A) \[3\times {{10}^{19}}\]
B) \[5\times {{10}^{22}}\]
C) \[2\times {{10}^{22}}\]
D) \[1.67\times {{10}^{18}}\]
Correct Answer: A
Solution :
[a] Let the number of photons hitting the photocell per second be n, then \[nhv=1.5mW\] Or \[n=\frac{(1.5\times {{10}^{-3}})\times 400\times {{10}^{-9}}}{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}\] \[\left[ \therefore v=\frac{c}{\lambda }=\frac{3\times {{10}^{8}}}{400\times {{10}^{-9}}} \right]\] \[\therefore n=3\times {{10}^{15}}\] Number of photoelectrons produced per second is \[3\times {{10}^{15}}\times \frac{0.1}{100}=(3\times {{10}^{12}})\] So, current in the photocell is \[(3\times {{10}^{12}})(1.6\times {{10}^{-19}})=4.8\times {{10}^{-7}}A\]
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