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

`:. upsilon_(0) gt upsilon'_(0)`
`:. upsilon_(0)=(c )/(lambda_(0))rArr(c )/(lambda_(0)) gt lambda_(0) lt lambda'_(0)`
Therefore, metal `X` has larger threshold wavelength
(ii) Since `E= h upsilon_(0) rArr K.E.= E-hupsilon_(0)`
Therefore, metal `x` has larger Kinetic energy
(iii) `K.E= eV_(0)`
`:.` Kinetic energy of the emitted photoelectron is independent of the intensity of the incident light, hence `K.E`. fo the emitted phot-electrons remains unchanged if the intensity of the incident radiation is halved.