Homework Solutions
Chapter 5
Todd B. Krause
September 30, 2008
53. A photon energy of 10 keV is equivalent to an energy of:
joule
= 1.60 · 10−15 joule.
eV
The frequency and wavelength of this photon are
10, 000 eV · 1.60 · 10−19
E
160 · 10−15 joule
1
=
= 2.41 · 1018 = 2.41 · 1018 Hz;
−34
h
6.626 · 10
joule · s
s
m
8
3 · 10 s
c
= 1.25 · 10−10 m = 0.125 nm.
λ= =
f
2.41 · 1018 1s
f=
54. (a) The energy of a singe photon with wavelength 600 nanometers is
m
(6.626 · 10−34 joule · s) · 3 · 108 hc
s
E=
=
= 3.31 · 10−19 joule.
−19
λ
600 · 10 m
(b) A 100-watt light bulb emits 100 joules of energy each second.
Thus, the number of 600-nanometers photons it emits each second
is
total energy emitted
number of photons =
energy emited per photon
100 joule
=
joule
3.31 · 10−19 photon
= 3 · 1020 photons.
(c) Because the light bulb emits more than 1020 photons each second,
there is no chance that we will notice these individual photons
acting like particles in our everyday life. Instead, we notice only
their collective wave effects.
1
57. (a) To solve this, we will set up a ratio using the Stefan-Boltzmann
law:
4
σThotter
energy emitted by hotter Sun
=
energy emitted by the Sun
σT4
Thotter 4
=
.
T
So if Thotter = 2T , we get
energy emitted by hotter Sun
=
energy emitted by the Sun
2
T
T
4
= (2)4
= 16.
If the Sun were twice as hot, it would emit 16 times as much
power per square meter.
(b) We will again use a ratio to work this problem, using Wein’s law
this time:
(
2,900,000
(
nm
peak wavelength of hotter Sun
Thotter = (
((( 2,900,000
peak wavelength of Sun
nm
T
((
=
T
Thotter ,
where we have canceled the 2, 900, 000 in the numerator and denominator and we have used some of the rules for fractions to
simplify the expression. Using the fact that Thotter = 2T , we
find
peak wavelength of hotter Sun
Thotter =
peak wavelength of Sun
T
1
= .
2
If the Sun were twice as hot, its peak wavelength would be at 1/2
the peak wavelength of the real Sun.
(c) It is doubtful that life could exist on Earth around this star.
There would be so much energy coming from the star that our
planet would probably be much too hot.
2
59. We use the Doppler shift formula to find the speeds of the stars:
Star A :
∆λ
·c
λ0
− 1, 216 120.5 nm
nm
km
=
· 300, 000
121.6 nm
s
km
= −2, 714
.
s
v=
The negative value indicates that Star A is moving toward us.
Star B :
∆λ
·c
λ0
− 1, 216 121.2 nm
nm
km
=
· 300, 000
121.6 nm
s
km
.
= −987
s
v=
The negative value indicates that Star B is moving toward us.
61. (a) For a coil with a temperature of 3, 000 K, the wavelength of maximun intensity is:
λmax =
2, 900, 000 nm · K
= 966 nm.
3, 000 K
Note that this wavelength is considerably longer than the 500nanometer wavelength of maximum emission from the Sun and
lies in the infrared portion of the spectrum. Thus, light bulbs
with coils at 3, 000 K emit much of their energy in the infrared,
rather than as visible light.
(b) Because light from standard light bulbs has a spectrum that peaks
in the infrared, it is generally redder in color than sunlight. Thus,
to record ‘true’ colors, film for indoor photography must compensate for the fact that the indoor light bulbs emit more red light
by having enhanced sensitivity to the less abundant blue light.
(c) Standard light bulbs emit thermal radiation, so they must emit
over a wide range of wavelengths. In fact, because their thermal
emission peaks in the infrared, they actually emit most of their
energy as infrared light, rather than visible light.
(d) Because a standard light bulb emits much of its light in the infrared, this light is ‘wasted’ as far as electrical lighting is concerned. In contrast, a fluorescent light would have no ‘wasted’
3
light. Thus, a fluorescent bulb of the same wattage as a standard
bulb can actually produce much more visible light.
(e) Despite the higher cost of a fluorescent light bulb compared to a
standard light bulb, the former can save money over the long run
for two principal reasons: (1) lower cost for operation because of
its lower energy usage, and (2) longer life, so it needs replacement
less often. The latter can particularly save money for businesses,
since businesses must pay for the labor involved in changing light
bulbs; this labor can be substantial if the light bulb is in a hardto-reach area.
4