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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