Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.36
NCERT Solutions for Class 12 Physics Chapter 11 Question-36
Additional Exercise
Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10⁻¹⁰ m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distintguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
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Temperature, T = 27C = 27º+ 273 =300K
Mean separation between two electrons , r = 2 x 10⁻¹⁰ m
De Broglie wavelength of an electron is given as :
λ = h/ √ [3mkT]
Where ,h = Planck’s constant = 6.6 x 10-34 Js
m = Mass of an electron = 9.11 x 10-31 kg
k= Boltzmann constant = 1.38 x 10⁻23 J mol-1 K-1
Therefore ,λ = (6.6 x 10-34)/ √ [3 x (9.11 x 10-31) x (1.38 x 10⁻23) x 300]
≈ 6.2 x 10 ⁻⁹ m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.