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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

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Question : 36
Total: 37
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.
Solution:  
Considering free electrons as gas. Kinetic energy at temperature T
1
2
mv2 =
3
2
 kT
m2v2 = 3 kmT
Wavelength associated with moving electron
λ =
h
√3kmT
 =
6.63×10−34
√3×9.1×10−31×1.38×10−23×300
m = 6.2 × 10−9 m
Given that mean separation between two electrons is about 2 × 10–10 m.
∴
λ
r
 =
6.2×10−9
2×10−10
 = 31
So, de-Broglie wavelength is much greater than the electron separation
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