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

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ε ε ε 2ε ε ε ε ε ε π μ ε ε π π π ε - Physics Powered By Docstoc
					                                        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
        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
                    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. 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 ,
                           57

                                                   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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