The following graph shows the variation of stopping potential `V_0` with the frequency v of the incident radiation for two photosensitive metals X and Y.
(i) Which metal has larger threshold wavelength ? Give reason.
(ii) Explain giving reason , which metal gives out electrons , having larger kinetic energy , for same wavelength of the incident radiation.
(iii) If the distance between the light source and metal X is halved , how will the kinetic energy of electrons emitted from it change ? Given reason.

Explanation

(i) Since , `lamda prop 1/v`
As the threshold frequency of metal Y is greater than of metal X.
Thus metal X has greater threshold wavelength.
(ii) The kinetic energy of the emitted electrons depends on the work function of metal Y is greater than that X The kinetic energy of electrons emitted from metal X will have greater kinetic energy .
(iii) If the distance between light source and metal is changed , the intensity of the light falling on the surface will decrease. But the kinetic energy of the emitted electron is independent of the intensity of the light falling and hence there will be no change in the kinetic energy of the emitted electrons.