The mean free path of electrons in a metal is 4 × 10⁻⁸ m. The electric field which can give on an average 2 eV

Question

The mean free path of electrons in a metal is 4 × 10⁻⁸ m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of V m⁻¹:

Accepted Answer

The energy ($W$) gained by an electron accelerating through a distance ($d$) under the influence of an electric field ($E$) is given by the work done: $W = Fd = (eE)d$. 
Given:
Energy ($W$) = 2 eV
Distance ($d$, mean free path) = $4 	imes 10^{-8}$ m

Substituting the values (noting that 2 eV energy corresponds to the work done on charge $e$ moving through a potential difference of 2 Volts, or simply equating $eV$ units):
$eEd = 2 	ext{ eV}$
$E 	imes d = 2 	ext{ V}$
$E = \frac{2 	ext{ V}}{4 	imes 10^{-8} 	ext{ m}}$
$E = 0.5 	imes 10^8 	ext{ V/m}$
$E = 5 	imes 10^7 	ext{ V/m}$

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