Radiation of frequency `10^(15)Hz` is incident separately on two photosensitive surfaces P and Q. The following observations were made:
(i) surface P: Photoemission occurs but the photoelectrons have zero kinetic energy
(ii) surface Q: Photoemission occurs and electrons have non-zero kinetic energy.
Which of these two has higher work function? If the frequency of incident light is reduced, what will happen to photoelectron emission in the two cases?

Explanation

Energy of incident photon, `hf = E_(k) + W_(0)`, here `W_(0)=` work function of the emitting surface, and `E_(k)=` kinetic energy of emitted photoelectron
Given, for the surface P, `E_(k) =0`, but for the surface `Q, E_(k) gt 0`
For the same incident light, hf = constant. So the work function `W_(0)` is higher for the surface P.
If incident light of a lower frequency is taken, hf would reduce. Then it would be less than the work function of the surface P, so, this surface would emit no photoelectron. The surface Q may emit photoelectrons, provided `hf gt W_(o)` for the surface.