A) doubled
B) less than double
C) more than double
D) nothing can be said
Correct Answer: C
Solution :
From Einstein's photoelectric equation, we have \[{{E}_{k}}=hv-W\] where \[{{E}_{k}}\] is maximum kinetic energy of photoelectrons and W the work function \[(=h{{v}_{0}})\]. \[\therefore \] \[{{E}_{k}}=h(v-{{v}_{0}})\] ?.(i) where \[{{v}_{0}}\] is threshold frequency. For \[hv'=2hv\] \[\therefore \] \[2{{E}_{k}}=2h(v-{{v}_{0}})\] ?..(ii) From Eqs. (i) and (ii), we observe that kinetic energy of emitted electron will become more than doubled.
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