A) 1 V
B) 2 V
C) 3 V
D) Zero
Correct Answer: C
Solution :
\[n=\frac{E\lambda }{hc}\]\[=\frac{1\times {{10}^{-7}}\times 200\times {{10}^{-9}}}{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}=1\times {{10}^{11}}\] Number of electrons ejected \[=\frac{{{10}^{11}}}{{{10}^{3}}}={{10}^{8}}\] \ \[V=\frac{q}{4\pi {{\varepsilon }_{0}}r}=\frac{({{10}^{8}}\times 1.6\times {{10}^{-19}})\times 9\times {{10}^{9}}}{4.8\times {{10}^{-2}}}=3\,V\]
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