Question

What is the velocity of the conduction electron of silver having Fermi energy \(5.52\ \text{eV}\)?

• A. \(1.39 \times 10^{6}\ \text{m/s}\)
• B. \(2.39 \times 10^{6}\ \text{m/s}\)
• C. \(0.89 \times 10^{6}\ \text{m/s}\)
• D. \(0\)

Hint:

Set the Fermi energy equal to \(\tfrac{1}{2}mv^2\) and solve for \(v\); remember to convert eV to joules.

Answer 1

The Correct Option is A

Step 1: In the free electron model the fastest conduction electrons sit at the Fermi level. Their speed is the Fermi velocity \(v_F\), found by equating the Fermi energy to kinetic energy: \[E_F = \tfrac{1}{2} m v_F^{2}.\]
Step 2: Solve for the velocity: \[v_F = \sqrt{\frac{2 E_F}{m}}.\]
Step 3: Convert the Fermi energy to joules. With \(1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}\), \[E_F = 5.52 \times 1.602 \times 10^{-19} = 8.84 \times 10^{-19}\ \text{J}.\]
Step 4: Put in the electron mass \(m = 9.11 \times 10^{-31}\ \text{kg}\): \[v_F = \sqrt{\frac{2 \times 8.84 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{1.94 \times 10^{12}}\ \text{m/s}.\]
Step 5: Take the square root: \[v_F \approx 1.39 \times 10^{6}\ \text{m/s}.\] This matches option (A). \[\boxed{v_F \approx 1.39 \times 10^{6}\ \text{m/s}}\]