The mean free path of conduction electrons in copper is about `4 xx 10^(-8)m`. Find the electric field which can give, on an avrage, `2 eV` energy to a conduction electron in a block of copper.

Options

A. `5 times 10^8 V//m`
B. `5 times 10^-8 V//m`
C. `5 times 10^-7 V//m`
D. `5 times 10^T V//m`

Answer

A

Explanation

Mean free path `d=4 times 10^-8m, E=?`
Energy of electron =2eV
F=eE
Work done on electron when it moves through distance `d=eED` This work done is equal to the energy transformed to the electron, so
`therefore dEd=2ev`
`Ed=2v`
`E=(2v)/d=(2 times v)/(4 times 10^-4)=5 times 10^7 v//m`