Solutions Manual – Modern Physics, 6th Edition by Tipler & Llewellyn |All 13 Chapters Covered

Modern Physics 6th Edition
by Tipler, & Llewellyn Chapter 1-13




SOLUTION MANUAL

, Table of Contents


Chapter 1 – Relativity I 1

Chapter 2 – Relativity II 31

Chapter 3 – Quantization of Charge, Light, and Energy 53

Chapter 4 – The Nuclear Atom 79

Chapter 5 – The Wavelike Properties of Particles 109

Chapter 6 – The Schrödinger Equation 127

Chapter 7 – Atomic Physics 157

Chapter 8 – Statistical Physics 187

Chapter 9 – 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

, Chapter 1 – 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  2.1108 m / s
 4.4 108 m / s
(b) No, since the droid is moving faster than light speed relative to Hoth.


2 L 2  2.74 10 m 
4

1-2. (a) t    1.83 104 s
c 3.00 10 m / s
8




2L v 2
(b) From Equation 1-6 the correction  t  
c c2

 t  1.83 104 s 104   1.83 1012 s
2




4 km / s c
(c) From experimental measurements  1.3 105

c 299, 796 km / s
No, the relativistic correction of order 10-8 is three orders of magnitude smaller than
the experimental uncertainty.


0.4 fringe 1.0 fringe 1.0
 29.9 km / s   2.22 103  v  47.1 km / s
2
1-3.   v2 
 29.8km / s   v km / s 
2 2
0.4



1-4. (a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20
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 2 12.5mi  20mi / h 
2

t  3   0.0023 h  8.2s
130mi / h 
3
c


(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.

, Chapter 1 – Relativity I


1-5. (a) In this case, the situation is analogous to Example 1-1 with L  3 108 m,

v  3 104 m / s, and c  3 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.


2 Lv 2
1-7. N  (Equation 1-10) Where λ = 590 nm, L = 11 m, and ΔN = 0.01 fringe
c2


N  c 2
  0.01 fringe   590 109 m  3.00 108 m / s  2 11m 
2
v2 
2L

v  4.91103 m / s  5 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 2. If by “mass”, the “rest mass” is implied,
then the answer is “yes”, because that is a fundamental property of protons.)



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