Fermi energy of copper at 0 K and 15°C

Given: $\rho = 8.96\ \text{g/cm}^3 = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol}$; copper gives one free electron per atom.

Fermi energy at 0 K

Electron concentration:

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

Free-electron Fermi energy:

$$\begin{aligned} E_{F0} &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\ &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \times \left(\frac{3 \times 8.50 \times 10^{28}}{\pi}\right)^{2/3} \\ &= 6.025 \times 10^{-38} \times 1.874 \times 10^{19} = 1.129 \times 10^{-18}\ \text{J} = 7.05\ \text{eV} \end{aligned}$$

Fermi energy at 15°C

The temperature dependence of the Fermi energy in a metal is

$$E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right]$$

At $T = 15 + 273.15 = 288.15\ \text{K}$: $kT = \frac{1.381 \times 10^{-23} \times 288.15}{1.602 \times 10^{-19}} = 0.02484\ \text{eV}$.

$$\begin{aligned} \frac{kT}{E_{F0}} &= \frac{0.02484}{7.0487} = 3.524 \times 10^{-3} \\ \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2 &= 0.8225 \times 1.242 \times 10^{-5} = 1.02 \times 10^{-5} \\ E_F(288\ \text{K}) &= 7.0487 \times (1 - 1.02 \times 10^{-5}) = 7.0486\ \text{eV} \end{aligned}$$

The decrease is only about $7.2 \times 10^{-5}$ eV (0.001%), so the Fermi energy of a metal is practically independent of temperature.

Answer: $E_{F0} = 7.05\ \text{eV}$; at 15°C, $E_F \approx 7.05\ \text{eV}$ (lower by only $\approx 7.2 \times 10^{-5}$ eV).