ch38 - Chapter 38 1. (a) With E = hc/min = 1240 eVnm/min =...

1455
Chapter 38
1. (a) With E = hc / 
min
= 1240 eV·nm/ 
min
= 0.6 eV, we obtain  = 2.1  10
3
nm = 2.1
 m.
(b) It is in the infrared region.
2. Let
1
2
2
m v E
hc
e
 
photon

and solve for v :
v 
2 hc
 m
e

2 hc
 m
e
c 2
c 2  c
2 hc
 m
e
c 2
 
 2.998  10
8
m/s
 
2 1240eV  nm
 
588nm
 
511  10
3
eV
 
 8.6  10
5
m/s.
Since v c  , the nonrelativistic formula K mv 
1
2
2
may be used. The m
e
c
2
value of
Table 37-3 and 1240eV nm hc   are used in our calculation.
3. Let R be the rate of photon emission (number of photons emitted per unit time) of the
Sun and let E be the energy of a single photon. Then the power output of the Sun is given
by P = RE . Now
E = hf = hc /  ,
where h = 6.626  10
–34
J·s is the Planck constant, f is the frequency of the light emitted,
and  is the wavelength. Thus P = Rhc /  and
 
 
  
26
45
34 8
550nm 3.9 10 W
1.0 10 photons/s.
6.63 10 J s 2.998 10 m/s
P
R
hc



   
  
4. We denote the diameter of the laser beam as d . The cross-sectional area of the beam is
A =  d
2
/4. From the formula obtained in Problem 38-3, the rate is given by

CHAPTER 38 1456
R
A

 P
hc  d 2 / 4
 

4 633nm
 
5.0  10
 3
W
 
 6.63  10  34 J  s
 
2.998  10 8 m/s
 
3.0  10  3 m
 
2
 2.3  10 21 photons/m 2  s .
5. The energy of a photon is given by E = hf , where h is the Planck constant and f is the
frequency. The wavelength  is related to the frequency by  f = c , so E = hc /  . Since h =
6.626  10
–34
J·s and c = 2.998  10
8
m/s,
hc 
  

 

 
6 626 10 2 998 10
1602 10 10
1240
34 8
19 9
. .
.
J s m / s
J / eV m / nm
eV nm.
c hc h
c hc h
Thus,
E 
 1240eV nm

.
With
 = (1, 650, 763.73)
–1
m = 6.0578021  10
–7
m = 605.78021 nm,
we find the energy to be
E
hc
 



1240
60578021
2 047
eV nm
nm
eV.
.
.
6. The energy of a photon is given by E = hf , where h is the Planck constant and f is the
frequency. The wavelength  is related to the frequency by  f = c , so E = hc /  . Since h =
6.626  10
–34
J·s and c = 2.998  10
8
m/s,
hc 
  

 

 
6 626 10 2 998 10
1602 10 10
1240
34 8
19 9
. .
.
J s m / s
J / eV m / nm
eV nm.
c hc h
c hc h
Thus,
E 
 1240eV nm

.
With 589 nm   , we obtain
1240eV nm
2.11eV.
589nm
hc
E


  
7. The rate at which photons are absorbed by the detector is related to the rate of photon
emission by the light source via
abs
abs emit
2
(0.80) .
4
A
R R
r 

Given that
6 2
abs
2.00 10 m A

  and 3.00 m, r  with
abs
4.000 photons/s, R  we find the
rate at which photons are emitted to be

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 
2 2
8
emit abs
6 2
abs
4 4 (3.00 m)
4.000 photons/s 2.83 10 photons/s
(0.80) (0.80)(2.00 10 m )
r
R R
A
 

   

.
Since the energy of each emitted photon is
ph
1240 eV nm
2.48 eV
500nm
hc
E


   ,
the power output of source is
 
8 8 10
emit emit ph
2.83 10 photons/s (2.48 eV) 7.0 10 eV/s 1.1 10 W. P R E

      
8. The rate at which photons are emitted from the argon laser source is given by R =
P / E
ph
, where P = 1.5 W is the power of the laser beam and E
ph
= hc /  is the energy of