InterviewSolution
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InterviewSolution
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Compute the typical de-Broglie wavelength of an electrons in a metal at `27^(@)C` and compare it with the mean separation between two electrons in a metal which is given to be about `2xx10^(-10)m`. |
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Answer» Given , T `= 27 + 273 = 300 K, r = 2xx 10^(-10)` m Momemtum , P = ` sqrt(3" m KT") = sqrt(3xx 9.11 xx10^(31) xx1.38 xx10^(-23) xx300 ) = 1.06xx10^(-25)` kg -m/s `lambda = h/p = (6.63 xx10^(-34))/(1.06 xx10^(-25)) = 62.6 xx10^(-10) ` m , Mean seperation `r = 2xx 10^(-10)` m ` lambda/r = (62.6 xx10^(-10))/(2 xx10^(-10))= 31.3` We can see that de - Brogile wavelength is much greater than the electron seperation . |
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