3-20. (a) mT  2.898  103 m  K
m 
(Equation 3-5)
2.898  103 m  K
 1.45  107 m  145nm
4
2 10 K
(b) m is in the ultraviolet region of the electromagnetic spectrum.
3-21. Equation 3-4: R   T 4
Pabs  1.36 103W / m 2  RE2 m 2  where RE = radius of Earth
Pemit   RW / m 2  4 RE2   1.36 103W / m 2  RE2 m 2 
  RE2
R  1.36  103W / m 2  
2
 4 RE
T4 
 1.36  103 W
 T 4

2
4
m

1.36  103W / m 2
 T  278.3K  5.3C
4  5.67  108W / m 2  K 4 
3-22. (a) mT  2.898  103 m  K  m 
f m  c / m 
2.898  103 m  K
 8.78 107 m  878nm
3300 K
3.00  108 m / s
 3.42 1014 Hz
8.78 107 m
(b) Each photon has average energy E  hf and NE  40 J / s.
N
40 J / s
40 J / s

 1.77 1020 photons / s
34
14
hf m
 6.63 10 J  s  3.42 10 Hz 
(c) At 5m from the lamp N photons are distributed uniformly over an area
A  4 r 2  100 m 2 . The density of photons on that sphere is  N / A  / s  m 2 .
The area of the pupil of the eye is   2.5  103 m  , so the number of photons
2
entering the eye per second is:
n   N / A    2.5  103 m 
2
1.77 10

20
/ s     2.5  103 m 
2
100 m 2
 1.77  1020 / s     2.5  103 m   1.10 1013 photons / s
2
3-50. (a) mT  2.898  103 mK
(b) Equation 3-18:
 T
2.898  103 mK
 3.50  104 K
9
82.8  10 m
 70nm  /  ehc / 70nmkT  1

u  82.8nm   82.8nm 5 /  e hc / 82.8 nm kT  1
5
u  70nm 



6.63  1034 J s 3.00  108 m / s
hc
where

 5.88 and
 70nm  kT 70  109 m 1.38  1023 J / K 3.5  104 K


5
100nm  /  e4.12  1

 0.924
u  82.8nm   82.8nm 5 /  e 4.97  1
5
u 100nm 
3-53. (a) Equation 3-18: u    
gives u    
(b)

 70nm  /  e5.88  1

 0.929
u  82.8nm   82.8nm 5 /  e 4.97  1
u  70nm 
hc
 4.97
82.8nm  kT
Similarly,

8 hc 5
Letting C  8 hc and a  hc / kT
e hc /  kT  1
C  5
ea /   1


  5  1 e a /   a 2

du
d  C  5 
5 6 


 a/

 C
2
a/
d  d   ea /   1
e  1
1
e



6
6 a / 
C
 a a/
 C e
a

e  5 ea /   1  

 5 1  ea /    0
2 
2 


ea /   1  
ea /   1  










The maximum corresponds to the vanishing of the quantity in brackets.
Thus, 5 1  e  a /   a


(c) This equation is most efficiently solved by trial and error; i.e., guess at a value


for a /  in the expression 5 1  e  a /   a , solve for a better value of a /  ;
substitute the new value to get an even better value, and so on. Repeat the
process until the calculated value no longer changes. One succession of values
is 5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations
repeat the same value (to seven digits), so we have

a
m

 4.965114 
6.63  1034 J s 3.00  108 m / s
hc
(d) mT 

 4.965114  k  4.965114  1.38  1023 J / K

Therefore, mT  2.898  103 mK

hc
m kT

Equation 3-5
1.24 103
1.24 103
3-34. Equation 3-24: m 
 0.016nm
nm 
80  103V
V
3-36.
6.63  1034 J s  1  cos110 

h
 3.26 1012 m
2  1 
1  cos   
8
31
mc
 9.1110 kg  3.00 10 m / s 
6.63  1034 J s  3 108 m / s 

hc

 2.43 1012 m
1 
6
19
E1  0.511 10 eV 1.60  10 J / eV 
2  1  3.26 1012 m   2.43  3.26  1012 m  5.69 1012 m
E2 
hc
2

1240eV nm
 2.18  105 eV  0.218MeV
3
5.69  10 nm
Electron recoil energy Ee  E1  E2 (Conservation of energy)
Ee  0.511MeV  0.218MeV  0.293MeV . The recoil electron momentum makes
an angle θ with the direction of the initial photon.
PE
θ
h/λ1
110°
h/λ2
20°
h
2
cos 20  pe sin   1/ c  E 2   mc 2  sin 
2
 3.00 10 m / s  6.63 10
8
sin  
 5.69 10
12
(Conservation of momentum)
34
J s  cos 20
2
2
m   0.804MeV    0.511MeV  


1/ 2
1.60 10
13
J / MeV 
 0.330 or   19.3
3-37.
  2  1   
1  100 
h
1  cos    100  0.00243nm 1  cos 90   0.243nm
mc
hc
3-38. (a) E1 
1

(b) 2  1 
(c) E2 
h
1  cos    0.011 Equation 3-25
mc
hc
2
1240eV nm
 1.747 104 eV
0.0711nm
h
1  cos    0.0711   0.00243nm 1  cos180  0.0760nm
mc

1240eV nm
 1.634 104 eV
0.0760nm
(d) Ee  E1  E2  1.128 103 eV
3-41. (a) Compton wavelength =
h
mc
h
6.63 1034 J s
electron:

 2.43  1012 m  0.00243nm
31
8
mc  9.1110 kg  3.00 10 m / s 
proton:
(b) E 
hc

h
6.63 1034 J s

 1.32 1015 m  1.32 fm
mc 1.67 1027 kg  3.00 108 m / s 
(i) electron: E 
(ii) proton: E 
4-1.
1
mn
1240eV nm
 5.10  105 eV  0.510 MeV
0.00243nm
1240eV nm
 9.39 108 eV  939 MeV
6
1.32 10 nm
1 
 1
 R  2  2  where R  1.097  107 m 1 (Equation 4-2)
n 
m
The Lyman series ends on m =1, the Balmer series on m =2, and the Paschen
series on
1
m =3. The series limits all have n  , so  0 .
n
1
1
 R  2   1.097  107 m 1
L
1 
L  limit   1.097  107 m 1  91.16  109 m  91.16nm
 1 
 R  2   1.097  107 m 1 / 4
B
2 
B  limit   4 / 1.097  107 m 1  3.646  107 m  364.6nm
1
1
 R  2   1.097  107 m 1 / 9
P
3 
1
P  limit   9 / 1.097  107 m 1  8.204  107 m  820.4nm
4-2.
1
mn
1 
 1
 R  2  2  where m  2 for Balmer series (Equation 4-2)
n 
m
1
1.097  107 m 1  1
1 

 2

9
2
379.1nm
10 nm / m  2
n 
1 1
109 nm / m


 0.2405
4 n 2 379.1nm 1.097  107 m 1


1
 0.2500  0.2405  0.0095
n2
n2 
1
0.0095

n  1 / 0.0095 
1/ 2
 10.3

n  10
n  10  n  2
1
4-3.
mn
1 
 1
 R  2  2  where m  1 for Lyman series (Equation 4-2)
n 
m
1
1.097  107 m 1 
1 

1 2 

9
164.1nm
10 nm / m  n 
1
109 nm / m
 1
 1  0.5555  0.4445
n2
164.1nm 1.097  107 m 1

n  1 / 0.4445 
1/ 2

 1. 5
No, this is not a hydrogen Lyman series transition because n is not an
integer.
4-4.
1
mn
1 
 1
 R 2  2 
n 
m
(Equation 4-2)
For the Brackett series m = 4 and the first four (i.e., longest wavelength lines have
n = 5,
6, 7, and 8.
1
45
1
 1
 1.097  107 m 1  2  2   2.468  105 m 1
5 
4
45 
1
 4.052  106 m  4052nm. Similarly,
2.68  105 m 1
46 
1
 2.625  106 m  2625nm
1
5
3.809  10 m
47 
1
 2.166  106 m  2166nm
1
5
4.617  10 m
48 
1
 1.945  106 m  1945nm
5.142  105 m 1
These lines are all in the infrared.