ch41-p035 - 35. (a) The probability that a state with...

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( E – E F )/ kT = (0.555 eV)/(0.02586 eV) = 21.46. Thus, PE e bg = + 1 1 479 10 21 46 10 . .. (b) For the doped semiconductor, ( E – E F )/ kT = (0.11 eV)/(0.02586 eV) = 4.254 and e = + 1 1 140 10 4254 2 . (c) The energy of the donor state, relative to the top of the valence band, is 1.11 eV – 0.15 eV = 0.96 eV. The Fermi energy is 1.11 eV – 0.11 eV = 1.00 eV. Hence, ( E – E F )/ kT = (0.96 eV – 1.00 eV)/(0.02586 eV) = – 1.547 and e = + = 1 1 0824 1547 . 35. (a) The probability that a state with energy
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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