A) 2.4 V
B) -1.2 V
C) 2.4 V
D) 1.2 V
Correct Answer: D
Solution :
Energy of incident light \[E(eV)=\frac{12375}{2000}=6.2eV(200nm=2000{\AA})\] According to the relation \[E={{W}_{0}}+e{{V}_{0}}\] \[\Rightarrow \] \[{{V}_{0}}=\frac{E-{{W}_{0}}}{e}\] \[=\frac{(6.2-5.01)e}{e}\] \[=1.2\,V\]You need to login to perform this action.
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