Q: Photons with energy 5eV are incident on a cathode C, on a photoelectric cell. The maximum energy of the 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

(A) 3 V

(B) –3 V

(C) –1 V

(D) 4 V

**Solution :** KE_{max} = 2 ev

E_{photon} = 5ev

φ = 5 ev – 2 ev = 3 ev

Now no current when

E_{photon} = 6ev

i.e. KE_{max} < 3ev

eV_{max} < 3ev

V_{max }< 3ev/e = 3V max

φ = 3eV

K.E. = 3eV in Second case

Correct option is (B)