The mean free path of electrons in a metal is `4 xx 10^(-8)m` The electric field which can give on an average `2eV` energy to an electron in the metal will be in the units `V//m`

Options

A. `5xx10^(7)`
B. `8xx 10^(7)`
C. `5xx10^(-11)`
D. `8xx10^(-11)`

Answer

A

Explanation

`qV=2eV`
`implies 1.6xx10^(-19)V=2xx 1.6xx10^(-19)V `
`implies V=2V`
`E=(V)/(d)`
`implies E=(2V)/(4 xx 10^(-8))=5xx10^(7)`