A) doubled
B) less than double
C) more than double
D) nothing can be said
Correct Answer: C
Solution :
From Einsteins 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.You need to login to perform this action.
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