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 x 10^–10 m.
[Note: Questions 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

Here,
$Temperature, T = 27 ° C \Rightarrow T = 273 + 27 = 300 K$

Mass of electron, $m = 9.1 \times 10^{- 31} kg$

Using formula,
$λ = \frac{h}{\sqrt{3 mkT}}$

$\Rightarrow$ $λ = \frac{6.63 \times 10^{- 34}}{\sqrt{3 \times 9.1 \times 10^{- 31} \times (\frac{8.31 \times 10^{3}}{6 \times 10^{23}}) \times 300}}$
$λ = 62.15 \times 10^{- 10} m$

Interelectronic separation, $r = 2 \times 10^{- 10} m$

Hence, $r < λ$ . That is, interelectronic spacing is less than the wavelength.
We find that wave-packets in metals strongly overlap with one another whereas, this is not the case in gas atoms.

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