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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
Energy `= 2eV`
`eV_(0)=2eVimplies V_(0)=2`
Now, electric field `E= (2)/(4xx10^(-8))=0.5xx10^(8)`
Hence `E= 5xx10^(7)Vm^(-1)`
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