Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearly

Options

A. `9.61xx10^(14)s^(-1)`
B. `4.12xx10^(13)s^(-1)`
C. `1.51xx10^(12)s^(-1)`
D. `2.13xx10^(11)s^(-1)`.

Answer

C

Explanation

Energy incident over `1cm^(2)=1.0xx10^(-4)J`
Energy required to produce photoelectrons
`=1.0xx10^(-4)xx10^(-2)=10^(-6)J`
As number of photoelectrons ejected=number of photons which can produce photoelectrons=energy required for producing electron/energy of photon.
`=(10^(-6))/(lc//lamda)`
`=(10^(-6)xx300xx10^(-9))/(6.6xx10^(-36)xx3xx10^(8))`
`=1.51xx10^(12)s^(-1)`.