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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 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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