A beam of light consists of four wavelength `4000Å , 4800 A, 6000 A` and `7000 Å`, each of intensity `1.5xx10^(-3)Wm^(-2)`. The beam falls normally on an area `10^(-4)m^(2)` of a clean metalic surface of work function `1.9 eV`. Assuming no loss of light energy (i.e. each capable photon emits one electron) calculate the number of photoelectrons liberated per second.

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)`