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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 36 of 37
Marks: +1, -0
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 × 10−1010^{-10} m.
Solution:  
Considering free electrons as gas. Kinetic energy at temperature T
12mv2\frac{1}{2} m v^2 = 32\frac{3}{2} kT
m2v2m^2 v^2 = 3 kmT
Wavelength associated with moving electron
λ = h3kmT\frac{h}{\sqrt{3 k m T}} =
6.63×10−343×9.1×10−31×1.38×10−23×300\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 9.1 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}
m = 6.2 × 10−910^{-9} m
Given that mean separation between two electrons is about 2 × 10−1010^{-10} m.
∴ λr\frac{\lambda}{r} = 6.2×10−92×10−10\frac{6.2 \times 10^{-9}}{2 \times 10^{-10}} = 31
So, de-Broglie wavelength is much greater than the electron separation
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