question_answer

Figure shows the variation of stopping potential \[{{V}_{0}}\] with the frequency \[\mathbf{v}\] of the incident radiation for two photosensitive metals P and \[Q\]: (i) Explain which metal has smaller threshold wavelength. (ii) Explain, giving reason, which metals emits photo electrons 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 charge?         

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.