Question

The mean free path of electrons in a metal is $4\times {10}^{-8}m$. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of V/m :

• A. $5\times {10}^7$
• B. $8\times {10}^7$
• C. $5\times {10}^{-11}$
• D. $8\times {10}^{-11}$

Answer 1

Answer: (A)

$5\times {10}^7$

qV = 2eV
$\Rightarrow 1.6\times {10}^{-19}V=2\times 1.6\times {10}^{-19}V$
$\Rightarrow V=2V$
$E=\frac{V}{d}$
$\Rightarrow E=\frac{2V}{4\times {10}^{-8}}=5\times {10}^7$