Radiation of frequency 10^{15} Hz is incident on two photosensitive surface P and Q. There is no photoemission from surface P. Photoemission occurs from surface Q but photoelectrons have zero kinetic energy. Explain these observations and find the value of work function for surface Q.

#### Solution

(a) Intensity of radiation is proportional to the number of energy quanta per unit area per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta and greater, therefore, is the number of electrons coming out of the metal (for ν > ν_{0}). This explains why, for ν > ν_{0}, photoelectric current is proportional to intensity.

(b) The basic elementary process involved in photoelectric effect is the absorption of a light quantum by an electron. This process is instantaneous. Thus, whatever may be the intensity i.e., the number of quanta of radiation per unit area per unit time, photoelectric emission is instantaneous. Low intensity does not mean delay in emission, since the basic elementary process is the same. Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron) and, therefore, the photoelectric current.

(ii) λ of radiation `= c/f=(3×10^8)/10^15=300 nm`

Energy of radiation`=(hc)/λ=1240/λ=1240/300=4.133 eV`

Work function of P is greater than 4.133 eV, therefore there is no photoemission from surface P.

For Q, KE_{max}= E - ϕ_{o}, as KE of photo electrons is zero, energy provided is equal to work function.

∴ Work function of Q = 4.133 eV