When a point light source, of power `W`, emitting 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 `W lambda S//4 pi hca^2`
B. The maximum energy of the emitted photoelectrons as `(1/lambda)(hc-lambda phi)`
C. The maximum potential needed to stop the most enegetic emitted photoelectrons as `(e//lambda)(hc-lambda phi)`
D. Photo-emission only if `lambda` lies in the range `0 le lambda le (hc//phi)`

Answer

A, B, D

Explanation

(a,b,d) : The energy of each photon is `hc//lambda` , so that the numner of photons released per unit time `hc// lambda`, These photons are spread out in all directions over an area `4 pia^2` , so that the share of an area S is a fraction `S//4 pia^2` of the total number of photons emitted.
The maximum energy of emitted photoelectrons is .
`E_(max) = (hc)/(lambda) - phi = (1)/(lambda)(hc-lambda phi)`
The stompping potential is given by `eV_(0) = E_(max)`
Hence , `v_(0) = (E_(max))/(e)= (1)/(e lambda)(hc-lambda phi)`
For photoemission to be possible ,
We have , `(hc)/(lambda) ge phi or lambda le (hc)/(phi)`
Thus , the permitted range of values of `lambda is 0 le lambda le (hc)/(phi)`
Hence , the correct choices are (a),(b) and (d).