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 =
3.6×10−3
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 =
(2.4×10−7)(4144×10−10)
(6.63×10−34)(3×108)
= 0.5 8 1012 Number of photons n2 due to the wavelength 4972 Å n2 =
(2.4×10−7)(4972×10−10)
(6.63×10−34)(3×108)
= 0.572 × 1012 Therefore total number of photo electrons liberated in 2s, N = n1+n2 = 0.5 × 1012 + 0.575 × 1012 = 1.075 × 1012