A beam of light has two wavelengths 4972 Ao and 6216 Ao with a total intensity of 3.6 × 10⁻⁷ W/m². The intensity is equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm² of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately:

The threshold wavelength is given as λ = hc/Φ = (6.6 × 10⁻³⁴ × 3 × 10⁸) / (2.3 × 1.6 × 10⁻¹⁹) = 5380 Ao

Thus the beam with wavelength 6216 Ao would not emit any electrons from the metal.

Now the intensity of each beam is I = (1/2) × 3.6 × 10⁻⁷ W/m² = 1.8 × 10⁻⁷ W/m²

Now, for wavelength 4972 Ao
1.8 × 10⁻⁷ = n h c / λ
Or, n = 4.5 × 10¹⁵ after putting the values (i)

Now number of photoelectrons emitted from the area A = 10⁻⁴ m² in time t = 2 s is given as
ne = n A t = 4.5 × 10¹⁵ × 10⁻⁴ × 2 = 9 × 10¹¹ electrons

Answer is B.