Given, λ1=4972Å and λ2=6216Å and I=3.6×10−3Wm−2 Intensity associated with each wavelength =
3.6×10−3
2
=1.8×10−3Wm−2 work function ϕ=hv =
hc
λ
=
(6.62×10−34)(3×108)
λ
=
12.4×103
λ
ev for different wavelengths ϕ1=
12.4×103
λ1
=
12.4×103
4972
=2.493eV=3.984×10−19J ϕ2=
12.4×103
λ2
=
12.4×103
6216
=1.994eV=3.184×10−19 J Work function for metallic surface ϕ=2.3eV (given) ϕ2<ϕ Therefore, ϕ2 will not contribute in this process. Now, no. of electrons per m2−s= no. of photons per m2−s no. of electrons per m2−s=
1.8×10−3
3.984×10−19
×10−4 (∵1cm2=10−4m2)=0.45×1012 So, the number of photo electrons liberated in 2 sec. =0.45×1012×2 =9×1011