When a point light source of power W emitting monochromatic light of wavelength `lamda` is kept at a distance `alpha` from a photo-sensitive surface of work function `phi` and area S, we will have

Options

A. number of photons striking the surface per unit time as `(WlamdaS)/(4)pihca^2`
B. the maximum energy of the emitted photoelectrons as `((1)/(lamda))(hc-lamda_phi)`
C. the stopping potential needed to stop the most energetic emitted photoelectrons as `((e)/(lamda))(hc-lamdaphi)`
D. photo emission only if `lamda` lies in the range `0lelamdale((hc)/(phi))`

Answer

A, B, D

Explanation

The energy of each photon is `(hc)/(lamda)`, so that the number of photons released per unit time is `(W)/(((hc)/(lamda)))`. These photons are spread out in all directions over an area `4pialpha^2`, so that the `share` of an area S is a fraction `(S)/(4pialpha^2)` of the total number of photons emitted.
The maximum energy of emitted photoelectrons is `E_(max)=hc-phi=(hc)/(lamda)-phi=(1)/(lamda)(hc-lamdaphi)`
The stopping potential is given by
`eV_S=E_(max)`
Hence, `V_S=(E_(max))/(e)=(1)/(elamda)(hc-lamdaphi)`
Hence, choice (c ) is incorrect.
For photoemission to be possible, we have `hcgephi`.
Hence, `(hc)/(lamda)gephi` or `lamdale(hc)/(phi)`
Thus, the permitted range of values of `lamda` is
`0lelamdale(hc)/(phi)`
Hence, the correct choices are (a), (b), (d)