1.

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

Answer»

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

Solution :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)


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