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