Answer:
(i) Threshold wavelength, \[{{\lambda }_{0}}=\frac{c}{{{v}_{0}}}\] As \[{{v}_{0}}(Q)>{{v}_{0}}(P)\] \[\therefore \] \[{{\lambda }_{0}}(Q)<{{\lambda }_{0}}(P)\] Thus the metal \[Q\]has smaller threshold wave length . (ii) According to Einstein?s photoelectric equation, \[\frac{hc}{\lambda }=\frac{hc}{{{\lambda }_{0}}}+\]K.E of photoelectron For the same \[\lambda \] of incident radiation, L.H.S. is constant. So metal \[\text{Q}\] with smaller value of \[{{\lambda }_{0}}\] will emit photo- electrons of smaller K.E. (iii) Stopping potential \[{{V}_{0}}\] will remain same because it is independent of intensity and hence of distance between the light source and the metal surface.
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