Solutions to problem set 6 (141A Sp07)
1
1. (ISSP 6.2)
(a)
p
=

∂
U
∂
V
S
,
N
At
T
=
0 the entropy is a constant as
V
is varied. Therefore, since
E
F
=
CV

2
/
3
,
p
=

3
5
NC
(

)
2
3
V

5
/
3
=
3
5
NE
F
2
3
V

1
=
2
3
U
V
(b) From part (a),
p
=
C V

5
/
3
where
C
is a constant. Then
B
=

V
∂
p
∂
V
=
5
3
p
(c)
B
=
10
9
U
0
V
=
2
3
nE
F
=
2
3
(1
.
4
×
10
28
)(2
.
12)(1
.
6
×
10

19
)
=
3
.
17
×
10
9
J
/
m
3
=
3
.
17
GPa
Wikipedia gives the bulk modulus of potassium as 3.1 GPa (presumably at room tem
perature).
2. (ISSP 6.4)
(a) The sun is 70% hydrogen and 30% helium; therefore about 85% of its mass comes from
protons, giving
Number of electrons
=
Number of protons
=
.
85(2
×
10
30
)
1
.
67
×
10

27
=
1
.
0
×
10
57
Therefore there will be 10
57
electrons in the star. The electron density is then
n
=
10
57
4
3
π
(2
×
10
7
)
3
=
3
32
π
2
×
10
36
=
1
.
0
×
10
34
m

3
Then
k
F
==
(3
π
2
n
)
1
/
3
(30
π
2
10
33
)
1
/
3
=
6
.
7
×
10
11
. Then
v
F
=
(
/
m
)
k
F
7
×
10
7
m
/
s. We’re
on the border of the relativistic region:
v
/
c
.
2. Therefore we can use the nonrelativistic
limit but only trust our answer up to (
v
/
c
)
2
.
01.
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 Spring '08
 SOUZA
 Electron, Solid State Physics, Entropy, 30%, 0k, 15 J, 11 J

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