 874–876colors of stars and, 875from human body, 875–876Thermometer, resistance, platinum,578–579Thermonuclear fusion reactions, 981Thin films, interference in, 792–796,792f, 793f, 794f, 795f, 811Thin-lens equation, 770–771, 779Thin lenses, 769–777, 769f, 770f,771f, 772f, 774f, 775f, 776fcombinations of, 776–777, 776fray diagrams for, 771–776, 772f,774f, 775f, 776fsign conventions for, 771tThomson, G. P., 888Thomson, Sir Joseph John, 904fmodel of, of atom, 904, 904fThree Mile Island, 979Threshold energy, 958, 966Threshold voltage, 881Timeproper, 852relativity of, simultaneity and,850–851, 851fTime constant, 605, 615for RL circuit, 680, 681–682, 684Time dilation, 851–855, 852f, 854f,868Ting, Samuel, 995Tokamak, 982–983, 983fTomographycomputed axial, 960–962, 961f, 962fpositron emission, 986Torque, on current loop, 634–636,634f, 635f, 650Total energy, 861, 869Tracing, radioactive, 960Track detectors, 964–965, 965fTransformer, AC, 705–707, 705f, 707fTransistor, junction, in semiconductordevices, 929–930, 930fIndex I.11Transmission axis, 805, 805fTransmission electron microscope,889–890, 890fTwin paradox, 855–856, 855fUUhlenbeck, George, 912, 915Ultraviolet catastrophe, 876Ultraviolet light, 716f, 717Uncertainty principle, 891–893, 892f,896Unified mass unit, definition of, 940Uranium fission, 973–975conversion of mass to kineticenergy in, 864–865VValence band, 925, 926, 926fholes for, 927, 927fVan de Graaff, Robert J.generator of, 516–517, 516fVan der Meer, Simon, 998Velocity, relativistic addition of,859–860, 859f, 868–869Virtual image, 754, 779Virtual photon, 987, 987fVisible light, 716f, 717Volt, 592electron, 542Voltage(s)induced, 660–692measurement of, in circuits,572–573notations for, 695topen-circuit, 593potential difference as, 549rms, 695, 718threshold, 881Von Laue, Max, 881, 883WWave(s)de Broglie, hydrogen atom and,912–913, 913felectromagnetic, 693, 708, 719Doppler effect for, 718intensity of, 712production of, by antenna,709–710, 710fproperties of, 710–715spectrum of, 715–717, 720

I.12 IndexWave(s) (Continued)infrared, 716–717, 716f, 717flightlinearly polarized, 805, 805fpolarization of, 804–811unpolarized, 805, 805fplane, 710–711radio, 716, 716f, 717fsound, diffraction of, 800Wave front, 728, 728fin Huygens’ construction,739–740, 740fWave function, 890–891, 896Wave optics, 786–818Wave properties, of particles,887–888Wave theory, of light, 726–727WavelengthCompton, 885, 886cutoff, 879de Broglie’s, 888, 896of baseball, 889of electron, 889of light source, measuring, 790–791Wavelets, 740Weak force, 984, 1002Weber, 661Weber per square meter, 629Weinberg, Steven, 997–998Wheeler, John, 867Wien’s displacement law, 875, 895Wilson, Robert W., 1000, 1000fWindshield wipers, timed, as RCcircuits, 606Wirelevitating, 644long, straightAmpère’s law and, 641–642,642f, 650magnetic field of, 640–643, 640f,641f, 642f, 650magnetic force on, origin of, 632,632fWire chamber, as track detector,965Work, electric potential energy and,531–535Work function, of metal, 879, 879t,895XX-rays, 716f, 717, 717f, 805–896,880–885, 881f, 882f, 883f,884fcharacteristic, 920–921, 920f,932X-rays (Continued)diffraction of, by crystals, 883–885,883f, 884f, 895–896energy of, estimating, 920scattering of, 886–887in study of work of master painters,882Xerography, 544, 545fYYerkes Observatory, 830Young, Thomas, 727double-slit interferenceexperiment of, 787–791,787f, 788f, 789f, 811Yukawa, Hideki, 986–987, 986fZZeeman effect, 912Zonules, in accommodation,821Zweig, George, 993

CreditsPhotographsThis page constitutes an extension of the copyright page.We have made every effort to trace the ownership of all copyrightedmaterial and to secure permission from copyrightholders. In the event of any question arising as to the use ofany material, we will be pleased to make the necessary correctionsin future printings. Thanks are due to the followingauthors, publishers, and agents for permission to use thematerial indicated.Chapter 15. 497: © Keith Kent/Science PhotoLibrary/Photo Researchers, Inc. 498: © American PhilosophicalSociety/AIP 501: top, © 1968 Fundamental Photographs;bottom, Photo courtesy of AIP Niels Bohr Library, E. ScottBarr Collection 510: Photo courtesy of Harold M. Waage,Princeton University 511: top, Photo courtesy of Harold M.Waage, Princeton University; bottom, Photo courtesy ofHarold M. Waage, Princeton University 513: top, Photo courtesyof Harold M. Waage, Princeton University; middle right,Photo courtesy of Harold M. Waage, Princeton UniversityChapter 16. 531: Courtesy of Resonance Research Corporation543: bottom middle, Riei O’Harra/Black Star/PNI; bottomright, Greig Cranna/Stock, Boston/PNI 546: Photo courtesyof Harold M. Waage, Princeton University 557: © LorenWinters/Visuals Unlimited 558: a. Paul Silverman/FundamentalPhotographs; b. George SempleChapter 17. 568: Telegraph Colour Library/FPG/GettyImages 572: Michael Dalton, Fundamental Photographs574: © Bettmann/CORBIS 575: Courtesy of Henry Leap andJim Lehman 578: Courtesy of Central Scientific Company580: Courtesy of IBM Research Laboratory 585: Courtesy ofMedtronic, Inc.Chapter 18. 592: © Lester Lefkowitz/Corbis 593: GeorgeSemple 602: AIP ESVA, W.F. Meggers Collection 610: GeorgeSemple 613: Juergen Berger, Max-Planck Institute/SciencePhoto Library/Photo Researchers, Inc. 616: middle left,Superstock; middle right, Courtesy of Henry Leap and JimLehmanChapter 19. 624: Johnny Johnson/Stone/Getty 625: Courtesyof Central Scientific Company 626: a–c. Henry Leap andJim Lehman 631: Courtesy of Henry Leap and Jim Lehman637: Courtesy of American Honda Motor Co., Inc. 640: middleleft, North Wind Picture Archives; bottom, © Richard Megna,Fundamental Photographs 641: Leonard de Selva/CORBIS645: © Richard Megna, Fundamental Photographs 646: Courtesyof Henry Leap and Jim Lehman 650: © Loren Winters/Visuals Unlimited 651: top right, Courtesy of Central ScientificCompany; middle right, Courtesy of Central ScientificCompanyChapter 20. 660: PhotoDisc/Getty Images 661: By kindpermission of the President and Council of the Royal Society666: Charles D. Winters 667: Courtesy of PedsLink PediatricHealthcare Resources, Newport Beach, CA 674: LuisCastaneda/The Image Bank/Getty Images 678: North WindPicture Archives 681: Courtesy of Saab 688: bottom left, Courtesyof CENCOChapter 21. 693: © Bettmann/Corbis 701: Bettmann/Corbis 704: Ryan Williams/International Stock Photography707: George Semple 708: top, North Wind Photo Archives;bottom, Bettmann/Corbis 713: John Neal/Photo Researchers,Inc. 715: Ron Chapple/Getty Images 717: a. NASA/CXC/SAO; b. Palomar Observatory; c. WM Keck Observatory;d, VLA/NRAO 720: George SempleChapter 22. 726: Peter Bowater/Photo Researchers, Inc.727: Courtesy of Rijksmuseum voor de Geschiedenis der Natuurwetenschappen.Courtesy AIP Niels Bohr Library 728: bottomleft and right, Photographs Courtesy of Henry Leap andJim Lehman 729: top left and right, Photos by Charles D.Winters 731: top right, Courtesy of Henry Leap and JimLehman 735: bottom left, Courtesy of Sony Disc Manufacturing737: top right, David Parker/SPL/Photo Researchers, Inc.;bottom right, Courtesy of Bausch & Lomb 738: Courtesy ofPASCO Scientific 739: Mark D. Phillips/Photo Researchers,Inc. 740: Courtesy of Sabina Zigman/Benjamin CardozoHigh School 742: Courtesy of Henry Leap and Jim Lehman744: bottom left, Dennis O’Clair/Tony Stone Images/GettyImages; bottom right, Hank Morgan/Photo Researchers, Inc.Chapter 23. 754: Don Hammond/CORBIS 757: Courtesyof Henry Leap and Jim Lehman 759: Courtesy of Jim Lehman,James Madison University 761: top right, middle right, bottomright, Photos courtesy of David Rogers 762: © Junebug Clark1988/Photo Researchers, Inc. 768: John M. Dunay IV, FundamentalPhotographs, NYC 770: top left, Photos Courtesy ofHenry Leap and Jim Lehman 778: a–c. Photos by NormanGoldberg 780: middle right, Richard Megna/FundamentalPhotographs, NYC 784: Photo © Michael Levin/Opti-GoneAssociatesChapter 24. 786: RO-MA/Index Stock Imagery787: M. Cagnet, M. Francon, J. C. Thierr 788: RichardMegna/Fundamental Photographs, NYC 789: John S. Shelton792: Dr. Jeremy Burgess/Science Photo Library/Photo Researchers,Inc. 793: top right, Peter Aprahamian/SciencePhoto Library/Photo Researchers, Inc.; bottom middle, Courtesyof Bausch & Lomb Optical Company; bottom right, FromC.1

C.2 CreditsPhysical Science Study Committee, College Physics, LexingtonMass., D. C. Heath and Co., 1968 794: Richard Megna, FundamentalPhotographs 796: Courtesy of Sony Disc Manufacturing798: top left, Courtesy of P. M. Rinard, from Am. J. Phys.,44:70, 1976; top right, From M. Cagnet, M. Francon, and J. C.Thierr, Atlas of Optical Phenomena, Berlin, Springer-Verlag,1962, plate 18 803: © Kristen Brochmann/Fundamental Photographs806: a–c. Photos courtesy of Henry Leap 809: bottomleft, Sepp Seitz 1981; bottom right, Peter Aprahamian/Science Photo Library/Photo Researchers, Inc.Chapter 25. 819: © Denis Scott/CORBIS 821: top right,From Lennart Nilson, in collaboration with Jan Lindberg,Behold Man: A Photographic Journey of Discovery Inside theBody, Boston, Little, Brown & Co., 1974 827: bottom right,© Tony Freeman/Photo Edit 829: NASA 830: top right,© Tony Freeman/Photo Edit; bottom, Courtesy of PalomarObservatory/California Institute of Technology 832: a–c. Photosfrom M. Cagnet, M. Francon, and J. C. Thierr 833: FromM. Cagnet, M. Francon, and J. C. Thierr, Atlas of Optical Phenomena,Berlin, Springer-Verlag, 1962, plate 34Chapter 26. 843: Courtesy of the Archives, California Instituteof Technology 849: AIP Niels Bohr Library 861: LawrenceBerkeley Laboratory/Science Photo Library/Photo Researchers,Inc.Chapter 27. 874: © Eye of Science/Science Source/PhotoResearchers, Inc. 878: © Bettmann/CORBIS 881: Courtesy ofGE Medical Systems 884: Science Source/Photo Researchers,Inc. 885: Courtesy of AIP Niels Bohr Library 887: AIP NielsBohr Library 890: top right, © David Parker/Photo Researchers,Inc. 891: top right, AIP Emilio Segré VisualArchives; bottom right, Courtesy of the University of Hamburg894: IBM Corporation Research Division 902: John Chumack/Photo Researchers, Inc.Chapter 28. 903: Dembinsky Photo Associates 904: StockMontage, Inc. 907: top right, Princeton University/Courtesy ofAIP Emilio Segré Visual Archives 918: CERN/Courtesy of AIPEmilio Segré Visual Archives 923: Courtesy of HRL LaboratoriesLLC, Malibu, CA 924: top right, Courtesy of CentralScientific Company; middle left, Philippe Plailly/Photo Researchers,Inc. 931: top left, Courtesy of Texas Instruments,Inc.; top right, Courtesy of Intel CorporationChapter 29. 939: Courtesy of Public Service Electric andGas Company 940: North Wind Picture Archives 942: Courtesyof Louise Barker/AIP Niels Bohr Library 945: FPG International946: © Richard Megna/Fundamental Photographs952: National Accelerator Laboratory 953: HenryPaul/Gamma Liaison 955: Santi Visali/The IMAGE Bank962: Jay Freis/The Image Bank 963: top left, SBHA/GettyImages; top right, Scott Camazine/Science Source/PhotoResearchers, Inc.; bottom right, © Hank Morgan/PhotoResearchers 965: top left, Photo Researchers, Inc./SciencePhoto Library; top right, Dan McCoy/RainbowChapter 30. 973: Sandia National Laboratories/Photo byRandy Montoya 977: Courtesy of Chicago Historical Society980: NASA 983: top left, Courtesy of Princeton Plasma PhysicsLaboratory; top right, Courtesy of Princeton University 985:Courtesy AIP Emilio Segré Visual Archives 986: UPI/Corbis-Bettmann 987: © Shelly Gazin/CORBIS 993: Photo courtesy ofMichael R. Dressler 996: Courtesy of Fermi National AcceleratorLaboratory 998: Courtesy of CERN 1000: top left, Courtesyof AIP Emilio Segré Visual Archives; bottom left, AT&T BellLaboratoriesTables and IllustrationsThis page constitutes an extension of the copyright page.We have made every effort to trace the ownership of all copyrightedmaterial and to secure permission from copyrightholders. In the event of any question arising as to the use ofany material, we will be pleased to make the necessary correctionsin future printings. Thanks are due to the following authors,publishers, and agents for permission to use the materialindicated.Chapter 25. 821: Art by Robert Demarest, from Cecie Starrand Beverly McMillan, Human Biology 5th ed. Copyright© 2003, Brooks/Cole, a division of Thomson Learning, Inc.Chapter 27. 894–895: This section was written by Roger A.Freedman, University of California, Santa Barbara 895: Basedon a drawing from P. K. Hansma, V. B. Elings, O. Marti, andC. Bracker, Science 242:209, 1988, Copyright 1988 by theAAASChapter 28. 905: K. W. Whitten, R. E. Davis, M. L. Peck,and G. G. Stanley, General Chemistry, 7th ed., Belmont, CA,Brooks/Cole, 2004Chapter 30.1008: © 2005 Sidney Harris

PEDAGOGICAL USE OF COLORDisplacement andposition vectorsVelocity vectors (v)Velocity component vectorsForce vectors (F)Force component vectorsAcceleration vectors (a)Acceleration component vectorsTorque (t) andangular momentum(L) vectorsLinear or rotationalmotion directionsSpringsElectric fieldsMagnetic fieldsPositive charges +Negative charges –ResistorsBatteries and otherDC power supplies – +SwitchesCapacitorsInductors (coils)VoltmetersAmmetersLightbulbsAC sourcesGround symbolVALight raysObjectsLenses and prismsImagesMirrors

CONVERSION FACTORSLength1 m 39.37 in. 3.281 ft1 in. 2.54 cm1 km 0.621 mi1 mi 5280 ft 1.609 km1 light year (ly) 9.461 10 15 m1 angstrom (Å) 10 10 mMass1 kg 10 3 g 6.85 10 2 slug1 slug 14.59 kg1 u 1.66 10 27 kg 931.5 MeV/c 2Time1 min 60 s1 h 3600 s1 day 8.64 10 4 s1 yr 365.242 days 3.156 10 7 sVolume1 L 1000 cm 3 3.531 10 2 ft 31 ft 3 2.832 10 2 m 31 gal 3.786 L 231 in. 3Angle180° rad1 rad 57.30°1° 60 min 1.745 10 2 radSpeed1 km/h 0.278 m/s 0.621 mi/h1 m/s 2.237 mi/h 3.281 ft/s1 mi/h 1.61 km/h 0.447 m/s 1.47 ft/sForce1 N 0.2248 lb 10 5 dynes1 lb 4.448 N1 dyne 10 5 N 2.248 10 6 lbWork and energy1 J 10 7 erg 0.738 ft lb 0.239 cal1 cal 4.186 J1 ft lb 1.356 J1 Btu 1.054 10 3 J 252 cal1 J 6.24 10 18 eV1 eV 1.602 10 19 J1 kWh 3.60 10 6 JPressure1 atm 1.013 10 5 N/m 2 (or Pa) 14.70 lb/in. 21 Pa 1 N/m 2 1.45 10 4 lb/in. 21 lb/in. 2 6.895 10 3 N/m 2Power1 hp 550 ft lb/s 0.746 kW1 W 1 J/s 0.738 ft lb/s1 Btu/h 0.293 W

PHYSICAL CONSTANTSQuantity Symbol Value SI unitSpeed of light in vacuum c 3.00 10 8 m/sPermittivity of free space 0 8.85 10 12 C 2 /N m 2Coulomb constant, 1/4 0 k e 8.99 10 9 N m 2 /C 2Permeability of free space 0 1.26 10 6 T m/A(4 10 7 exactly)Elementary charge e 1.60 10 19 CPlanck’s constant h 6.63 10 34 J sh/2 1.05 10 34 J sElectron mass m e 9.11 10 31 kg5.49 10 4 uProton mass m p 1.672 65 10 27 kg1.007 276 uNeutron mass m n 1.674 95 10 27 kg1.008 665 uAvogadro’s number N A 6.02 10 23 mol 1Universal gas constant R 8.31 J/mol KBoltzmann’s constant k B 1.38 10 23 J/KStefan-Boltzmann constant 5.67 10 8 W/m 2 K 4Molar volume of ideal gas at STP V 22.4 L/mol2.24 10 2 m 3 /molRydberg constant R H 1.10 10 7 m 1Bohr radius a 0 5.29 10 11 mElectron Compton wavelength h/m e c 2.43 10 12 mGravitational constant G 6.67 10 11 N m 2 /kg 2Standard free-fall acceleration g 9.80 m/s 2Radius of Earth (at equator) R E 6.38 10 6 mMass of Earth M E 5.98 10 24 kgRadius of Moon R M 1.74 10 6 mMass of Moon M M 7.36 10 22 kgThe values presented in this table are those used in computations in the text. Generally, the physical constants areknown to much better precision.

PERIODIC TABLE OF THE ELEMENTSGroupHI1.00801GroupIITransition elements1s 1 1s 1 1s 23d 7 4s 2 6 2 2s 1 2s 2Atomic mass †2p40.082p 2 2p 3 2p 4 2p 5 2p 6Li 3 Be 4B 5 C 6 N 7 O 8 F 9 Ne 106.94 Na 22.99 3s 1 9.012Mg 24.313s 2Symbol Ca 4s 220 Atomic numberElectron configuration10.81 Al 26.98 3p 1 12.011 Si 28.09 3p 2 14.007 P 30.97 3p 3 15.999 S 32.06 3p 4 18.998 Cl 35.453 3p 5 20.18Ar 39.9483p 611 1213 14 15 16 17 184s 1 4s 2 3d 1 4s 2 3d 2 4s 2 3d 3 4s 2 3d 5 4s 1 3d 5 4s 2 3d 6 4s 2 3d 8 4s 2 3d 10 4s 1 3d 10 4s 2 4p 1 4p 2 4p 3 4p 4 4p 5 4p 6K 19 Ca 20 Sc 21 Ti 22 V 23 Cr 24 Mn 25 Fe 26 Co 27 Ni 28 Cu 29 Zn 30 Ga 31 Ge 32 As 33 Se 34 Br 35 Kr 3639.102 40.08 44.96 47.90 50.94 51.996 54.94 55.85 58.93 58.71 63.54 65.37 69.72 72.59 74.92 78.96 79.91 83.805s 1 5s 2 4d 1 5s 2 4d 2 5s 2 4d 4 5s 1 4d 5 5s 1 4d 5 5s 2 4d 7 5s 1 4d 8 5s 1 e 4d 10 4d 10 5s 1 4d 10 5s 2 5p 1 5p 2 5p 3 5p 4 5p 5 5p 6Rb 37 Sr 38 Y 39 Zr 40 Nb 41 Mo 42 Tc 43 Ru 44 Rh 45 Pd 46 Ag 47 Cd 48 In 49 Sn 50 Sb 51 T 52 I 53 Xe 5485.47 87.62 88.906 91.22 92.91 95.94 (99) 101.1 102.91 106.4 107.87 112.40 114.82 118.69 121.75 127.60 126.90 131.306s 1 6s 25d 2 6s 2 5d 3 6s 2 5d 4 6s 2 5d 5 6s 2 5d 6 6s 2 5d 7 6s 2 5d 9 6s 1 5d 10 6s 1 5d 10 6s 2 6p 1 6p 2 6p 3 6p 4 6p 5 6p 6Cs 55 Ba 56 57-71* Hf 72 Ta 73 W 74 Re 75 Os 76 Ir 77 Pt 78 Au 79 Hg 80 Tl 81 Pb 82 Bi 83 Po 84 At 85 Rn 86132.91 137.34178.49 180.95 183.85 186.2 190.2 192.2 195.09 196.97 200.59 204.37 207.2 208.98 (210) (218) (222)7s 1 7s 26d 2 7s 2 6d 3 7s 25d 1 6s 2 5d 1 4f 1 6s 2 4f 3 6s 2 4f 4 6s 2 4f 5 6s 2 4f 7 6s 2 5d 1 4f 7 6s 2 5d 1 4f 8 6s 2 4f 10 6s 2 4f 11 6s 2 4f 12 6s 2 4f 13 6s 2 4f 14 6s 2 5d 1 4f 14 Fr 87 Ra 88 89-103** Rf 104 Db 105 Sg 106 Bh 107 Hs 108 Mt 109 Ds 110 †† 111 †† 112†† 114(223) (226)(261) (262) (263) (262) (265) (266) (271) (272) (285)(289)*Lanthanide series La 138.91 57 Ce 140.12 58 Pr 140.91 59 Nd 144.24 60 Pm (147) 61 Sm 150.4 62 Eu 152.0 63 Gd 157.25 64 Tb 158.92 65 Dy 162.50 66 Ho 164.93 67 Er 167.26 68 Tm 168.93 69 Yb 173.04 70 Lu 174.97716d 1 7s 2 6d 2 7s 2 5f 2 6d 1 7s 2 5f 3 6d 1 7s 2 5f 4 6d 1 7s 2 5f 6 6d 0 7s 2 5f 7 6d 0 7s 2 5f 7 6d 1 7s 2 5f 8 6d 1 7s 2 5f 10 6d 0 7s 2 5s 11 6d 0 7s 2 5f 12 6d 0 7s 2 5f 13 6d 0 7s 2 6d 0 7s 2 6d 1 7s 2**Actinide series Ac 89 Th 90 Pa 91 U 92 Np 93 4f 6 6s 2 Pu 94 Am 95 Cm 96 Bk 97 Cf 98 Es 99 Fm 100 Md 101 No 102 Lr 6s 2103(227) (232) (231) (238) (239) (239) (243) (245) (247) (249) (254) (253) (255) (255) (257)GroupIIIGroupIVGroupVGroupVIGroupVIIH1.00801Group0He4.00262Note: Atomic mass values given are averaged over isotopes in the percentages in which they exist in nature.† For an unstable element, mass number of the most stable known isotope is given in parentheses.†† Elements 111–114 have not yet been named.

Quantum Physics

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