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

Answer

D

Explanation

According to the Einstein.s photoelectric equation maximum kinetic energy of photoelectrons,
`KE_(max) = E_v - phi " or " 2 = 5 - phi :.phi = 3eV`
When `E_v = 5eV` then
`KE_(max)=6-3eV " or " e(V_("cathode")-V_("anode"))=3eV`
or `V_("cathode")-V_("anode")=3 V=-V_("stopping"):. V_("stopping") =-3V`