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: Understanding the Concept:
In the free electron model of a metal, conduction electrons fill up available momentum states from the lowest energy up, obeying the Pauli exclusion principle. At absolute zero, all states are filled up to a maximum energy called the Fermi energy, \(E_F\).

Step 2: Key Formula or Approach:
For a free electron gas with number density \(n\), the Fermi energy is:
\[ E_F = \frac{\hbar^2}{2m_e}\left(3\pi^2 n\right)^{2/3} \]
This comes from counting the allowed k-states inside a sphere of radius \(k_F\) in momentum space and setting that count equal to the total number of electrons.

Step 3: Detailed Explanation:
First evaluate the bracket:
\[ 3\pi^2 n = 3\pi^2 \times 8.3\times 10^{28} = 2.458\times 10^{30}\text{ m}^{-3} \]
Take the two-thirds power:
\[ (2.458\times 10^{30})^{2/3} = 1.822\times 10^{20}\text{ m}^{-2} \]
Now put in \(\hbar\) and \(m_e\):
\[ E_F = \frac{(1.06\times 10^{-34})^2}{2\times 9.10\times 10^{-31}}\times 1.822\times 10^{20} \]
\[ E_F = \frac{1.1236\times 10^{-68}}{1.82\times 10^{-30}}\times 1.822\times 10^{20} = 6.174\times 10^{-39}\times 1.822\times 10^{20} \]
\[ E_F = 1.125\times 10^{-18}\text{ J} \]
Convert to eV by dividing by the electron charge:
\[ E_F = \frac{1.125\times 10^{-18}}{1.60\times 10^{-19}} = 7.03\text{ eV} \]

Final Answer:
The Fermi energy of copper is about 7.0 eV. \[ \boxed{E_F = 7.0\text{ eV}} \]