Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition

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vTable of ContentsCHAPTER 2–Special Theory of Relativity1CHAPTER 3–The Experimental Basis of Quantum Physics28CHAPTER 4–Structure of the Atom46CHAPTER 5–Wave Properties of Matter and Quantum Mechanics I62CHAPTER 6–Quantum Mechanics II78CHAPTER 7–The Hydrogen Atom100CHAPTER 8–Atomic Physics114CHAPTER 9–Statistical Physics123CHAPTER 10–Molecules, Laser, and Solids142CHAPTER 11–Semiconductor Theory and Devices160CHAPTER 12–The Atomic Nucleus169CHAPTER 13–Nuclear Interactions and Applications187CHAPTER 14–ParticlePhysics201CHAPTER 15–General Relativity212CHAPTER 16–Cosmology and Modern Astrophysics–The Beginning and the End221

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Chapter 2Special Theory of Relativity1Chapter 21.For a particle Newton’s second law says222222ˆˆˆd xd yd zFmamijkdtdtdt.Take the second derivative of each of the expressions in Equation (2.1):222222222222d xd xdd ydd zdtdtdtdtdtztyd . Substitution into the previous equation gives222222ˆˆˆ.d xddFmamijkyzFdtdtdt2.From Equation (2.1)ˆˆˆdxdydzpmijkdtdtdt.In a Galilean transformationdxdxdydydzdzvdtdtdtdtdtdt.Substitution into Equation (2.1) givesˆˆˆdxdydzpmv ijkpdtdtdt.However, becauseˆˆˆdxdydzpmijkdtdtdt the same form is clearly retained, giventhe velocity transformationdxdxvdtdt .3.Using the vector triangle shown, the speed of light coming toward the mirror is22cvand the same on the return trip. Therefore the total time is22222distancespeedtcv.Notice thatsinvc , so1sinvc.

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2Chapter 2Special Theory of Relativity4.As in Problem 3,12sin/vv , so11120.350 m/s1.25sin(/)sin1m/.s6 3vvand222221(1.25/s)(0.35mm/s)1.20 m/svvv.5.When the apparatus is rotated by 90°, the situation is equivalent, except that we haveeffectively interchanged1and2. Interchanging1and2in Equation (2.3) leads toEquation (2.4).6.Letn= the number of fringes shifted; thendn. Becausedctt , we have2122cttvnc . Solving forvand noting that1+2= 22 m,98120.00558910m3.0010m/s3.47 km/s.22 mnvc7.Letting2111(where/vc ) the text equation (not currently numbered) for1tbecomes2111221211tccwhich is identical to2twhen12so0t as required.8.Since the Lorentz transformations depend onc(and the fact thatcis the same constantfor all inertial frames), different values ofcwould necessarily lead two observers todifferent conclusions about the order or positions of two spacetime events, in violation ofpostulate 1.9.Let an observer in K send a light signal along the +x-axis with speedc. According to theGalilean transformations, an observer inKmeasures the speed of the signal to bedxdxvcvdtdt . Therefore the speed of light cannot be constant under the Galileantransformations.10.From the Principle of Relativity, we know the correct transformation must be of the form(assumingyyandzz):;xaxbtxaxbt.The spherical wave front equations (2.9a) and (2.9b) give us:;()()ctacb tctacb t.Solve the second wave front equation fortand substitute into the first:

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Chapter 2Special Theory of Relativity3()()acbacb tctc or2222()().cacbacba cbNowvis the speed of the origin of thex-axis. We can find that speed by setting0x which gives0axbt, or//vx tba, or equivalentlyb=av. Substituting this intothe equation above for2cyields22222222ca ca vacv. Solving fora:2211./vacThis expression, along withb = av, can be substituted into the original expressions forxandxto obtain:;xxvtxxvtwhich in turn can be solved fortandtto complete the transformation.11.Whenvcwe find211, so:21xctxxctxvt ;2//1tx cttx ct ;21xctxxctxvt;2//1txcttxct.12.(a) First we convert to SI units: 95 km/h = 26.39 m/s, so88/(26.39 m/s) / 3.0010m/s8.810vc(b)87/(240m/s) / 3.0010m/s8.010vc(c)sound2.32.3330 m/svvso86/2.3330 m/s / 3.0010m/s2.510v c(d) Converting to SI units, 27,000 km/h = 7500 m/s, so85/(7500 m/s) / 3.0010 m/s2.510v c(e) (25 cm)/(2 ns) =81.2510 m/sso88/(1.2510 m/s) / 3.0010m/s0.42vc (f) 142281 10m/ 0.3510s2.85710 m/s, so88/(2.85710m/s) / 3.0010 m/s0.95vc 

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4Chapter 2Special Theory of Relativity13.From the Lorentz transformations2/ttv x c  . But0tin this case, sosolving forvwe find2/vctx. Inserting the values21/ 2tttac  and21xxxa, we find2/ 2/ 2cacvca . We conclude that the frameKtravelsat a speedc/2 in thex-direction. Note that there is no motion in the transversedirection.14.Try setting0xxv t  . Thus0/ 2xv tavac    . Solving forvwe find2vc , which is impossible. There is no such frameK.15.For the smaller values ofβwe use the binomial expansion1/22211/ 2.(a)2151/ 213.8710(b)2131/ 213.210(c)2121/ 213.1 10(d)2101/ 213.1 10(e)1/21/222110.421.10(f)1/21/222110.953.2016.There is no motion in the transverse direction, so3.5yzm.22115 / 3110.85 2m0.8010 / 3 m3xxvtc2295/00.82 m /8.910s3ttvxccc17.(a)222222883 m(5 m)(10 m)3.8610s3.0010m/sxyztc(b) With0.8 we find5 / 3 . Then5yy m,10zz m,8-85 3m2.4010m/s (3.8610s)10.4 m3xxvt  228885/3.8610s2.4010m/s3 m / 3.0010m/s51.0 ns3ttvx c 

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Chapter 2Special Theory of Relativity5(c)2222228910.4 m5 m10 m2.99410m/s51.010sxyztwhich equalscto within rounding errors.18.At the point of reflection the light has traveled a distance11Lv tc t . On the returntrip it travels22Lv tc t . Then the total time is12222222/1/LcLctttcvvc   .But from time dilation we know (with0proper time2/tLc) that0222/1/Lcttvc . Comparing these two results fortwe get022222/2/1/1LcLcvvccwhich reduces to22001/LLLvc. This is Equation (2.21).19.(a) With a contraction of 1%,220/0.991/LLvc. Thus221(0.99)0.9801.Solving for, we find0.14 or0.14 .vc(b) The time for the trip in the Earth-based frame is6185.0010 m1.1910s0.143.0010 m/sdtv .With the relativistic factor1.01 (corresponding to a 1% shortening of the ship’s length), the elapsed time on the rocketship is 1% less than the Earth-based time, or a difference of130.01 1.210s = 1.210s.20.The round-trip distance isd= 40 ly. Assume the same constant speedvcfor theentire round trip. In the rocket’s reference frame the distance is only21.dd Thenin the rocket’s frame of reference2240ly1distance1.time40 y40dvcyRearranging21vc. Solving forwe find0.5 , or0.50.71vcc.To find the elapsed timeton Earth, we know40t y, so2140y56.6 y.1tt21.In the muon’s frame02.2s.μTIn the lab frame the time is longer; see Equation (2.19):0TT . In the lab the distance traveled is009.5cmvTv Tc T , sincevc.

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6Chapter 2Special Theory of RelativityTherefore209.5 cm1cT, so29.5 cm12.2μsvcc. Now all quantities areknown except β. Solving for β we find41.410or41.410vc.22.Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distanceis8m3.8410. In the earth’s frame of reference the time is the distance divided byspeed, or83.8410m34,359 s11,176 m/sdtv. In the astronauts’frame the time elapsed is2/1ttt . The time difference is22111tttttt .Evaluating numerically25811,176 m/s34,359 s 112.410s.3.0010m/st 23.0TT , so we know that2215 / 31/vc . Solving forvwe find4 / 5.vc24.0/LLso clearly2 in this case. Thus22121/vcand solving forvwe find3.2cv25.The clocks’rates differ by a factor of221/1/vc . Becauseis very small we willuse the binomial theorem approximation21/ 2. Then the time difference is1tttttt . Using21/ 2and the fact that the time for the tripequals distance divided by speed,8262375 m/s3.0010m/s810m/ 2375 m/s2tt 81.6710s16.7 ns.t 26.(a)22424/1/3.5810km10.941.2210kmLLLvc (b)Earth’s frame:783.5810m/0.127 s0.943.0010m/stLvGolf ball’s frame:2/0.127 s10.940.0433 stt 

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Chapter 2Special Theory of Relativity727.Spacetime invariant (see Section 2.9):222222ctxctx  . We know4xkm,0t , and5xkm. Thus222221022285000 m4000 m1.010s3.0010m/sxxtc and51.010t  s.28.(a) Convertingv= 120 km/h = 33.3 m/s. Now withc= 100 m/s, we have/0.333vc and22111.061110.333. We conclude that the movingperson ages 6.1% slower.(b)/(1 m) /(1.061)0.942 m.LL 29.Convertingv= 300 km/h = 83.3 m/s. Now withc=100 m/s, we have/0.833vc and22111.81.110.833So the length is0/40 /1.8122.1LLm.30.Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can seethat11xm,2121xm, and13tns.21distance120 m3 ns +3 ns + 408 ns = 411 ns.speed0.98ttcTo find the four primed quantities we can use the Lorentz transformations with theknown values of1x,2x,1t, and2t. Note that with0.8vc,221/5 / 3vc .21112222111222/0.56 ns/147 ns0.47 m37.3 mttvxcttvxcxxvtxxvt   31.Start from the formula for velocity addition, Equation (2.23a):21/xxxuvuvuc .(a)20.620.841.460.961(0.62 )(0.84 ) /1.52xcccucccc(b)20.620.840.220.461( 0.62 )(0.84 ) /0.48xcccucccc 32.Velocity addition, Equation (2.24):21/xxxuvuvuc with0.8vc and0.8 .xuc20.8( 0.8 )1.60.9761( 0.8 )(0.8 ) /1.64xcccucccc   

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8Chapter 2Special Theory of Relativity33.Conversion: 110 km/h = 30.556 m/s and 140 km/h = 38.889 m/s. Let30.556xum/sand38.889v m/s. Our premise is that100cm/s. Then by velocity addition, 2230.556 m/s38.889 m/s62.1 m/s.1/138.889 m/s30.556 m/s / 100 m/sxxxuvuvuc   By symmetryeach observer sees the other one traveling at the same speed.34.From Example 2.5 we have1/1/cnvcunvnc. For light traveling in opposite directions1/1/1/1/cnvcnvcunvncvnc. Because/vcis very small, use the binomial expansion:11/1/1/1/1/1//1/nvcnvcvncnvcvncnvcvncvnc, where wehave dropped terms of order22/.vcSimilarly1/1//.1/nv cnv cv ncv ncThus221//1//11/211/.cvunv cv ncnv cv ncnvnnnEvaluatingnumerically we find212(5 m/s) 14.35 m/s.1.33u35.Clearly the speed of B is just0.60c. To find the speed of C use0.60xucand0.60vc :220.60( 0.60 )0.88 .1/1( 0.60 )(0.60 ) /xxxuvccucvucccc   36.We can ignore the 400 km, which is small compared with the Earth-to-moon distance83.8410m. The rotation rate is122rad100 s210rad/s. Then the speedacross the moon’s surface is2811210rad/s3.8410m2.41 10m/s.vR37.Classical:54205 m1.4310s0.98tc. Then01/2ln 2exp14.6tNNtor about 15 muons.Relativistic:561.4310s/2.8610s5tt so01/ 2ln 2exp2710 muons.tNNtBecause of the exponential nature of the decaycurve, a factor of five (shorter) in time results in many more muons surviving.38.The circumference of the fixed point’s rotational path isE2cos(39 )R, whereEREarth’s radius= 6378 km. Thus the circumference of the path is 31,143 km. The

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Chapter 2Special Theory of Relativity9rotational speed of that point is31,143 km / 24 h1298 km/h360.m s5/v. Theobservatory clock runs slow by a factor of213211/ 217.22101. In41.2 h the observatory clock is slow by131141.2 h7.22102.974610h = 107 ns.In 48.6 h it is slow by131148.6 h7.22103.508910h = 126 ns. The Eastward-moving clock has a ground speed of 31,143 km/41.2 h = 755.9 km/h = 210.0 m/s and thushas a net speed of 210.0 m/s + 360.5 m/s = 570.5 m/s. For this clock212211/ 211.81 101and in 41.2 hours it runs slow by121141.2 h1.81 107.457210h = 268 ns. The Westward-moving clock has aground speed of 31,143 km/48.6 h = 640.8 km/h = 178.0 m/s and thus has a net speed of360.5 m/s−178.0 m/s = 182.5 m/s. For this clock213211/ 211.85101and in 48.6 hours it runs slow by131248.6 h1.85108.991 10h = 32 ns. So our prediction is that the Eastward-moving clock is off by 107 ns–269 ns162 ns, while the Westward-moving clock isoff by 126 ns−32 ns = 94 ns. These results are correct for special relativity but do notreconcile with those in the table in the text, because general relativistic effects are of thesame order of magnitude.39.The derivations of Equations (2.31) and (2.32) in the beginning of Section 2.10 willsuffice. Mary receives signals at a rateffor1tand a rateffor2t. Frank receivessignals at a rateffor1tand a rateffor2t.40.122LLLLLTttvcvcv; Frank sends signals at ratef,so Mary receives2/f Tf Lvsignals.122LTttv; Mary sends signals at ratef, so Frank receives2/fTf Lv signals.41.222222sxyzc t; Using the Lorentz transformation22222222222222222222222222()(/)1/1/svtyzctvcvcyzc tvcyxxxzcxst

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition - Page 12 preview image

10Chapter 2Special Theory of Relativity42.For a timelike interval20sso222xct. We will prove by contradiction. Supposethat there is a frameKis which the two events were simultaneous, so that0t. Thenby the spacetime invariant2222222xctctxx . But because222xct,this implies20xwhich is impossible becausexis real.43.As in Problem 42, we know that for a spacelike interval20sso222xct. Then ina frameKin which the two events occur in the same place,0xand22222222xctxctct  . But because222xctwe have220ct,which is impossible becauset is real.44.In order for two events to be simultaneous inK, the two events must lie along thexaxis, or along a line parallel to thexaxis. The slope of thexaxis is/vc , so/slope =c tvcx. Solving forv, we find2/vctx. Since the slope of thexaxismust be less than one, we see thatxc tso22220sxct is required.45.parts (a) and (b) To find the equation of the lineuse the Lorentz transformation. With0t wehave20/ttvx c or, rearranging,/ctvxcc. Thus the graph ofctvs.xis astraight line with a slope.(c) Now withtconstant, the Lorentz transformationgives2/ttvx c . Again we solve forct:/constantctxctx. This line isparallel to the0t line we found earlier butshifted by the constant.(d) Here both thexandctaxes are shiftedfrom their normal (x,ct) orientation and theyare not perpendicular.

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition - Page 13 preview image

Chapter 2Special Theory of Relativity1146.The diagram is shown here. Note that there is only one worldline for light, and it bisectsboth thex,ctaxes and the,x ctaxes. Thexandctaxes are not perpendicular. This canbe seen as a result of the Lorentz transformations, since0x defines thectaxis and0t defines thexaxis.47.The diagram shows that the events A and B that occur at the same time in K occur atdifferent times inK.48.The Doppler shift gives011. With numerical values0650 nm and540 nm, solving this equation forgives0.183 . The astronaut’s speed is75.5010vcm/s. In addition to a red light violation, the astronaut gets a speedingticket.49.According to the fixed source (K) the signal and receiver move at speedscandv,respectively, in opposite directions, so their relative speed iscv. The time intervalbetween receipt of signals is0/ ()1/tcvf . By time dilation()ttcv.Using0/cv and2211/vc we find2220011/()(1)cvctfcvfand002(1)1111ffft .

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition - Page 14 preview image

12Chapter 2Special Theory of Relativity50.For a fixed source and moving receiver, the length of the wave train is.cTvTSincenwaves are emitted during timeT,cTvTnand the frequency/fciscnfcTvT.As in the text00nTfand0/.TT Therefore20001/111fcf TffcTvT.51.0110.951400 kHz224 kHz110.95ff52.The Doppler shift function011ff is the rate at which #1 and #2 receive signals from each other and the rate at which #2and #3 receive signals from each other. But for signals between #1 and #3 the rate is01111fff.53.The Doppler shift function011ff is the rate at which #1 and #2 receive signalsfrom each other and the rate at which #2 and #3receive signals from each other. As for #1 and#3 we will assume that these plumbing vans arenon-relativisticvc. Otherwise it would benecessary to use the velocity addition law and applythe transverse Doppler shift. From the figure we seethat0211()fttt . Now001/ftand0212cos2vtxttcc. With an angle of45,cos(45)1/2and000011/(2 cos) /12 cos/12/ffffvcfvcv c .54.The Doppler shift to higher wavelengths is (with0589 nm)01700 nm1.Solving forwe find0.171 . Then8620.1713.0010m/s1.7510s29.4 m/svtawhich is 20.25 days. One problem with this analysis is that we have only computed the

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition - Page 15 preview image

Chapter 2Special Theory of Relativity13time as measured by Earth. We are not prepared to handle the non-inertial frame of thespaceship.55.Let the instantaneous momentum be in thex-direction and the force be in they-direction.ThendpF dtmdvanddvis also in they-direction. So we havedvFmmadt.56.The magnitude of the centripetal force is2vmam rfor circular motion. For a chargedparticleFqvB, so2vqvBm ror, rearrangingqBrmvp. Therefore.prqBWhen the speed increases the momentum increases, and thus for a given value ofBtheradius must increase.57.221/mvpmvvcanddpFdt. The momentum is the product of two factors thatcontain the velocity, so we apply the product rule for derivatives:222222322322232231//11/1/1221dmvFm dtvcdvdtdmv dtvcvcvdvmamvcdtvmamacvvmaccma58.From the preceding problem3Fma. We have1910am/s2and271.6710mkg.(a)2223271928111.000051/10.011.000051.6710kg10m/s1.6710NvcF(b)As in (a)1.005 and81.7010FN

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition - Page 16 preview image

14Chapter 2Special Theory of Relativity(c)As in (a)2.294 and72.0210FN(d)As in (a)7.0888 and65.9510FN59.pmvwith222112.55161/10.92vc ;1625810kg·m / s1.4210kg2.55160.923.0010m/spmv60.The initial momentum is0210.50.5773510.5pmvmcmc.(a)0/1.01pp1.010.577351.010.577350.58312mvmcvccSubstituting forand solving forv,1/ 222110.504 ..58312vccc(b) Similarly1/222110.536.63509vccc(c) Similarly1/222110.7561.1547vccc61.6.3 GeV protons have36.3 10KMeV and07238EKEMeV. Then2207177 MeV/EEpcc. Converting to SI units131881.6010J7177 MeV/3.8310kg·m/sMeV3.0010m/scpcFrom Problem 56 we have18193.8310kg m/s1.57 T1.6010C15.2 mpBqr.62.Initially Mary throws her ball with velocity (primes showing the measurements are inMary’s frame):0.0xyMMuuu After the elastic collision, the signs on the aboveexpressions are reversed, so the change in momentum as measured by Mary is00022222200021/1/1/Mmumumupucucuc.

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