A beam of light has two wavelengths of 4972 Å and 6216 Å with a total intensity of 3.6 10⁻³ Wm ⁻² equally distributed among the two wavelengths. The beam falls normally on an area of 1 ~cm ^2 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:

Step-by-step solution

Given, ₁=4972 Å and ₂=6216 Å and I =3.6 10⁻³ Wm ⁻² Intensity associated with each wavelength = 3.6 10⁻³ 2 =1.8 10⁻³ Wm ⁻² Work function = hv = hc aligned &= (6.62 10⁻³⁴ ) (3 10^8 ) &= 12.4 10^3 ev aligned for different wavelengths array r ₁= 12.4 10^3 ₁ = 12.4 10^3 4972 =2.493 eV =3.984 10⁻¹⁹ ~J ₂= 12.4 10^3 ₂ = 12.4 10^3 6216 =1.994 eV =3.184 10⁻¹⁹ ~J array Work function for metallic surface =2.3 eV (given) ₂ < Therefore, ₂ will not contribute in this process. Now, no. of electrons per m ^2- s = no. of photons per