When violet light of λ = 4000 Au strikes the cathode of a photocell a retarding potential of 0.4 V is required to stop emission of electrons.Calculate (i) light frequency(ii) photon energy(iii)work function (iv) threshold frequency and (v) net energy after the electron leaves the surface.
(ii) E = hf = 6.625 × 10–34 × 7.5 × 1014
= 4.95 × 10–19 J = 3.1 eV
(iii) W0 = hf – K.E. = hf – V0 = 3.1 – 0.4 = 2.7 eV
(v) Net energy hf – W0 = 3.1 – 2.7 = 0.4 eV = 6.4 × 10–20 J
