The mean free path of electrons in a metal is 4×10−8 m. The electric field which can give on an average 2eV energy to an electron in the metal will be in units of V/m ?

The mean free path of electrons in a metal is 4×10−8 m. The electric field which can give on an average 2eV energy to an electron in the metal will be in units of V/m ?

(a) \( 5 \times 10^{-11} \)

(b) \( 8 \times 10^{-11} \)

(c) \( 5 \times 10^{7} \)

(d) \( 8 \times 10^{7} \)

Answer :  (c) \( 5 \times 10^{7} \)

Mean free path: \( d = 4 \times 10^{-8} \, \text{m} \)

Desired energy gain: \( 2 \, \text{eV} \)

\( E = \frac{V}{d} \)

\( V = 2\,\text{V} \Rightarrow E = \frac{2}{4 \times 10^{-8}} = 0.5 \times 10^8 = 5 \times 10^7 \,\text{V/m} \)

\( 1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J} \)

\( E = 5 \times 10^7\,\text{V/m} \)