A) 4.81 eV
B) 3.74 eV
C) 2.66 eV
D) 1.07 eV
Correct Answer: C
Solution :
Energy of incident light \[E\,(eV)=\frac{12375}{3320}=3.72\,eV\] \[(332\,nm=3320\,{\AA})\] According to the relation \[E={{W}_{0}}+e{{V}_{0}}\] \[\Rightarrow \,\,{{V}_{0}}=\frac{(E-{{W}_{0}})}{e}=\frac{3.72\,eV-1.07\,eV}{e}\]\[=2.65\,\,Volt.\]You need to login to perform this action.
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