P42_014 - 14(a Eq 42-6 leads to = EF kT ln(P 1 1 = 7.0 eV(8.62 105 eV/K(1000 K ln = E 6.8 eV 1 1 0.90(b n(E = CE 1/2 =(6.81 1027 m3 eV3/2(6.8

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14. (a) Eq. 42-6 leads to E = E F + kT ln( P 1 1) =7 . 0eV+(8 . 62 × 10 5 eV / K)(1000 K) ln µ 1 0 . 90 1 =6 . 8eV . (b) n ( E )= CE 1 / 2 =(6 . 81 × 10 27 m 3 · eV 3 / 2 )(6 . 8eV) 1 / 2 =1 . 77 × 10 28 m 3 · eV 1 . (c) n 0 ( E P ( E ) n
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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