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4 eV is the energy of the incident photo...

4 eV is the energy of the incident photon and the work function is 2 eV. The stopping potential will be

A

2V

B

4V

C

6V

D

`2 sqrt(2)` V

Text Solution

Verified by Experts

The correct Answer is:
A
`eV_(0) = hv - W_(0) = 4eV - 2eV = 2eV`
`therefore" "V_(0) = (2eV)/(e) = 2V`
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