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 $V / m$
AIPMT 2009, Medium
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Average energy $=$ workdone $=q E \lambda$
$2 eV =e E \times 4 \times 10^{-8}$
$E=\frac{2}{4 \times 10^{-8} m }=5 \times 10^7 \;Vm ^{-1}$
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