Question

The mean free path of conduction electrons in copper is about $4 \times 10^{-8} m .$ Find the electric field which can give, on an average, $2 eV$ energy to a conduction electron in a block of copper.

• A. $5 \times 10^{7} V / m$
• B. $5 \times 10^{6} V / m$
• C. $5 \times 10^{-6} V / m$
• D. $5 \times 10^{-7} V / m$

Answer 1

Answer: (A)

Mean free path $d=4 \times 10^{-8} m , \quad E =?$
Energy of electron $=2 eV$
$F = e E$
Work done on electron when it moves through distance $d = eED$.
This work done is equal to the energy transfered to the electron, so
$\therefore eEd =2 eV $
$Ed =2 V$
$E=\frac{2 V}{d}=\frac{2}{4 \times 10^{-8}}$
$=5 \times 10^{7} V / m$