Metals — Question 4

Question 4

Counting occupied wave vectors gives the Fermi energy

A three-dimensional free-electron metal at zero temperature contains two spin states per wave vector. A volume V supplies one spatial state per reciprocal-space volume $(2\pi)^3/V$.

1. Derive $k_F$ and $E_F$ in terms of electron density n.

2. For $n=8.00\times10^{28}\,\mathrm{m^{-3}}$, $\hbar=1.055\times10^{-34}\,\mathrm{J\,s}$, $m=9.11\times10^{-31}\,\mathrm{kg}$, and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$, evaluate them.

Question 4 - Solution

Counting occupied wave vectors gives the Fermi energy

(a) Method and calculation

$$N=2\frac{V}{(2\pi)^3}\frac{4\pi k_F^3}{3}\Rightarrow\boxed{k_F=(3\pi^2n)^{1/3}}.$$$$\boxed{E_F=\hbar^2k_F^2/(2m)}.$$

(b) Interpretation and further calculation

$$k_F=(3\pi^2\times8.00\times10^{28})^{1/3}=1.333\times10^{10}\,\mathrm{m^{-1}}.$$$$\boxed{E_F=1.085\times10^{-18}\,\mathrm J=6.78\,\mathrm{eV}}.$$

Physical check

The spin factor is essential. Even at zero temperature the electrons fill a sphere of nonzero radius because Pauli exclusion forbids placing all electrons at k=0.