Light of wavelength `5000Å` and intensity `39.8Wm^(-1)` is incident on a metal surface. If only 1% photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly :

Options

A. `10^(18)`
B. `10^(20)`
C. `10^(22)`
D. `10^(24)`

Answer

A

Explanation

`E=nh rArr E=(nhc)/(lambda)`
`rArr n=(E lambda)/(hc)`
But `n=(n)/(100)=(E lambda)/(100 hc)=10^(18)`