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Homework #8
1. Pressure and bulk modules of an electron gas. (a) Derive a relation connecting the
3
pressure and volume of an electron gas at 0K. Hint: Use the result of U 0 = Nε F and
5
the relation between ε F and electron concentration. The result may be written as
∂p
2U 0
. (b) Show that the bulk modules B = −V
of an electron gas at 0K is
p=
3V
∂V
5 p 10U 0
=
. (c) Estimate for potassium, using Table 1 of the textbook, the value
B=
3
9V
of the electron gas contribution to B.
Solution:
(a) At T=0K, p = −
∂U 0 3N ∂ε F
=
.
∂V
5 ∂V
∂ε F
2ε
⎛N⎞
Note ε F ∝ n = ⎜ ⎟ . Thus
=− F .
∂V
3V
⎝V ⎠
3 N ∂ε F 3N 2ε F 2U 0
=
=
p=−
5 ∂V
5 3V
3V
2U 0
(b) From (a), p =
.
3V
2∂U 0 2U 0 2 p
∂p
5 p 10U 0
Thus B = −V
=−
+
=
+p=
=
.
∂V
3∂V
3V
3
3
9V
3
(c) From U 0 = Nε F ,
5
10U 0 2nε F 2
B=
=
= × 1.4 ×10 22 cm −3 × 2.12 ×1.6 × 10 −12 erg = 3.16 ×1010 erg / cm 3 .
9V
3
3
2/3
2/3
2. Chemical potential in two dimensions. Show that the chemical potential of a Fermi
gas in two dimensions is given by:
μ (T ) = k BT ln[exp(πnh 2 / mk BT ) − 1] ,
for n electrons per unit area. Note: The density of orbitals of a free electron gas in
two dimensions is independent of energy: D(ε ) = m / πh 2 , per unit area of specimen.
Solution:
For 2D free particle system, the density of state is
S 2πpdp
Sπdp 2 4mSπdε
=
=
. Here S is the area of the system.
Ddε = 2
2
h2
h2
h2
The total number of particles is
N =∫
Ddε
∞
0
e
( ε − μ ) / k BT
+1
=
4mSπ
h2
∫
dε
∞
0
e
( ε − μ ) / k BT
+1
4mSπ
k BT ln[e −(ε − μ ) / k BT + 1]
2
0
h
4mSπ
mS
=
k BT ln(e μ / k BT + 1) = 2 k BT ln(e μ / k BT + 1)
2
h
πh
2
⎛ πnh ⎞
⎟⎟ = e μ / kBT + 1 with n being the area density.
Then exp⎜⎜
mk
T
⎝ B ⎠
=−
∞
Finally, we have μ = k BT ln[exp(πnh 2 / mk BT ) − 1] .
3. Fermi gases in astrophysics. (a) Given M ⊕ = 2 × 1033 g for the mass of the Sun,
estimate the number of electrons in the Sun. In a white dwarf star this number of
electrons may be ionized and contained in a sphere of radius 2 × 109 cm ; find the
Fermi energy of the electrons in electron volts. (b) The energy of an electron in the
relativistic limit ε >> mc 2 is related to the wavevector as ε ≅ pc = hkc . Show that
the Fermi energy in this limit is ε F ≈ hc( N / V )1 / 3 , roughly. (c) If the above number
of electrons were contained within a pulsar of radius 10km, show that the Fermi
energy would be ≈ 108 eV . This value explains why pulsar are believed to be
composed largely of neutrons rather than of protons and electrons, for the energy of
release in the reaction n → p + e − is only 0.8 × 106 eV , which is not large enough to
enable many electrons to form a Fermi sea. The neutron decay proceeds only until
the electron concentration build up enough to create a Fermi level of 0.8 × 106 eV , at
which point the neutron, proton, and electron concentrations are in equilibrium.
Solution:
(a) The hydrogen atom mass is mH = 1.67 × 10 −24 g . Assuming each hydrogen atom
contributes one electron, the total number of electrons in the white draft is
M
2 × 10 33
= 1.20 × 10 57 .
N= ⊕ =
− 24
mH 1.67 × 10
The
density
of
the
electron
is
57
N
3N
3 × 1.2 × 10
n=
=
=
= 3.58 × 10 34 m −3 .
3
3
21 3
V 4πr
4 × 3.14 × 2 × 10 m
2
For ε = p / 2m and electrons in a 3D system,
2/3
2/3
(1.05 × 10−34 ) 2
h2
(
(
3π 2 n ) =
3π 2 × 3.58 × 1034 )
− 31
2m
2 × 9.1 × 10
−15
= 6.30 × 10 J = 3.9 × 104 eV
εF =
(b) For ε ≅ pc = hkc in 3D, the density of states is
Ddε = 2
V 4πp 2 dp
V 4πε 2 dε Vε 2 dε
=
= 2 3 3.
2
π hc
h3
h 3c 3
N=∫
εF
0
Ddε = ∫
εF
0
1/ 3
Vε F3
Vε 2 dε
=
.
π 2 h 3c 3 3π 2 h 3c 3
1/ 3
⎛ 3π 2 h 3 c 3 N ⎞
⎛ 3π 2 N ⎞
⎟⎟ = hc⎜⎜
⎟⎟
ε F = ⎜⎜
V
⎝
⎠
⎝ V ⎠
(c) For N = 1.20 × 1057 , V ≈ (10km) 3 = 1012 m 3 ,
1/ 3
ε F = 10
−34
⎛ 3π 2 1.2 × 10 57 ⎞
⎟⎟
× 3 ×10 ⎜⎜
1012
⎝
⎠
8
1
eV ≈ 6 ×108 eV .
−19
1.6 ×10
4. Liquid He3. The atom He3 has spin ½ and is a Fermion. The density of liquid He3 is
0.081gcm −3 near absolute zero. Calculate the Fermi energy ε F and the Fermi
temperature TF .
Solution:
N
0.081g / cm 3
=
= 0.0162 × 10 24 cm −3 = 1.62 ×10 28 m −3
V 3 × 1.67 ×10 −24 g
2/3
2/3
(1.05 × 10 −34 ) 2
h2
3π 2 n ) =
3π 2 ×1.62 × 10 28 ) J
εF =
(
(
− 27
2m
2 × 3 ×1.67 × 10
− 23
= 6.74 ×10 J = 4.21× 10 −2 eV
ε
6.74 × 10 −23 J
TF ≡ F =
= 4.9 K .
k B 1.38 × 10 −23 J / K