42. (a) Measured from the top of the valence band, the energy of the donor state isE=1.11 eV−0.11 eV =1.0eV. We solveEFfrom Eq. 42-6:EF=E−kTln£P−1−1¤).0eV−(8.62×10−5eV/K)(300 K) ln£(5.00×10−5)−1−1¤=0.744 eV
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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.