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# K t ns b f 2 2 d 1 e 0 f 2 d 2 k

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Unformatted text preview: h this change of variables, we find: ∞ ⎛ m* ⎞ k BTdη nS = ⎜ gV 2∫ η −η ⎝ π ⎟ 0 1+ e F ⎠ the integral can be done analytically: ECE- 656 3 Fall 2013 Mark Lundstrom 8/24/2013 ECE 656 Homework 2: (Week 2) (continued) ∞ dη ∫ 1+ e η −η F ( = ln 1 + eηF ) but we also recognize it as a Fermi- Dirac integral of order 0: 0 F 0 (η F ) = ln 1 + eηF ( ) so the answer is: nS = N 2 DF 0 (η F ) N2D m*k BT = gv π 2 2b) graphene ∞ ⎛ 2E ⎞ 1 D2 D ( E ) f0 ( E ) dE = ∫ ⎜ 2 2 ⎟ ∫ E − EF ) ( EC 0 ⎝ π υF ⎠ 1+ e ∞ nS = kB T dE define: ηF = EF k BT η= E k BT E = k BT η dη = dE k BT dE = k BTdη ⎛ 2 ⎞∞ ⎛ 2 ⎞ ∞ k BT η E 2 ⎛ k BT ⎞ nS = ⎜ 2 2 ⎟ ∫ ( E − EF ) k B T dE = ⎜ π 2υ 2 ⎟ ∫ 1 + eη −η F k BTdη = π ⎜ υ ⎟ ⎝ π υF ⎠ 0 1+ e ⎝ ⎝ F⎠ F⎠ 0 The integral is recognized as a Fermi- Dirac integral of order 1: 2⎛ k T⎞ nS = ⎜ B ⎟ π ⎝ υ F ⎠ 2 2 ∞ η dη ∫ 1+ e 0 η −η F 2 η dη 2 ⎛ k BT ⎞ ∫ 1 + eη−ηF = π ⎜ υ ⎟ F 1 (η F ) ⎝ F⎠ 0 ∞ nS = N 2 DF 1 (η F ) N2D 2 2 ⎛ k BT ⎞ =⎜ π ⎝ υ F ⎟ ⎠ ECE- 656 4 Fall 2013...
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