The correct option is (C) 1.64 V.
Explanation:
ɸ = 2.5eV = 2.5 × 1.6 × 10–19J.
λ = 3000 Å = 3000 × 10–10 m
h = 6.62 × 10–34 Js
(Kinetic energy)max = (1/2)mVmax2 = eVo = (hc / λ) – ɸ
∴ Vo = (hc / eλ) – (ɸ/e)
stopping potential = Vo = [{6.62 × 10–34 × 3 × 108} / {1.6 × 10–19 × 3000 × 10–10}] – (2.5) --- (ɸ/e) = ɸ in eV
= 4.137 × 10° – 2.5 V
= (4.137 – 2.5) V
= 1.637 V
= 1.64 V