ch44_p31 - 31. (a) From Eq. 41-29, we know that N 2 N1 = e...

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31. (a) From Eq. 41-29, we know that NN e EkT 21 = −∆ . We solve for E : () 5 1 2 4 1 0.25 ln 8.62 10 eV K 2.7K ln 0.25 2.56 10 eV 0.26 eV. N Ek T N m §· ∆= = × ¨¸ ©¹
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This note was uploaded on 01/23/2011 for the course PHY 214 taught by Professor Timothybolton during the Fall '10 term at Kansas State University.

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