Ch41-p049 - 49(a The derivative of P(E is F GH ce 1 E E F kT Ide 1h J dE K 2 E E F kT = F GH ce 1 E EF kT I1e 1h J kT K 2 E EF kT Evaluating this

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49. (a) The derivative of P ( E ) is + F H G I K J = + F H G I K J 1 1 1 1 1 22 e d dE e e kT e EE k T T T T F F F F () / / / / . c h c h Evaluating this at E = E F we readily obtain the desired result. (b) The equation of a line may be written y = m ( x – x o ) where m is the slope (here: equal to – 1/ kT , from part (a)) and x o is the x -intercept (which is what we are asked to solve for).
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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