Ncert Exercises & Solutions

Question 11.36:
Compute the typical de Broglie wavelength of an electron in metal at 27 QC and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10^-10m. [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 that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishability has many fundamental implications which you will explore in more advanced physics courses.]

Hint: \(\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}\)

Step 1: Find the de Broglie wavelength of an electron.
de Broglie wavelength of an electron is given as:

λ = \frac{h}{\sqrt{3 mkT}}

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

\begin{aligned} ∴ λ & = \frac{6 .6 \times 10^{- 34}}{\sqrt{3 \times 9 .11 \times 10^{- 31} \times 1 .38 \times 10^{- 23} \times 300}} \\ = 6 .2 \times 10^{- 9} m \end{aligned}

Step 2: Compare the de Broglie wavelength with the inter-electron separation.
The inter-electron separation = 2 x 10^-10m
de Broglie wavelength of an electron = 6.2 x 10^-9m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

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