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The threshold energy of a surface is 2 e...

The threshold energy of a surface is 2 eV. When irradiated with some radiations, the stopping potential is found to be 3.6 V. The energy of the incident photon is

A

`4.48xx10^(-19)`J

B

`8.96xx10^(-19)`J

C

`15xx10^(-18)`J

D

`17.92xx10^(-19)`J

Text Solution

Verified by Experts

The correct Answer is:
B
Threshold energy `=W_(0)=2eV`
`hv=hv_(0)+`max K.E.
`hv=hv_(0)+eV_(s)=2eV+3.6eV=5.6 eV`
`=5.6xx1.6xx10^(-19)J=8.96xx10^(-19)J`
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