A) 0.48\[\mu A\]
B) resistance value not given
C) zero
D) 0.96\[\mu A\]
Correct Answer: D
Solution :
\[n\frac{hc}{\lambda }=3\times {{10}^{-3}}\] \[n=6.04\times {{10}^{2}}\] The electron which have been ejected per second are 0.1% of \[n=6.04\times {{10}^{12}}.\] So, rate of flow of charge \[I=\frac{dq}{dt}=0.1%\,\text{of}\,n\] \[6.04\times {{10}^{12}}\times 1.6\,\times {{10}^{-19}}=0.96\,\mu A\]
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