Homework #6 Solutions
Question 1)
Kittel+Kroemer, Chapter 4, Problem 1. The total number of
thermal photons is the sum of photons in each mode:
N
=
X
n
< s
n
>
=
X
n
1
e
~
ω
n
/τ

1
(1)
Replace the sum over n with an integral, similar to the calcualtion in the text
on pages 93 and 94.
X
n
→
2
×
1
8
Z
∞
0
d
n
4
πn
2
(2)
The factor of 2 takes care of the two polarizations of light. The factor of 1/8
comes from integrating only over one quadrant (in 3 dimensional nspace). In
the integrand
ω
n
=
nπc/L
. To bring the integral to a dimensionless form we
make the change of variables
x
=
π
~
c
Lτ
n
. Thus the integral becomes:
N
=
π
±
Lτ
~
πc
²
3
Z
∞
0
d
x
x
2
e
x

1
(3)
We note that the integral is a known one and equals 2
ζ
(3) = 2
.
404. So the ﬁnal
answer reads
N
= 2
.
404
V
π
2
±
τ
~
c
²
3
(4)
Question 2a)
Kittel+Kroemer, Chapter 4, Problem 2. The solar constant
gives the power per unit area, which is placed a distance
R
e

s
= 1
.
5
×
10
23
cm
away from the sun (the distance from the sun to earth). To get the total power
output of the sun we have to integrate over the area over which this radiation
is spread: 4
πR
2
e

s
.
P
sun
= 4
πR
2
e

s
×
0
.
136
J
s
·
cm
2
= 3
.
85
×
10
26
J/s
(5)
b)
To use the StefanBoltzmann Law we need the power radiated per area.
This can be easily obtained using the radius of the sun
r
s
= 7
×
10
10
cm
.
J
sun
=
P
sun
4
πr
2
s
= 6
.
2
×
10
3
J
s
·
cm
2
(6)
The StefanBoltzmann Law reads
J
sun
=
σ
B
T
4
sun
. Solving for the temperature
we get
T
sun
=
±
J
sun
σ
B
²
1
/
4
= 5750
K
(7)
1
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View Full DocumentQuestion 3)
Kittel+Kroemer, Chapter 4, Problem 5. For this question we
could readily use some results from the previous problem. But we prefer to do
it from scratch in order to get familiarity with the methods. The Earth is ﬁrst
assumed to be a black body absorber in equilibrium with the radiation from
the sun. This means that the earth will radiate the same amount of energy it
receives from the sun. We will start by calculating the power of the radiation
from the sun using StefanBoltzmann law
P
±
=
J
±
×
4
πR
2
±
= (
σ
B
T
4
±
)
×
4
πR
2
±
(8)
Note that the circle with a dot in it (
±
) is a symbol for the sun. The ﬂux
density at earth’s orbit is
J
E
=
P
±
4
πR
2
e

s
=
J
±
±
R
±
R
e

s
²
2
(9)
The ﬂux absorbed by earth is this value multiplied by the cross section of the
Earth (which is just
πR
2
E
): Φ
in
=
J
E
×
πR
2
E
. In equilibrium this has to be equal
to the energy radiated by earth, which again is given by the StefanBoltzmann
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