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. `2.13xx10^(11)`
B. `1.51x10^(12)`
C. `4.12xx10^(13)`
D. `9.61xx10^(14)`

Answer

b

Explanation

No. of incident photons falling per sq. cm.
`n=I/(hc//lambda)=(Ilambda)/(hc)`
`=((1.0//10^(4))xx(300xx10^(-9)))/((6.6xx10^(-34))xx(3xx10^(8)))=(10^(15))/6.6`
As `1%` of the incident photons produces photoelectrons, therefore no. of photoelectrons produced per sec. `=1/100n=1/100xx(10^(15))/6.6`
`=1.51xx10^(12)`