Fig. shows the variation of stopping potential `V_0` with the frequency v of the incident radiation for two photosensitive metal P and Q:
(i) Explain which metal has smaller threshold wavelength

(ii) Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy, for the same wavelength of incident radiation.
(iii) If the distance between the light source and metal P is doubled, how will the stopping potential change.

Explanation

(i) Threshold wavelength,
`lambda_0=c/(v_0) or lambda_0prop 1/(v_0)`
`:. ((lambda_0)_P)/((lambda_0)_Q)=((V_0)_Q)/((V_0)_P)gt1 [:' (V_0)_Qgt(V_0)_P]`
So, `(lambda_0)_P gt (lambda_0)_Q or (lambda_0)_Q lt (lambda_0)_P`
It mens metal Q has smaller threshold wavelength.
(ii) From Einstein's photoelectric equation
`K.E. =(hc)/lambda-(hc)/(lambda_0)`
For given value of `lambda, (hc//lambda)` is constant. From above relation, we note that if `lambda_0` is small, then K.E. emitted is small. Thus metal Q with smaller `lambda_0` will emit photoelectrons of smaller K.E.
(iii) If the distance between light source and metal P is doubled then intensity of light becomes one-fourth (as `Iprop1//r^2`)but depends upon the frequency of the incident light, hence, Stopping potential of metal plate P remains unchanged by changing the distance between source of light and metal plate.