Question

Copper has an electron number density of \(8.3 \times 10^{28}\text{ m}^{-3}\). Its Fermi energy in eV (rounded off to one decimal place) is ______.
(\(\hbar = 1.06 \times 10^{-34}\text{ J.s}\), mass of electron \(m_e = 9.10\times 10^{-31}\text{ kg}\), charge of electron \(= 1.60\times 10^{-19}\text{ C}\))

Hint:

Hint:
Use \(E_F = \dfrac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3}\), then convert the answer from joules to electron volts by dividing by the electron charge.

Answer 1

Correct Answer: 7

Step 1: Find the Fermi wave vector first.
Instead of jumping straight to the energy formula, first find the radius of the Fermi sphere in k-space, $k_F$, which is fixed by how many electron states must fit inside it:
\[ k_F = (3\pi^2 n)^{1/3} \]

Step 2: Evaluate k_F numerically.
With $n = 8.3\times 10^{28}$ m$^{-3}$:
\[ 3\pi^2 n = 2.458\times 10^{30}\ \text{m}^{-3} \]
\[ k_F = (2.458\times 10^{30})^{1/3} = 1.350\times 10^{10}\ \text{m}^{-1} \]

Step 3: Get the Fermi energy from k_F.
A free electron with wave vector $k_F$ has energy $E_F = \hbar^2 k_F^2/(2m_e)$, treating the electron like a free particle carrying momentum $\hbar k_F$.
\[ E_F = \frac{(1.06\times 10^{-34})^2 \times (1.350\times 10^{10})^2}{2\times 9.10\times 10^{-31}} \]
\[ E_F = \frac{1.1236\times 10^{-68}\times 1.822\times 10^{20}}{1.82\times 10^{-30}} = 1.125\times 10^{-18}\ \text{J} \]

Step 4: Convert to electron volts.
$1$ eV $= 1.60\times 10^{-19}$ J, so:
\[ E_F = \frac{1.125\times 10^{-18}}{1.60\times 10^{-19}} = 7.03\ \text{eV} \]

Final Answer:
Rounded to one decimal place, the Fermi energy of copper is 7.0 eV. \[ \boxed{E_F = 7.0\text{ eV}} \]