p42_029 - energy and zero for energies above. Thus, E avg =...

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29. The average energy of the conduction electrons is given by E avg = 1 n Z 0 EN ( E ) P ( E ) dE where n is the number of free electrons per unit volume, N ( E ) is the density of states, and P ( E )isthe occupation probability. The density of states is proportional to E 1 / 2 ,sowemaywr ite N ( E )= CE 1 / 2 , where C
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Unformatted text preview: energy and zero for energies above. Thus, E avg = C n Z E F E 3 / 2 dE = 2 C 5 n E 5 / 2 F . Now n = Z ∞ N ( E ) P ( E ) dE = C Z E F E 1 / 2 dE = 2 C 3 E 3 / 2 F . We substitute this expression into the formula for the average energy and obtain E avg = µ 2 C 5 ¶ E 5 / 2 F Ã 3 2 CE 3 / 2 F ! = 3 5 E F ....
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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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