Fermi energy of copper at 0 K

The Fermi energy at 0 K for a free-electron metal is

$$E_{F0} = \frac{h^2}{8 m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

where $n$ is the free electron concentration.

Origin of the formula: at 0 K all states up to $E_{F0}$ are filled. Integrating the density of states $g(E) = \frac{8\pi\sqrt{2}m_e^{3/2}}{h^3}E^{1/2}$ from 0 to $E_{F0}$ gives $n = \frac{8\pi}{3}\left(\frac{2m_e E_{F0}}{h^2}\right)^{3/2}$; solving for $E_{F0}$ gives the expression above.

Given: $\rho = 8.96\ \text{g/cm}^3 = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol} = 0.0635\ \text{kg/mol}$, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$. Copper is monovalent, so each atom gives one free electron.

Electron concentration

$$n = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.0635} = 8.50 \times 10^{28}\ \text{m}^{-3}$$

Fermi energy

$$\begin{aligned} \left(\frac{3n}{\pi}\right)^{2/3} &= \left(\frac{3 \times 8.50 \times 10^{28}}{\pi}\right)^{2/3} = 1.874 \times 10^{19}\ \text{m}^{-2} \\ \frac{h^2}{8m_e} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2 \\ E_{F0} &= 6.025 \times 10^{-38} \times 1.874 \times 10^{19} = 1.13 \times 10^{-18}\ \text{J} \\ &= \frac{1.13 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.05\ \text{eV} \end{aligned}$$

The corresponding Fermi speed is $v_F = \sqrt{2E_{F0}/m_e} \approx 1.57 \times 10^{6}\ \text{m/s}$, showing that even at 0 K the top electrons move very fast.

Answer: $n = 8.50 \times 10^{28}\ \text{m}^{-3}$, $E_{F0} \approx 1.13 \times 10^{-18}\ \text{J} = 7.05\ \text{eV}$.