A) \[V-{{V}_{0}}\]
B) \[V\]
C) \[\sqrt{V-{{V}_{0}}}\]
D) \[{{V}_{0}}\]
Correct Answer: A
Solution :
When light falls on a metallic surface, ejection of photoelectron results. In this process, conservation of energy hold. Thus, from law of conservation of energy, energy imparted by the photon = maximum kinetic energy of the emitted electron + work function of the metal or \[hv={{(KE)}_{\max }}+\phi \] but \[\phi =h{{v}_{0}},\,{{v}_{0}}\] being threshold frequency. \[\therefore \] \[{{(KE)}_{\max }}=hv-h{{v}_{0}}\] or \[{{(KE)}_{\max }}\propto v-{{v}_{0}}\] Note: Photoelectric emission is an instantaneous process without any apparent time lag \[(\sim {{10}^{-9}}s)\]. Even when the incident radiation is made exceedingly dim.
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