The beam of light has three wavelengths 4144 Ã, 4972 Ã and 6216 Ã with a total intensity of 3.6 10⁻³ Wm ² equally distributed amongst the three wavelengths. The beam falls normally on the area 1 ~cm ² of a clean metallic surface of work function 2.3 eV . Assume that there is n 0 loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated

Step-by-step solution

As we know, threshold wavelength array c ( ₀ )= hc ₀= (6.63 10⁻³⁴ ) 3 10⁸ 2.3 (1.6 10⁻¹⁹ ) =5.404 10⁻⁷ ~m array ₀=5404 Ã… Hence, wavelength 4144 Ã… and 4972 Ã… 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 ( lcm )²=1.2 10⁻⁷ joule Therefore, energy incident on the surface for each wavelength in 2 ~s E = (1.2 10⁻⁷ ) 2=2.4 10⁻⁷ ~J Number of photons n ₁ due to wavelength 4144 Ã… n₁= (2.4 10⁻⁷ )