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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.
 Fall '08
 SPRUNGER
 Physics, Energy

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