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The mean free path of electrons in a met...

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`

A

`8xx10^(7)`

B

`5xx10^(-11)`

C

` 8xx10^(-11)`

D

`5xx10^(7)`

Text Solution

Verified by Experts

The correct Answer is:
D
Kinetic energy (k) = eV = eEl
`implies 2eV = eEl implies E = 2/l (V//m)`
`= 2/(4xx10^(-8))=5xx10^7 V//m`
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