As we know, threshold wavelength (λ0)=hcϕ
⇒λ0=(6.63×10−34)×3×1082.3×(1.6×10−19)=5.404×10−7m.
⇒λ0=5404∘A
Hence, wavelength 4144∘A and 4972∘A will emit electron from the metal surface.
For each wavelength energy incident on the surface per unit time
= intensity of each × area of the surface wavelength
=3.6×10−33×(1 cm)2=1.2×10−7joule
Therefore, energy incident on the surface for each wavelengths is 2s
E=(1.2×10−7)×2=2.4×10−7J
Number of photons n1 due to wavelength 4144∘A
n1=(2.4×10−7)(4144×10−10)(6.63×10−34)(3×108)=0.5×1012
Number of photon n2 due to the wavelength 4972∘A
n2=(2.4×10−7)(4972×10−10)(6.63×10−34)(3×108)=0.572×1012
Therefore total number of photoelectrons liberated in 2s,
N=n1+n2
=0.5×1012+0.575×1012
=1.075×1012.