Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

(N/A) The temperature is $T = 27^{\circ} C = 27 + 273 = 300 \; K$.
The mean separation between two electrons is $r = 2 \times 10^{-10} \; m$.
The de Broglie wavelength $\lambda$ of an electron at temperature $T$ is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$, where $k$ is the Boltzmann constant.
Given constants:
$h = 6.63 \times 10^{-34} \; J \cdot s$
$m = 9.11 \times 10^{-31} \; kg$
$k = 1.38 \times 10^{-23} \; J/K$
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{11.30 \times 10^{-51}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{3.36 \times 10^{-25}} \approx 1.97 \times 10^{-9} \; m$.
Comparing this with the mean separation $r = 2 \times 10^{-10} \; m$, we see that $\lambda \approx 19.7 \times 10^{-10} \; m$, which is about $10$ times larger than the mean separation between electrons.