Given, λ1=4972A˚ and λ2=6216A˚ and I=3.6×10−3Wm−2 Intensity associated with each wavelength =23.6×10−3=1.8×10−3Wm−2 Work function ϕ=hv=λhc =λ(6.62×10−34)(3×108)=λ12.4×103ev for different wavelengths ϕ1=λ112.4×103=497212.4×103=2.493eV=3.984×10−19 Jϕ2=λ212.4×103=621612.4×103=1.994eV=3.184×10−19 J Work function for metallic surface ϕ=2.3 eV (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 =3.984×10−191.8×10−3×10−4(∵1 cm2=10−4 m2)=0.45×1012 So, the number of photo electrons liberated in 2 sec. =0.45×1012×2 =9×1011