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Physics 237 Spring 2005
Solution to Homework Assignment #3
314.
We are given:
u
!
( ) =
8
"
hc
5
1
exp
hc
kT
#
$
%
’
(
)
1
. By definition,
u
(
λ
)
d
λ
gives the energy density
in the wavelength interval (
λ
,
λ
+d
λ
). Hence,
u
( )
d
=
u f
( )
df
"
u f
( ) =
u
( )
d
df
.
This then yields:
u f
( ) =
8
hc
5
1
exp
hc
kT
#
$
%
’
(
)
1
c
f
2
=
8
hcf
5
c
5
1
exp
hf
kT
#
$
%
’
(
)
1
c
f
2
=
8
f
2
c
3
hf
exp
hf
kT
#
$
%
’
(
)
1
.
(Here, we used
f
λ
=
c
, and just took the absolute value of the derivative.)
3.15.
(a)
From Chapter 3, we know that
λ
m
T
= 2.898 mm. K.
Hence, for
T
= 2.7 K,
λ
m
=
1.07 mm.
(b)
Using
f
λ
=
c
, we get
f
= 0.28 THz.
(c)
Treating the universe as a blackbody, the total power radiated per unit area =
=
1
4
cU
=
1
4
c u
( )
d
0
"
#
=
2
$
5
k
4
15
h
3
c
2
T
4
.
(You could also just use Stefan’s Law
directly, but make sure you know where this comes from!) The total power incident
on Earth is then obtained by multiplying this by the surface area of Earth:
P
earth
=
2
5
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 Spring '08
 STEPHONALEXANDER
 Energy, Work

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