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.

Explanation

`E_(1)=(12400)/(4000)=3.1 eV, E_(2)=(124000)/(4800)=2.58 eV E_(3)(12400)/(6000)=2.06 eV`
and `E_(4)=(12400)/(7000)=1.77eV`
Therefore, light of wavelengths `4000 Å, 4800 Å` and `6000Å` can only emit photoelectrons.
`:.` Number of photoelectrons emitted per second `=` No. of photons incident per second)
`=(I_(1)A_(1))/(E_(1))+(I_(2)A_(2))/(E_(2))+(I_(3)A_(3))/(E_(3))=IA((1)/(E_(1))+(1)/(E_(2))+(1)/(E_(3)))`
`=((1.5xx10^(-3))(10^(-4)))/(1.6xx10^(-19))((1)/(3.1)+(1)/(2.58)+(1)/2.06)`
`=1.12xx10^(12)`