Question

For a metal, electron density is $6.4\times10^{28}\ \mathrm{m}^{-3}$. The Fermi energy is ................. eV. (Specify answer up to one digit after the decimal point.)
 

Hint:

Higher electron density → larger Fermi energy, since $E_F \propto n^{2/3}$.

Answer 1

Correct Answer: 5.6

Step 1: Use formula for Fermi energy.
$E_F = \frac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3}$.

Step 2: Substitute values.
$n = 6.4\times10^{28}$, $\hbar = 1.055\times10^{-34}$ J·s, $m_e = 9.11\times10^{-31}$ kg.

Step 3: Compute inside term.
$(3\pi^2 n)^{2/3} = (3\pi^2 \times6.4\times10^{28})^{2/3} \approx (1.89\times10^{30})^{2/3} \approx 1.51\times10^{20}$.

Step 4: Compute Fermi energy.
$E_F = \frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})}(1.51\times10^{20})$.
$E_F \approx 1.68\times10^{-18}$ J.

Step 5: Convert to eV.
$1\ \text{eV} = 1.6\times10^{-19}$ J.
$E_F = \frac{1.68\times10^{-18}}{1.6\times10^{-19}} = 10.5\ \text{eV}$.