# ε ε ε 2ε ε ε ε ε ε π μ ε ε π π π ε - Physics

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```					                                        Homework #8

1. Pressure and bulk modules of an electron gas. (a) Derive a relation connecting the
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pressure and volume of an electron gas at 0K. Hint: Use the result of U 0 = Nε F and
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the relation between ε F and electron concentration. The result may be written as
2U 0                                             ∂p
p=       . (b) Show that the bulk modules B = −V          of an electron gas at 0K is
3V                                              ∂V
5 p 10U 0
B=      =       . (c) Estimate for potassium, using Table 1 of the textbook, the value
3      9V
of the electron gas contribution to B.

Solution:
∂U 0 3N ∂ε F
(a) At T=0K, p = −         =        .
∂V    5 ∂V
∂ε F      2ε
2/3
⎛N⎞
Note ε F ∝ n = ⎜ ⎟ . Thus
2/3
=− F .
⎝V ⎠             ∂V        3V
3 N ∂ε F 3N 2ε F 2U 0
p=−            =         =
5 ∂V       5 3V      3V
2U 0
(b) From (a), p =       .
3V
∂p      2∂U 0 2U 0 2 p             5 p 10U 0
Thus B = −V        =−        +       =      +p=        =         .
∂V       3∂V     3V       3         3       9V
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(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. 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
Ddε = 2           =2         =          . Here S is the area of the system.
h2          h2        h2
The total number of particles is
∞            Ddε                    4mSπ       ∞              dε
N =∫           ( ε − μ ) / k BT
=          ∫           ( ε − μ ) / k BT
0   e                      +1        h2    0       e                      +1
4mSπ                                                ∞
=−    2
k BT ln[e −(ε − μ ) / k BT + 1]
h                                     0

4mSπ                                 mS
=        k BT ln(e μ / k BT + 1) = 2 k BT ln(e μ / k BT + 1)
h 2
πh
⎛ πnh ⎞2
μ / k BT
Then exp⎜ ⎜ mk T ⎟ = e
⎟                 + 1 with n being the area density.
⎝ 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
N= ⊕ =               − 24
= 1.20 × 10 57 .
mH 1.67 × 10
The             density                    of           the         electron   is
N    3N             3 × 1.2 × 10 57
n=     =       =                              = 3.58 × 10 34 m −3 .
V 4πr   3
4 × 3.14 × 2 × 10 m
3       21 3

For ε = p / 2m and electrons in a 3D system,
2

−34 2
εF =
h2
2m
(3π 2n )2 / 3 = (2 ×05.× 10 −) (3π 2 × 3.58 × 1034 )2 / 3
1.
9 1 × 10 31
= 6.30 × 10−15 J = 3.9 × 104 eV

(b) For ε ≅ pc = hkc in 3D, the density of states is
V 4πp 2 dp    V 4πε 2 dε Vε 2 dε
Ddε = 2               =2           = 2 3 3.
h3            h 3c 3  π hc
εF                 εF        Vε 2 dε      Vε 3
N=∫         Ddε = ∫                          = 2 F 3.
0                 0          π 2 h 3c 3 3π h 3c
1/ 3              1/ 3
⎛ 3π 2 h 3 c 3 N ⎞     ⎛ 3π 2 N ⎞
εF = ⎜
⎜                ⎟ = hc⎜
⎟     ⎜ V ⎟    ⎟
⎝      V         ⎠     ⎝        ⎠
(c) For N = 1.20 × 10 , V ≈ (10km) 3 = 1012 m 3 ,
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1/ 3
⎛ 3π 2 1.2 × 10 57 ⎞              1
ε F = 10    −34
× 3 ×10 ⎜
8
⎜                  ⎟
⎟                 −19
eV ≈ 6 ×108 eV .
⎝      1012        ⎠         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
10 −34 ) 2
εF =
h2
2m
(3π 2 n)2 / 3 = 2 ×13.051×67 ×10 −27 (3π 2 ×1.62 ×10 28 )2 / 3 J
(
× .
− 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

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