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Question
Compute the typical de Broglie wavelength of an electron 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 2 × 10−10 m.
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Solution
Temperature, T = 27°C = 27 + 273 = 300 K
Mean separation between two electrons, r = 2 × 10−10 m
De Broglie wavelength of an electron is given as:
`lambda = "h"/sqrt(3"mkT")`
Where,
h = Planck’s constant = 6.6 × 10−34 Js
m = Mass of an electron = 9.11 × 10−31 kg
k = Boltzmann constant = 1.38 × 10−23 J mol−1 K−1
∴ `lambda = (6.6 xx 10^(-34))/sqrt(3 xx 9.11 xx 10^(-31) xx 1.38 xx 10^(-23) xx 300)`
≈ 6.2 × 10−9 m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
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