p42_043 - E E F ) /kT = (0 . 11 eV) / (0 . 02586 eV) = 4 ....

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43. (a) The probability that a state with energy E is occupied is given by P ( E )= 1 e ( E E F ) /kT +1 where E F is the Fermi energy, T is the temperature on the Kelvin scale, and k is the Boltz- mann constant. If energies are measured from the top of the valence band, then the energy associated with a state at the bottom of the conduction band is E =1 . 11 eV. Furthermore, kT =(8 . 62 × 10 5 eV / K)(300 K) = 0 . 02586 eV. For pure silicon, E F =0 . 555 eV and ( E E F ) /kT = (0 . 555 eV) / (0 . 02586 eV) = 21 . 46. Thus, P ( E )= 1 e 21 . 46 +1 =4 . 79 × 10 10 . For the doped semiconductor, (
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Unformatted text preview: E E F ) /kT = (0 . 11 eV) / (0 . 02586 eV) = 4 . 254 and P ( E ) = 1 e 4 . 254 + 1 = 1 . 40 10 2 . (b) The energy of the donor state, relative to the top of the valence band, is 1 . 11 eV . 15 eV = . 96 eV. The Fermi energy is 1 . 11 eV . 11 eV = 1 . 00 eV. Hence, ( E E F ) /kT = (0 . 96 eV 1 . 00 eV) / (0 . 02586 eV) = 1 . 547 and P ( E ) = 1 e 1 . 547 + 1 = 0 . 824 ....
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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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