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The mean free path of electrons in a metal is $4\times10^{-8}$ 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
$5\times10^7$
B
$8\times10^7$
C
$5\times10^{-11}$
D
$8\times10^{-11}$
Detailed Solution
Between two collisions the electron moves freely over the mean free path d and gains energy from the field.
Energy gained = work done by the field = force × distance = eEd
$eEd = 2$ eV, so $Ed = 2$ V
$E = \frac{2}{d} = \frac{2}{4\times10^{-8}}$
$E = 5\times10^7$ V/m
Energy gained = work done by the field = force × distance = eEd
$eEd = 2$ eV, so $Ed = 2$ V
$E = \frac{2}{d} = \frac{2}{4\times10^{-8}}$
$E = 5\times10^7$ V/m