A beam of light has three wavelengths `4144 A`, `4972 A`, and `6216 A` with a total intensity of `3.6xx10^(-3) Wm^(-2)` equall distributed aming the three wavelengths. The beams fall normally on an area `1.0cm^2` of a clean metallic surface of work function `2.3eV`. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in 2 s.

Answer

A, B

Explanation

Energy of photon having wavelength `4144 Å`,
`E_1 = (12375)/(4144) eV`
`=2.99 eV`
Similarly, `E_2 = (12375)/(4972) eV`
= 2.49 eV and
`E_3 (12375)/(6216) eV`
=1.99 eV
Since, only `E_1` and `E_2` are greater than the work function `W = 2.3 ev`, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, instensity corresponding to each wavelength is
`(3.6xx10^(-3))/(3) = 1.2 xx10^(-3) W/m^2`
Or energy incident per second in the given area (A = 1.0 cm^2 = 10^(-4) m^2)` is
`P=1.2xx10^(-3)xx10^(-4)`
`= 1.2 xx10^(-7) J/s`
Let `n_1` be the number of photons incident per unit time in the ginven area corresponding to first
wavelength. Then, `n_1 = (p)/(E_1)`
`= (1.2xx10^(-7)/(2.99xx1.6xx10^(-19)`
`=2.5xx10^11`
Similarly, `n_2=(P)/(E_2)`
`=1.2xx10^(-7)/(2.49xx1.6xx10^(-19)`
`=3.0xx10^11`
Since each energetically capable photon ejects one electron, total number of photoelectrons
liberated in 2 s.