42. (a) Measured from the top of the valence band, the energy of the donor state is E =1.11 eV-0.11 eV =
1.0eV. We solve EF from Eq. 42-6:

-1
EF = E - kT ln P - 1 )

= 1.0eV- (8.62 × 10-5 eV/K)(300 K) ln (5.00 × 10-5)-1 - 1
= 0.744 eV .
(b) Now E =1.11 eV, so
1 1
P (E) = =
F
e(E-E )/kT +1 e(1.11 eV-0.744 eV)/[(8.62×10-5 eV/K)(300 K)] +1
= 7.13 × 10-7 .

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