A parallel beam of ligth of intensity `I` and cross section area `S` is incident on a plate at normal incidence. The photoelectric emission efficiency is `100%`, the frequency of beam is `v` and the work function of the plate is `phi(hv gt phi)`. Assuming all the electrons are ejected normal to the plane and with same maximum possible speed. The net force exerted on the plate only due to striking of photons and subsequent emission of electrons is

Options

A. `(IS)/(hv)((2h)/lambda+sqrt(2m(hv-phi)))`
B. `(2IS)/(hv)((h)/lambda+sqrt(2m(hv-phi)))`
C. `(IS)/(hv)((h)/lambda+sqrt(2m(hv-phi)))`
D. `(2IS)/(hv)((h)/lambda+sqrt(m(hv-phi)))`

Answer

C

Explanation

`I=((dn)/(dt))(hv)/S rArr(dn)/(dt)=(IS)/(hv)`=number of photon falling per second =number of electrons emitted per second from the metal (`because` efficiency of emission is `100%`)
Force on the plate due to the absorption of photos is `F_(1)=(dnxxp)/(dt)=(dn)/(dt)xx(hv)/c=(IS)/c`
Force on the plane due to the ejection of electrons from it is `F_(2)=(dn)/(dt)sqrt(2mKE)=(IS)/(hv)sqrt(2m(hv-phi))`
`therefore` Total force on the metal =`F_(1)+F_(2)=(IS)/c+(IS)/(hv)sqrt(2m(hv-phi))`
=`(hIS)/(hvlambda)+(IS)/(hv)sqrt(2m(hv-phi)) =(IS)/(hv)[h/lambda+sqrt(2m(hv-phi))]`