Question:

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 distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]

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:

$\lambda=\frac{h}{\sqrt{3 m k T}}$

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

$\therefore \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$

$\approx 6.2 \times 10^{-9} \mathrm{~m}$

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.