Question

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 $2\times {10}^{–10}m$.

Answer 1

Given,
temperature, $T={27}^{\circ }C=27+273=300K$
Mean separation between two electrons,
$r=2\times {10}^{-10}m$
As the de Broglie wavelength of an electron is given as:
$\lambda =\frac{h}{\sqrt{2mK}}$
Average energy of a gas at temperature T, is given as:
$K=\frac{3}{2}kT$
$\lambda =\frac{h}{\sqrt{3kTm}}$
Where,
h = Planck’s constant $6.626\times {10}^{-34}Js$
m = Mass of an electron $=9.11\times {10}^{-31}\text{kg}$
k = Boltzmann constant
$=1.38\times {10}^{-23}\text{J /mol K}$
Therefore,
$\lambda =\frac{6.626\times {10}^{-34}}{\sqrt{3\times 1.38\times {10}^{-23}\times 300\times 9.11\times {10}^{-31}}}$
$\lambda =6.2\times {10}^{-9}m$
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
Final Answer: $\lambda =6.2\times {10}^{-9}m$, Which is much greater than the given inter-electron separation.