Fermi energy from electron concentration

For free electrons in a metal at 0 K, all states up to $E_{F0}$ are filled, giving

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

Given: $n = 8.43 \times 10^{28}\ \text{m}^{-3}$, $h = 6.626 \times 10^{-34}\ \text{J s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$.

Calculation

$$\begin{aligned} \left(\frac{3n}{\pi}\right)^{2/3} &= \left(\frac{3 \times 8.43 \times 10^{28}}{\pi}\right)^{2/3} = (8.050 \times 10^{28})^{2/3} = 1.864 \times 10^{19}\ \text{m}^{-2} \\ \frac{h^2}{8m_e} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2 \\ E_{F0} &= 6.025 \times 10^{-38} \times 1.864 \times 10^{19} = 1.123 \times 10^{-18}\ \text{J} \end{aligned}$$

In electron-volts:

$$E_{F0} = \frac{1.123 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.01\ \text{eV}$$

This agrees with the measured value for copper (about 7 eV). The Fermi temperature $T_F = E_{F0}/k \approx 8.1 \times 10^{4}$ K, far above any working temperature, so $E_F$ stays almost the same at room temperature.

Answer: $E_{F0} \approx 1.12 \times 10^{-18}\ \text{J} = 7.01\ \text{eV}$.