Photons with energy `5 eV` are incident on a cathode `C` in a photoelectric cell . The maximum energy of emitted photoelectrons is `2 eV`. When photons of energy `6 eV` are incident on `C` , no photoelectrons will reach the anode `A` , if the stopping potential of `A` relative to `C` is

Options

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

Answer

B

Explanation

`e V_(3) = (1)/(2) m v_("max")^(2) = h v - phi_(0)`
`2 = 5 - phi_(0) rArr phi_(0) = 3 eV`
In second case
`eV_(3) = 6-3 = 3eV rArr V_(0) = 3V`
`therefore V_(AC) = -3V`