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

Name: Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition
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Author: Sofia Garcia

Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition gives you all the tools you need to solve your textbook problems effectively.

Sofia Garcia

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Table of Contents
CHAPTER 2 – Special Theory of Relativity 1
CHAPTER 3 – The Experimental Basis of Quantum Physics 28
CHAPTER 4 – Structure of the Atom 46
CHAPTER 5 – Wave Properties of Matter and Quantum Mechanics I 62
CHAPTER 6 – Quantum Mechanics II 78
CHAPTER 7 – The Hydrogen Atom 100
CHAPTER 8 – Atomic Physics 114
CHAPTER 9 – Statistical Physics 123
CHAPTER 10 – Molecules, Laser, and Solids 142
CHAPTER 11 – Semiconductor Theory and Devices 160
CHAPTER 12 – The Atomic Nucleus 169
CHAPTER 13 – Nuclear Interactions and Applications 187
CHAPTER 14 – Particle Physics 201
CHAPTER 15 – General Relativity 212
CHAPTER 16 – Cosmology and Modern Astrophysics – The Beginning and the End 221

Chapter 2 Special Theory of Relativity 1
Chapter 2
1. For a particle Newton’s second law says2 2 2
2 2 2
ˆˆ ˆd x d y d z
F ma m i j k
dt dt dt
 
    
  .
Take the second derivative of each of the expressions in Equation (2.1):2 2 2 2 2 2
2 2 2 2 2 2
d x d x d d y d d z
dt dt dt dt dt
z
t
y
d
   
 
. Substitution into the previous equation gives2 2 2
2 2 2
ˆˆ ˆ .
d x d d
F ma m i j k
y z F
dt dt dt
        
 

2. From Equation (2.1)ˆˆ ˆdx dy dz
p m i j k
dt dt dt
 
     .
In a Galilean transformationdx dx dy dy dz dz
v
dt dt dt dt dt dt
  
    .
Substitution into Equation (2.1) givesˆˆ ˆdx dy dz
p m v i j k p
dt dt dt
           .
However, becauseˆˆ ˆdx dy dz
p m i j k
dt dt dt
         the same form is clearly retained, given
the velocity transformationdx dx v
dt dt
   .
3. Using the vector triangle shown, the speed of light coming toward the mirror is2 2
c v
and the same on the return trip. Therefore the total time is2
2 2 2
2distance
speed
t c v
   .
Notice thatsin v
c
  , so1
sin v
c
   
  
  .

2 Chapter 2 Special Theory of Relativity
4. As in Problem 3,1 2sin /v v
  , so1 1
1 2
0.350 m/s
1.25
sin ( / ) sin 1
m/ .
s 6 3v v
    
    
  and2 2 2 2
2 1 (1.25 /s) (0.35m m/s) 1.20 m/sv v v    
.
5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have
effectively interchanged1 and2 . Interchanging1 and2 in Equation (2.3) leads to
Equation (2.4).
6. Let n = the number of fringes shifted; thend
n


 . Because d c t t     , we have   2
1 2
2
c t t v
n c


   
 
. Solving for v and noting that1 +2 = 22 m,    9
8
1 2
0.005 589 10 m
3.00 10 m/s 3.47 km/s.
22 m
n
v c
 

   

7. Letting2
1 1 1
  (where/v c
  ) the text equation (not currently numbered) for1t
becomes 
2
1 1
1 2
2 1 2
1 1
t c c




 
 
which is identical to2t when1 2 so0t  as required.
8. Since the Lorentz transformations depend on c (and the fact that c is the same constant
for all inertial frames), different values of c would necessarily lead two observers to
different conclusions about the order or positions of two spacetime events, in violation of
postulate 1.
9. Let an observer in K send a light signal along the + x-axis with speed c. According to the
Galilean transformations, an observer inK measures the speed of the signal to bedx dx v c v
dt dt
    
. Therefore the speed of light cannot be constant under the Galilean
transformations.
10. From the Principle of Relativity, we know the correct transformation must be of the form
(assumingy y andz z ):;x ax bt x ax bt     
.
The spherical wave front equations (2.9a) and (2.9b) give us:;( ) ( )ct ac b t ct ac b t    
.
Solve the second wave front equation fort and substitute into the first:

Chapter 2 Special Theory of Relativity 3( )( )ac b ac b t
ct c
  
   
or2 2 2 2
( )( ) .c ac b ac b a c b    
Now v is the speed of the origin of thex -axis. We can find that speed by setting0x 
which gives0 ax bt  , or/ /v x t b a  , or equivalently b = av. Substituting this into
the equation above for2
c yields 2 2 2 2 2 2 2 2
c a c a v a c v    . Solving for a:2 2
1
1 .
/v
a c


 
This expression, along with b = av, can be substituted into the original expressions for x
andx to obtain:   ;x x vt x x vt

     
which in turn can be solved for t andt to complete the transformation.
11. Whenv c we find2
1 1
  , so:2
1
x ct
x x ct x vt
 


     

;2
/ /
1
t x c
t t x c t
 


    

;2
1
x ct
x x ct x vt
 

         

;2
/ /
1
t x c
t t x c t
 

       

.
12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so 8 8
/ (26.39 m/s) / 3.00 10 m/s 8.8 10v c
 
    
(b) 8 7
/ (240m/s) / 3.00 10 m/s 8.0 10v c
 
    
(c) sound2.3 2.3 330 m/sv v   so   8 6
/ 2.3 330 m/s / 3.00 10 m/s 2.5 10v c
 
     
(d) Converting to SI units, 27,000 km/h = 7500 m/s, so 8 5
/ (7500 m/s) / 3.00 10 m/s 2.5 10v c
 
    
(e) (25 cm)/(2 ns) =8
1.25 10 m/s so 8 8
/ (1.25 10 m/s) / 3.00 10 m/s 0.42v c
     
(f)   14 22 8
1 10 m / 0.35 10 s 2.857 10 m/s 
    , so 8 8
/ (2.857 10 m/s) / 3.00 10 m/s 0.95v c
     

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

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