The occupancy probability function can be applied to semiconductorsas well as to

Question

The occupancy probability function can be applied to semiconductorsas well as to metals. In semiconductors the Fermi energy is closeto the midpoint of the gap between the valence band and theconduction band. Consider a semiconductor with an energy gap of0.74 eV, at T = 280 K. What is the probability that(a) a state at the bottom of the conduction bandis occupied and (b) a state at the top of thevalence band is not occupied? (Note: In a puresemiconductor, the Fermi energy lies symmetrically between thepopulation of conduction electrons and the population of holes andthus is at the center of the gap. There need not be an availablestate at the location of the Fermi energy.)

Explanation / Answer

Given :
Eg = 0.74 eV
EF = 1/2 Eg = 0.37 eV (for pure metal)
T = 280 K

a )
Occupied below
P(E) = 1/(e(E-Ef)/kT + 1)
P(E) = 1/(e(.37/(8.62e-5*280)) + 1)
P(E) = ------
b:
For unoccupied above
P(E) = 1/(e(.37/(8.62e-5*280))+1)
P(E) = -------
   Solve the above
   I hope it helps you

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