Here λ = 5000 Å = 5 x 10-7 m
W0 = 1.9 ev
(i) Energy of a photon,
E = 2.475 eV
(ii) K.E of a photoelectron,
K.E = hυ – W0 = 2.475 – 1.9 = 0.575 eV
(iii) Let V0 be the stopping potential. Then
eV0 = \(\frac{1}{2}\)mv2 = K.E of a photoelectron
V0 = \(\frac{0.575}{e}\) eV
V0 = 0.575 V