1.

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

Answer»

2V
4V
6V
`2 sqrt(2)` V

Solution :`eV_(0) = hv - W_(0) = 4EV - 2eV =2eV`
`therefore""V_(0) = (2eV)/(e) = 2V`


Discussion

No Comment Found

Related InterviewSolutions