ch41-p015 - 15. We use N O ( E ) = N ( E ) P ( E ) = CE1/ 2...

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15. We use 1 () / 1/2 O ()() 1 F EE k T NE NEPE C E e == + , where (see Problem 41-3) 3/2 31 3/2 56 3/2 3 3 33 4 3 27 3 8 2 8 2 (9.109 10 kg) 1.062 10 kg / J s (6.626 10 J s) 6.81 10 m (eV) . m C h ππ −− × = × ×⋅ (a) At E = 4.00 eV, 27 3 1/ 2 28 3 1 O 5 6.81 10 m (eV) (4.00eV) 1.36 10 m eV . exp (4.00eV 7.00eV) /[(8.62 10 eV / K)(1000K)] 1 N × −× + (b) At E = 6.75 eV, 27 3 3/ 2 28 3 1 O 5 6.81 10 m (eV) (6.75eV) 1.68 10 m eV . exp (6.75eV 7.00eV) /[(8.62 10 eV / K)(1000K)]
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