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quiz5slns.pdfName: Physics 403 Quiz #5 1)
quiz5slns.pdfName: Physics 403 Quiz #5 1)
quiz5slns.pdfName: Physics 403 Qui...
quiz5slns.pdfName: Physics 403 Quiz #5 1)
Page 1
Name:
Physics 403 Quiz #5
1) The FermiDirac distribution is
1
e
β
(
E

ζ
)
+1
, the BoseEinstein distribution is
1
e
β
(
E

ζ
)

1
, and
the MaxwellBoltzmann distribution is
e

β
(
E

ζ
)
. In three dimensions, the density of states
for a spinless nonrelativistic particle is
g
(
E
)=
1
4
π
2
(
2
m
~
2
)
3
2
E
1
2
.
All problems below are in
three dimensions.
a) What is the mean number of electrons at zero temperature?
The electron is a spin 1/2 particle and has spin degeneracy 2
S
+ 1 = 2. It is a fermion
as it has halfintegral spin; thus
<N>
=
V
Z
∞
0
2
g
(
E
)
dE
e
β
(
E

ζ
)
+1
As
T
= 0,
ζ
=
E
F
and
1
e
β
(
E

E
F
)
+1
= 1 if
E<E
F
and
1
e
β
(
E

E
F
)
+1
= 0 if
E>E
F
. Thus
<N>
=
V
Z
E
F
0
2
g
(
E
)
dE
=
2
3
V
1
2
π
2
2
m
~
2
3
2
E
3
2
F
b) What is mean energy of a spin 1 particle at temperature
T
and zero chemical potential?
The particle has spin degeneracy 2
S
+ 1 = 3. It is a boson as it has integral spin; thus
<U>
=
V
Z
∞
0
3
Eg
(
E
)
dE
e
βE

1
Evaluating
<U>
=
V
3
4
π
2
2
m
~
2
3
2
1
β
5
2
Z
∞
0
x
3
2
dx
e
x

1
=
V
3
4
π
2
2
m
~
2
3
2
1
β
5
2
Γ(
5
2
)
ζ
(
5
2
)
The mean energy per particle is given by
<U>
<N>
where
<N>
=
V
Z
∞
0
3
g
(
E
)
dE
e
βE

1
evaluated similarly;
<N>
=
V
3
4
π
2
2
m
~
2
3
2
1
β
3
2
Γ(
3
2
)
ζ
(
3
2
)
c) What is the chemical potential of low density gas of helium at room temperature
T
r
consisting of
N
atoms in a container of volume
V
?
Helium is a spin zero boson; at low density at room temperature, its distribution is the
MaxwellBoltzmann distribution.
1
Page 2
<n>
=
Z
∞
0
g
(
E
)
e

β
(
E

ζ
)
dE
Evaluating
<n>
=
1
4
π
2
2
m
~
2
β
3
2
e
βζ
√
π
2
Thus
ζ
=
1
β
(ln
<n>

ln(
1
4
π
2
2
m
~
2
β
3
2
√
π
2
))
Alternately,
ζ
=
∂F
∂N
where
F
=

1
β
ln
Z
and
Z
=
Z
N
1
N
where
Z
1
=
V
Z
∞
0
g
(
E
)
e

βE
dE
yields, after calculation, the same answer.
d) What is the mean energy of a spin 3/2 particle at zero temperature?
A spin 3/2 particle has spin degeneracy 2
S
+ 1 = 4. It is a fermion as it has halfintegral
spin; thus
<U>
=
V
Z
∞
0
4
Eg
(
E
)
dE
e
β
(
E

ζ
)
+1
As
T
= 0,
ζ
=
E
F
and
1
e
β
(
E

E
F
)
+1
= 1 if
E<E
F
and
1
e
β
(
E

E
F
)
+1
= 0 if
E>E
F
. Thus
<U>
=
V
Z
E
F
0
4
Eg
(
E
)
dE
=
2
5
V
1
π
2
2
m
~
2
3
2
E
5
2
F
The mean energy per particle is given by
<U>
<N>
where
<N>
=
V
Z
E
F
0
4
g
(
E
)
dE
evaluated similarly;
<N>
=
2
3
V
1
π
2
2
m
~
2
3
2
E
3
2
F
thus
<U>
<N>
=
3
5
E
F
.
2
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