A) \[4.8\mu A\]
B) \[48\mu A\]
C) \[1.8\mu A\]
D) \[0.48\mu A\]
Correct Answer: D
Solution :
[d] The number of photons per second directed at the cell is \[n=\frac{\text{Power}}{hv}=\frac{P\lambda }{hc}=\frac{1.5\times {{10}^{-3}}\times 400\times {{10}^{-9}}}{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}\] \[n=\frac{1}{3.31}\times {{10}^{16}}(\text{photons/s})\] But it is given that only 0.1 % of these photons produce photoelectrons. Therefore number of photoelectrons produced per second is \[{{n}_{e}}=\frac{0.1}{100}\times \frac{{{10}^{16}}}{3.31}=\frac{{{10}^{13}}}{3.31}\] Therefore, the current is \[I={{n}_{e}}e\] \[=\frac{{{10}^{13}}\times 1.6\times {{10}^{-19}}}{3.31}=0.48\times {{10}^{-6}}=0.48\mu A\]
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