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What fraction (in per cent) of free electrons in a metal at T=0 has kinetic energy exceeding half the maximum energy? |
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Answer» SOLUTION :The fraction is `eta= int_(1/2E_(max))^(E_(max))E^(1//2)dE// int_(0)^(E_(max))E^(1//2)dE= 1-2^(-3//2)= 0.646 or 64.6%` |
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