When cathode C of photocell is incident with photon of 5 eV energy ,maximum KE of electron emitted is 2 eV.For which value of stopping potential on anode (A) photo incident with energy of 6 eV,number of electron reaching to anode will become zero?

Options

A. `-1 V`
B. `-3V`
C. `+3V`
D. `+4V`

Answer

B

Explanation

`(1)/(2)m_(max)^(2)=hf-phi_(0)`
2=5-`phi_(0)`
`therefore phi_(0)=3 eV`
Again `eV_(0)=hf-phi_(0)` 3 eV
`therefore V_(0)=3eV`
`therefore V_(C)-V_(A)=3V`
`therefore 0-V_(A)=3V`
`therefore V_(A)=-3V`