Q. 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) $8 \times 10^{7}$
B) $5 \times 10^{- 11}$
C) $8 \times 10^{- 11}$
D) $5 \times 10^{7}$
Explanation
(d) $K = eV = eEl$
$\Rightarrow 2(eV) = eEl \Rightarrow E = \frac{2}{l} = \frac{2}{4 \times 10^{- 8}} \Rightarrow E = 5 \times 10^{7}V/m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.