Question

At some temperature \( T \), two metals A and B, have Fermi energies \( \epsilon_A \) and \( \epsilon_B \), respectively. The free electron density of A is 64 times that of B. The ratio \( \frac{\epsilon_A}{\epsilon_B} \) is ............ (Round off to two decimal places).

Hint:

The relationship between Fermi energy and electron density for a free electron gas is \( n \propto \epsilon_F^{3/2} \).

Answer 1

Correct Answer: 16

Step 1: Relate the Fermi energy to electron density.
The Fermi energy \( \epsilon_F \) is related to the electron density \( n \) by the following equation for a free electron gas: \[ n = \frac{4 \pi (2m)^{3/2} \epsilon_F^{3/2}}{3 h^3}. \] Since the free electron density of metal A is 64 times that of metal B, we have \[ n_A = 64 n_B. \] Step 2: Express the ratio of Fermi energies.
Substituting the relation for the electron density in terms of Fermi energy, we get \[ \frac{n_A}{n_B} = \frac{64}{1} = \left( \frac{\epsilon_A}{\epsilon_B} \right)^{3/2}. \] Taking the cube root of both sides, \[ \left( \frac{\epsilon_A}{\epsilon_B} \right) = 4. \] Step 3: Final Answer.
Thus, the ratio \( \frac{\epsilon_A}{\epsilon_B} = 4.00 \).