Instructor Solutions Manual for Modern Physics Sixth Edition Paul A Tipler Ralph A Llewellyn Prepared by Mark J Llewellyn Department of Electrical Engineering and Computer Science Computer Science Division University of Central Florida W H Freeman and Company New York www.elsolucionario.org Instructor Solutions Manual to Accompany Tipler & Llewellyn Modern Physics, Sixth Edition © 2012, 2008, 2003 by W.H Freeman and Company All rights reserved Published under license, in the United States by W H Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com Preface This book is an Instructor Solutions Manual for the problems which appear in Modern Physics, Sixth Edition by Paul A Tipler and Ralph A Llewellyn This book contains solutions to every problem in the text and is not intended for class distribution to students A separate Student Solutions Manual for Modern Physics, Sixth Edition is available from W H Freeman and Company The Student Solutions Manual contains solutions to selected problems from each chapter, approximately one-fourth of the problems in the book Figure numbers, equations, and table numbers refer to those in the text Figures in this solutions manual are not numbered and correspond only to the problem in which they appear Notation and units parallel those in the text Please visit W H Freeman and Company’s website for Modern Physics, Sixth Edition at www.whfreeman.com/tiplermodernphysics6e There you will find 30 More sections that expand on high interest topics covered in the textbook, the Classical Concept Reviews that provide refreshers for many classical physics topics that are background for modern physics topics in the text, and an image gallery for Chapter 13 Some problems in the text are drawn from the More sections Every effort has been made to ensure that the solutions in this manual are accurate and free from errors If you have found an error or a better solution to any of these problems, please feel free to contact me at the address below with a specific citation I appreciate any correspondence from users of this manual who have ideas and suggestions for improving it Sincerely, Mark J Llewellyn Department of Electrical Engineering and Computer Science Computer Science Division University of Central Florida Orlando, Florida 32816-2362 Email: markl@cs.ucf.edu www.elsolucionario.org Table of Contents Chapter – Relativity I Chapter – Relativity II 31 Chapter – Quantization of Charge, Light, and Energy 53 Chapter – The Nuclear Atom 79 Chapter – The Wavelike Properties of Particles 109 Chapter – The Schrödinger Equation 127 Chapter – Atomic Physics 157 Chapter – Statistical Physics 187 Chapter – Molecular Structure and Spectra 209 Chapter 10 – Solid State Physics 235 Chapter 11 – Nuclear Physics 259 Chapter 12 – Particle Physics 309 Chapter 13 – Astrophysics and Cosmology 331 www.elsolucionario.org Chapter – Relativity I 1-1 (a) Speed of the droid relative to Hoth, according to Galilean relativity, uHoth , is uHoth uspaceship udroid 2.3 108 m / s 2.1108 m / s 4.4 108 m / s (b) No, since the droid is moving faster than light speed relative to Hoth 1-2 L 2.74 10 m (a) t 1.83 104 s c 3.00 10 m / s (b) From Equation 1-6 the correction t 2L v c c2 t 1.83 104 s 104 1.83 1012 s c km / s 1.3 105 c 299, 796 km / s No, the relativistic correction of order 10-8 is three orders of magnitude smaller than (c) From experimental measurements the experimental uncertainty 0.4 fringe 1.0 fringe v2 1.0 29.9 km / s 2.22 103 v 47.1 km / s 0.4 1-3 29.8km / s 1-4 (a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20 v km / s mph Calling the plane flying perpendicular to the wind plane #1 and the one flying parallel to the wind plane #2, plane #1 win will by Δt where Lv 12.5mi 20mi / h t 0.0023 h 8.2s c 130mi / h (b) Pilot #1 must use a heading sin 1 20 /130 8.8 relative to his course on both legs Pilot #2 must use a heading of 0 relative to the course on both legs www.elsolucionario.org Chapter – Relativity I 1-5 (a) In this case, the situation is analogous to Example 1-1 with L 108 m, v 104 m / s, and c 108 m / s If the flash occurs at t = 0, the interior is dark until t =2s at which time a bright circle of light reflected from the circumference of the great circle plane perpendicular to the direction of motion reaches the center, the circle splits in two, one moving toward the front and the other moving toward the rear, their radii decreasing to just a point when they reach the axis 10-8 s after arrival of the first reflected light ring Then the interior is dark again (b) In the frame of the seated observer, the spherical wave expands outward at c in all directions The interior is dark until t = 2s at which time the spherical wave (that reflected from the inner surface at t = 1s) returns to the center showing the entire inner surface of the sphere in reflected light, following which the interior is dark again 1-6 Yes, you will see your image and it will look as it does now The reason is the second postulate: All observers have the same light speed In particular, you and the mirror are in the same frame Light reflects from you to the mirror at speed c relative to you and the mirror and reflects from the mirror back to you also at speed c, independent of your motion 1-7 N Lv (Equation 1-10) Where λ = 590 nm, L = 11 m, and ΔN = 0.01 fringe c2 v2 N c 0.01 fringe 590 109 m 3.00 108 m / s 11m 2L v 4.91103 m / s km / s 1-8 (a) No Results depends on the relative motion of the frames (b) No Results will depend on the speed of the proton relative to the frames (This answer anticipates a discussion in Chapter If by “mass”, the “rest mass” is implied, then the answer is “yes”, because that is a fundamental property of protons.) Chapter – Relativity I (Problem 1-8 continued) (c) Yes This is guaranteed by the 2nd postulate (d) No The result depends on the relative motion of the frames (e) No The result depends on the speeds involved (f) Yes Result is independent of motion (g) Yes The charge is an intrinsic property of the electron, a fundamental constant 1-9 The wave from the front travels 500 m at speed c + (150/3.6) m/s and the wave from the rear travels at c – (150/3.6) m/s As seen in Figure 1-14, the travel time is longer for the wave from the rear t tr t f 500m 500m 3.00 10 m / s 150 / 3.6 m / s 3.00 10 m / s 150 / 3.6 m / s 108 150 / 3.6 108 150 / 3.6 500 108 150 / 3.6 108 150 / 3.6 2 500 150 / 3.6 10 4.63 1013 s * 1-10 A * B * v C While the wavefront is expanding to the position shown, the original positions of A, B, and C have moved to the * marks, according to the observer in S (a) According to an S observer, the wavefronts arrive simultaneously at A and B (b) According to an S observer, the wavefronts not arrive at A and C simultaneously (c) The wavefront arrives at A first, according to the S observer, an amount Δt before arrival at C , where Chapter 13 – Astrophysics and Cosmology (Problem 13-8 continued) or R 3600 " 1AU 1" 180 rad 3.086 1016 m 9.45 1015 m / c y pc 3.26c y 0.01", R 100 pc and the volume of a sphere with that radius is (b) When R V 3.086 1016 m pc 4.19 106 pc3 If the density of stars is 0.08 / pc3, then the number of stars in the sphere is equal to 0.08 / pc3 4.19 106 pc3 13-9 L r2 f m1 m2 rp2 f p and LB Thus, Lp rp2 f p rB2 f B rB2 12 pc, Because rp 2.5 log f1 / f rB2 f B and Lp 0.30 rp f p / f B rB f p / fB 1/ 13-10 (a) M 0.3M Te 3300K (b) M 3.0M Te 13, 500K (c) R M R R M Similarly, R3.0 3.0 R tL M (d) tL 0.3 or tL 0.3 tL 0.3M 2.00 12 17.0 pc 10 L L L 102 L 1.93 1025W 3.85 1028W R /M M 0.3M R0.3 LB rp2 f p / f B 1.16 0.41 2.5 log f p / f B 3.4 105 stars 2.09 108 m 0.3R 2.09 109 m M 0.3M tL / M tL M tL M 0.3M 37tL Similarly, tL 3.0 333 0.3 0.04tL tL www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology S R 13-11 Angular separation 100 106 km 100c y distance between binaries distance Earth 1011 m 100c y 3.15 107 s / y 1.057 10 rad 6.06 10 degrees 1.68 10 arcseconds 13-12 Equation 13-18: 56 26 13 24 He 4n m56 Fe Fe Energy required: 13 m4 He 1u 4mn 931.49432 MeV / c Equation 13-19: He Energy required: 2m1H 13-13 55.939395u, m4 He 2n 2mn m4 He 0.020277u (a) r 1.5 c y ; assuming constant expansion rate, m1H Age of Shell (b) Lstar R M Te M 1/ R mstar , M Te star or Rstar Lstar L 28.3MeV 1.5 c y / 2.4 104 m / s Te M 1/ Te L star 2.95 1011 s 9400 y 1.4Te M4 L M4 L /M4 Te 1/ M star , 1/ M Using either the Te or L relations, Rstar 1.007825 12L Te / M / , R /M , Rstar 120 MeV 1H 1.5 c y M 1.008665u 0.129104u m56 Fe 0.129104u 24 km / s R 4.002603u, mn Lstar M star R M 1/ 1.86 R 334 L M star M Te star Te R 1.4 R 1.96 R Chapter 13 – Astrophysics and Cosmology 2GM / c2 13-14 RS (Equation 13-24) 6.67 10 (a) Sun RS (b) Jupiter mJ 318mE 1.99 1030 / c2 2.9 103 m 3km 2.8m RS 8.86 10 m (c) Earth RS 13-15 M 11 9mm ! 2M (a) (Equation 13-22) R 1.6 1014 M (b) 0.5rev / s I 2 MR I d 2 2M 1.01 104 d / day 108 where 8.0 1038 J d 1.99 1030 1011 (b) Its radius would be RS 11 1.8 1042 kg 2GM / c (Equation 13-24) 1.8 1042 / c2 2.6 1015 m 17, 000 AU 72, 000 km / s (a) v Hr r v H 72, 000km / s 21.2 km / s 106 c y 3.40 109 c y (b) From Equation 13-29 the maximum age of the galaxy is: 1/ H 4.41 1017 s L 1.85 1025W 10% of total (a) Mass of a central black hole = 1.99 1041 13-17 v 8.0 1038 J 1011 stars of average mass M , therefore the visible mass = 1.99 10 41 kg 6.67 10 1.85 1025 J / s 10 d 8.64 10 s / d 13-16 Milky Way contains RS 1.01 10 m where for a sphere I d 1/ 1.6 1014 M rad / s I (c) d 1/ 1.4 1010 y 335 www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology (Problem 13-17 continued) 1/ H r/v 1/ H 1/ H r/v 1/ H r r 10% so the maximum age will also be in error by 10% 13-18 The process that generated the increase could propagate across the core at a maximum rate of c, thus the core can be at most 1.5 y 3.15 107 s / y 3.0 108 m / s 1.42 1016 m 9.45 104 AU in diameter The Milky Way diameter is 60, 000c y 3.8 109 AU 13-19 Combining Hubble’s law (Equation 13-28) and the definition of the redshift (Equation 1327) yields z Hor c r 106 c y (a) Hor c 21.7 km 106 c y 656.3nm s 106 c y 108 m/s 656.5nm (b) r 50 10 c y Similarly, (c) r (d) r 658.7nm 500 106 c y Similarly, 680.0nm 109 c y Similarly, 893.7nm 336 Chapter 13 – Astrophysics and Cosmology 13-20 Equation 13-33: 3H G c 1/ H G c 1.5 1010 y 3.15 107 s / y 8.02 10 6.67 10 11 27 kg / m3 Nm / kg (This is about hydrogen atoms/m3 !) 13-21 Present size 1010 c y Sp T with T T Sp 2.7 K 2.7 1010 c yK (a) 2000 years ago, S = Sp (b) 106 years ago, S = Sp (c) 10 seconds after the Big Bang, S 2.7 1010 c yK / 109 K 2.7 10 S p 2.7 1010 c yK / 109 K 5.4 10 (d) second after the Big Bang, S 10 Sp 25c y 5c y (e) 10-6 seconds after the Big Bang, S 13-22 2.7 1010 c yK / 1012 K 5.4 10 proton pl 10 1.67 10 10 osmium 5.5 10 kg m pl Planck time 15 27 kg m 35 m 13 Sp 0.005c y 6.4 10 AU 5.5 1097 kg / m3 1.67 1018 kg / m3 2.45 104 kg / m3 13-23 Wien’s law (Equation 3-11): max 2.898mm K T 2.898mm K 2.728K 1.062mm (this is in the microwave region of the EM spectrum) 13-24 Muon rest energy 208me 106MeV / c2 The universe cooled to this energy (average) at about 10-3s (see Figure 13-34) 2.728K corresponds to average energy = 10-3 eV Therefore, m 10 eV 1.6 10 c2 19 J / eV 337 1.8 10 39 kg www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology 13-25 M / 4/3 r t0 r t R t r t0 t (Equation 13-37) \ M 0 4/3 R3 t r3 t M / 4/3 r t /R t M R3 t 4/3 r2 t t 13-26 B If Hubble’s law applies in A, then vBA FBC vCA HrCA From mechanics, FBA vBC C FCA HrBA , vBA vCA H rBA rCA HrBC and Hubble’s lab applies in C, as well, and A (Milky Way) by extension in all other galaxies 13-27 At a distance r from the Sun the magnitude of the gravitational force acting on a dust GM m where m (4 / 3) a The force acting on r2 the particle due to the Sun’s radiation pressure at r is given by: (See Equation RP-9.) article of radius a is: F grav Frad a Prad a2 U where a2 is the cross sectional area of the particle and U is the energy density of solar radiation at r U is given by: (See Equation 3-6.) U R c L c r2 Therefore, Frad a2 4L r 2c The minimum value of a is obtained from the condition that Fgrav 338 Frad : Chapter 13 – Astrophysics and Cosmology (Problem 13-27 continued) GM m 4L a2 r r 2c GM (4 / 3) a 4L a2 r r 2c Simplifying this expression yields: L cGM a 3.84 1026 W (3.00 108 m/s)(6.67 10 11 N m2 / kg )(1.99 1030 kg)(5500 kg/m3 ) a 1.40 10 m or 1.40 10 cm Note that (i) a is very small and (ii) the magnitude of a is independent of r a 13-28 (Equation 13-31) Robs M / 4/3 Z Remit Z r3 Substituting for r0 in the 0 M / 4/3 Z / Z or 13-29 (a) H available for fusion = M (b) Lifetime of H fuel = r Z M / 4/3 r03 equation: r Z r0 M / 4/3 Z Z r3 Z 3 0.75 0.13 2.0 1030 kg 0.75 0.13 2.0 1029 kg 2.0 1029 kg 6.00 101 kg / s 3.3 1017 s 3.3 1017 s / 3.15 107 s / y 1.03 1010 y (c) Start being concerned in 1.03 1010 y 0.46 1010 y 5.7 109 y 13-30 SN1987A is the Large Magellanic cloud, which is 170,000c•y away; therefore (a) supernova occurred 170,000 years BP 339 www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology (Problem 13-30 continued) (b) E K mo c mo c 109 eV , v2 c2 mo2 9.28 108 eV 109 9.38 108 9.38 108 v2 c 0.875c or v Therefore, the distance protons have traveled in 170,000y =v 170, 000 y 149, 000c y No, they are not here yet 1.99 1030 kg 13-31 M (a) When first formed, mass of H = 0.7 M , m 1H number of H atoms 0.7 M 1.007825u 1.66 10 He; 1H (b) If all H produced = 27 1.007825u 1.66 10 kg / u 27 8.33 1056 He 26.72eV The number of He atoms 8.33 106 Total energy produced = 8.33 106 5.56 1057 MeV 26.72MeV 8.89 1044 J (c) 23% of max possible = 0.23 8.89 1044 J 0.23 8.89 1044 L tL 13-32 (a) F Gm1m2 / r v2 / r ac m2 5.53 1017 s 1.7 1010 y 3.85 1026W L v / r m2 Gm1 / r and orbital frequency f v/2 r Substituting for f and noting that the period T or, T kg / u, thus 1/ f , 4 r / Gm1, which is Kepler’s third law (b) Rearranging Kepler’s third law in part (a), 340 f2 Gm1 / r Chapter 13 – Astrophysics and Cosmology (Problem 13-32 continued) moon mE r / GT 6.67 10 11 3.84 108 m Nm2 / kg 27.3d 8.64 104 s / d 6.02 1024 kg (c) T 11 T 12d 2 1.97 107 r2 2 12 24 3600 r2 6.06 10 / s 2 v2 r G m1m2 and r m2 r G from the graph v1 200km / s and v2 3.3 1010 m and, similarly, r2 100km / s 1.6 1010 m 4.9 1010 m Assuming circular orbits, m1 1.48 1022 kg m1m2 , then m1 m2 5.44 103 s 1.5h 24 6.46d 3.1 10 s / d , and 1/ 6.02 10 200 103 m / s 6.06 10 / s r1 11 Gm1m2 or m1 r2 r r r1 1, v2 r 6.67 10 m2 : reduced mass m1m2 m1 m2 r1 6.67 106 6.67 10 (b) For m1 (c) v1 1/ (d) mcomb 13-33 (a) T rsh3 GmE m1v12 r1 6.63 1030 kg and m2 m2v22 and m1 r2 1.37 1031 kg 341 r1v22 m2 Substituting yields, r2v12 www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology 13-34 E mv or GM m / r GmM / r mv mv GM m r E FG GM m / r mv / r GM m / r GM m r GM m r 13-35 dV 20km / s 106 c y H universe V Current average density = 1H atom / m3 R V dV R dR dR The current expansion rate at R is: v dV HR 20km / s 1010 c y 10 c y dR 20 107 m / s 3.16 107 s / y R dR 1010 7.07 1074 m3 106 c y Current volume V 20 104 km / s 20 107 m / s 106 y 106 y 9.45 1015 m / c y 20 107 m / s 3.16 107 s / y 106 y 106 y # of H atoms to be added 106 c y "new" H atoms = 1010 8.4 1077 m3 7.07 1074 atoms / 106 c y 8.4 1077 m3 342 0.001 "new" H atoms / m3 106 c y ; no Chapter 13 – Astrophysics and Cosmology 13-36 (a) Equation 8-12: vrms 3RT / M is used to compute vrms vs T for each gas ® = gas constant M Gas ( 10 kg ) vrms (m/s) at T = : 3R / M 50K 200K 500K 750K 1000K H2O 18 37.2 263 526 832 1020 1180 CO2 44 23.8 168 337 532 652 753 O2 32 27.9 197 395 624 764 883 CH4 16 39.5 279 558 883 1080 1250 H2 111.6 789 1580 2500 3060 3530 He 78.9 558 1770 1770 2160 2500 The escape velocities vsc gR 2GM / R , where the planet masses M and radii R, are given in table below Planet Earth Venus Mercury Jupiter Neptune Mars vesc (km/s) 11.2 10.3 4.5 60.2 23.4 5.1 vesc/6 (m/s) 1870 1720 750 10,000 3900 850 On the graph of vrms vs T the vesc/6 points are shown for each planet 343 www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology (Problem 13-36 continued) 2GM / R (b) vesc 2GM Pl / RPl vPl M Pl / RPl M E / RE vPl vE 2GM E / RE vE M E / RE M E / RE vPl vE M Pl / M E RPl / RE (c) All six gases will still be in Jupiter’s atmosphere and Netune’s atmosphere, because vesc for these is so high H2 will be gone from Earth; H2 and probably He will be gone from Venus; H2 and He are gone from Mars Only CO2 and probably O2 remain in Mercury’s atmosphere 13-37 (a) α Centauri d in pc d 1AU sin 0.742 " (b) Procyon d Earth's orbit radius (in AU) sin p 2.78 105 pc 1AU sin 0.0286 " 9.06 105 c y 7.21 105 pc 2.35 106 c y 13-38 Earth is currently in thermal equilibrium with surface temperature T and I Earth radiates as a blackbody I is f 102 L then f 0.338 1.36 103 102 away vrms 459W / m2 The solar constant 1.36 103W / m currently, so Earth absorbs 459/1360 = 0.338 of incident solar energy When L I 300 300K Assuming However, 3RT / M 102 f If the Earth remains in equilibrium T or T the vrms 8.31 949 18 10 solution to problem 14-26) 994 K for 676 C sufficient to boil the oceans H2 O molecules 1146m / s 1.15km / s at 994K is The vesc = 11.2km/s (see Because vrms ≈ 0.1 vesc, the H2O will remain in the atmosphere 344 Chapter 13 – Astrophysics and Cosmology 13-39 (a) a n = grains/cm3 a total scattering area = dust grain R which is R dx photons N/mrs N R na dx R2 a n dx a2 R2 n dx of the total area = fraction scattered = dN/N d dN N N0 n R dx or N N 0e n R2d From those photons that scatter at x = (N0), those that have not scattered again after traveling some distance x = L is N L N0e n R2 L The average value of L (= d0) is given by: L d0 (b) I dN L dL dL dN L dL dL I 0e d / d0 n R2 Note: near the Sun d0 n n R N 0e n R2 L R 10 cm 3000c y 3000c y 9.45 1017 cm / c y (c) dN L dL 10 n 1.1 10 12 / cm3 gm / cm3 grains mgrains cm of space mass in 300c y 10 1.1 10 9.41 10 27 gm / cm3 M 0.0012 12 9.41 10 27 9.45 1017 cm / c y 0.1%M 345 / cm3 gm / cm3 300 www.elsolucionario.org Chapter 13 – Astrophysics and Cosmology 13-40 56 11H 14 142 He 56 26 Fe 2e 14 4.002603 m56Fe 14 m4He 2e 2.04MeV / c 55.939395u 56.036442u 14 26.72 MeV 2.04MeV Net energy difference (release) = 90.40MeV 466.5MeV 56 26 Fe 2m 56 Fe 112 48 Cd 4e 55.939395u m Net energy required = 2m56Fe 13-41 (a) dt 4.08MeV / c 4e 112 111.902762u Cd 0.023972u 4.08MeV m112Cd 18.25MeV 1.024 104 2G M dM hc hc 1.024 1024 dM dt rearranging, the mass rate of change is G2M Clearly, the larger the mass M, the lower the rate at which the black hole loses mass (b) t t (c) t 1.024 104 6.62 10 2.0 1030 hc 3.35 1044 s 1.024 104 1.06 1037 y far larger than the present age of the universe 6.67 10 6.63 10 t 11 34 11 2.0 1030 1012 3.00 108 3.35 1068 s 1.06 1061 y 346 ... observer, the wavefronts arrive simultaneously at A? ?? and B (b) According to an S observer, the wavefronts not arrive at A? ?? and C simultaneously (c) The wavefront arrives at A? ?? first, according... v / c VA Space station vAy uy vB ? ?A v A vB / A tan AB where A vBx v vA vBx v / c 1 vA2 / c 2 vB v B A v A A vA (c) The situations are not symmetric... www.elsolucionario.org Chapter – Relativity I 1-61 1-62 (a) Apparent time A B T / t A tB and apparent time B A T / t A tB where tA = light travel time from point A to Earth and tB = light travel