p42_039

# p42_039 - e ( E c E F ) /kT 1 and e ( E v E F ) /kT 1....

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39. (a) The number of electrons in the valence band is N ev = N v P ( E v )= N v e ( E v E F ) /kT +1 . Since there are a total of N v states in the valence band, the number of holes in the valence band is N hv = N v N ev = N v · 1 1 e ( E v E F ) /kT +1 ¸ = N v e ( E v E F ) /kT +1 . Now, the number of electrons in the conduction band is N ec = N c P ( E c )= N c e ( E c E F ) /kT +1 , Hence, from N ev = N hc ,weget N v e ( E v E F ) /kT +1 = N c e ( E c E F ) /kT +1 . (b) In this case,
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Unformatted text preview: e ( E c E F ) /kT 1 and e ( E v E F ) /kT 1. Thus, from the result of part (a), N c e ( E c E F ) /kT N v e ( E v E F ) /kT , or e ( E v E c +2 E F ) /kT N v /N c . We solve for E F : E F 1 2 ( E c + E v ) + 1 2 kT ln N v N c ....
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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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