As we know, threshold wavelength (λ0) = Φhc ⇒ λ0 = 2.3×(1.6×10−19)(6.63×10−34)×3×108 = 5.404 × 10−7 m ⇒ λ0 = 5404 A0 Hence, wavelength 4144 Å and 4972 Å 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 = 33.6×10−3 × (1cm)2 = 1.2 × 10−7 joule Therefore, energy incident on the surface for each wavelength in 2s E = (1.2 × 10−7) × 2 = 2.4 × 10−7 J Number of photons n1 due to wavelength 4144 Å n1 = (6.63×10−34)(3×108)(2.4×10−7)(4144×10−10) = 0.5 8 1012 Number of photons n2 due to the wavelength 4972 Å n2 = (6.63×10−34)(3×108)(2.4×10−7)(4972×10−10) = 0.572 × 1012 Therefore total number of photo electrons liberated in 2s, N = n1+n2 = 0.5 × 1012 + 0.575 × 1012 = 1.075 × 1012