1.

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?

Answer»

`-1 V`
`-3V`
`+3V`
`+4V`

SOLUTION :`(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`


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