When a point light source, of power `W`, emiting monochromatic light of wavelength `lambda` is kept at a distance `a` from a small photosensitive surface of work function `phi` and area `S`. Then

Options

A. number of photons striking the surface per unit time as `WSlambda//4pihca^(2)`
B. the maximum energy of the emitted photoelectrons is `(hc-lambdaphi)/lambda`
C. the stopping potential needed to stop the most energetic emitted photoelectrons as `e((hc-lambdaphi))/lambda`
D. photoemission occurs only if `lambda` lies in the range `0 le lambda le hc//phi`.

Answer

A, B, D

Explanation

The energy of each photon is `hc//lambda` so that the number of photons released per unit time is `W//(hc//lambda)`.These photons are spread out in all directions over an area `4pia^(2)` so that the share of an area `S` is a fraction `S//4pia^(2)` of the total number of photons emitted.The maximum kinetic energy of the emitted photocelectrons is `KE_(max)=hv-phi=1/lambda(hc-lambdaphi)`
The stopping potential is given by `eV_(S)=KE_(max)` hence `V_(S)=1/eKE_(max)=1/(elambda)(hc-lambdaphi)`
For photoemission to be possible, we must have `hv gephi(hc)/lambdarArrgephirArr lambda le(hc)/phi`
Thus the permitted range of values of `lambda` is `0lelambdalehc//phi.36. KE_(max)=(5-phi)eV`
when these electrons are accelerated through `5V`,they will reach the anode with maximum energy =`(5-phi+5)eV rArr 10-phi=8 rArr phi=2eV`
Just after emission kinetic energies of electrons range from `0` to `3 eV`.If all the electrons reach the anode,photo current reaches saturation and in this situation range of `KE s` of electrons reaching anode should be `5eV` to `8eV`.As given `KE` range is `6eV` to `8eV`,some of the emitted electrons are not reaching the anode hence the current is less than the saturation value.