Answer:
Option B
Explanation:
Given : work function Φ of metal = 2.28 eV
Wavelength of light λ = 500 nm = 500 x 10-9m
KEmax=hcλ−ϕ
KEmax=6.6×10−34×3×1085×10−7−2.82
KEmax=2.48−2.28=0.2eV
λmin=hp=h√2m(KE)max
203×10−34√2×9×10−31×0.2×1.6×10−19
λmin=259×10−9 = 2.80 x 10-9nm
.'. λ ≥ 2.8 x 10-9m