InterviewSolution
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InterviewSolution
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Employing Eq. (6.4g), find at T=0 (a) the maximum kinetic energy of free electrons in a metal if their concentration is qual to n, (b) the mean kinetic energy of free electrons if their maximum kinetic energy T_(max) is known. |
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Answer» SOLUTION :(a) From the fromula `DN=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3))E^(1//2)dE` the maximum value `E_(max)` of `E` is determined in terms of `n` by `n=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3)) int_(0)^(E_(max))E^(1//2)dE` `=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3))(2)/(3)E_(max)^(3//2)` or `E_(max)^(3//2)=( ħ^(2)/(2M))^(3//2)(3 pi^(2)n)` `E_(max)=( ħ^(2))/(2m)(3PI^(2)n)^(2//3)` (b) Mean `K.E lt E gt` is `lt E geint_(0)^(E_(max))Edn//int_(0)^(E_(max))dn` `=int_(0)^(E_(max))E^(3//2)//int_(0)^(E_(max))E^1//2dE(2)/(5)E_(max)^(5//2)//(2)/(3)E_(max)^(3//2)=(3)/(5)E_(max)` |
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