A) 1.2 volts
B) 2.24 volts
C) 3.6 volts
D) 4.8 volts
Correct Answer: A
Solution :
Energy of incident light \[E=\frac{12375}{2000}=6.18\,eV\] According to relation \[E={{W}_{0}}+e{{V}_{0}}\] \[\Rightarrow \,\,{{V}_{0}}=\frac{(E-{{W}_{0}})}{e}=\frac{(6.18\,eV-5.01\,eV)}{e}\]\[=\,1.17\,V\,\approx \,1.2\,V\]
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