The correct option is D 15140 ∘A
Energy of Incident radiation according to planck's quantum theory E = hcλ
where λ = wavelength of incident radiation
E = 6.626×10−34×3×108400×10−9
= 4.96×10−19J
According to Photo electric effect:
K.Emax=E−W
where E = Energy of incident radiation
W = work function of the metal
W=2.28eV=2.28×1.6×10−19
= 3.65×10−19J
So, K.Emax=4.96×10−19−3.65×10−19 = 1.31×10−19 J
= 1.31×10−191.6×10−19
= 0.819 eV
Using relation between kinetic energy and de-broglie wavelength
E (eV) = 12400λ ∘A
So, λ = 124000.819∘A
λ = 15140.42 ∘A = 15140∘A(approx)