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

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31. (a) From Eq. 41-29, we know that N N e E kT 2 1 = −∆ . 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 E kT N m § · = = × ¨ ¸ © ¹ = × (b) Using the result of problem 83 in Chapter 38,
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