Question

The mean free path of conduction electrons in copper is about \(4 \times 10^{-8} \, \text{m}\). For a copper block, the electric field which can give on average \(1 \, \text{eV}\) energy to a conduction electron is:

• A. \(F = 2.5 \times 10^7 \, \text{V/m}\)
• B. \(F = 2.5 \times 10^8 \, \text{V/m}\)
• C. \(F = 2.5 \times 10^9 \, \text{V/m}\)
• D. \(F = 2.5 \times 10^6 \, \text{V/m}\)

Hint:

When dealing with electric fields, remember E = Energy e·d where d is the mean free path.

Answer 1

The Correct Option is A

The energy gained by an electron in an electric field $E$ over a distance $d$ is:
\[eE \cdot d = \text{Energy}\]
Given $d = 4 \times 10^{-8} \text{ m}$, $e = 1.6 \times 10^{-19} \text{ C}$, and $\text{Energy} = 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$:
\[E = \frac{\text{Energy}}{e \cdot d} = \frac{1.6 \times 10^{-19}}{1.6 \times 10^{-19} \times 4 \times 10^{-8}} \approx 2.5 \times 10^{7} \text{ V/m}\]