Dual Nature Of Light Question 388
Question: 1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.10 per cent of the incident photons produce photoelectrons, then find the current in the cell.
Options:
A) $ 4.8\mu A $
B) $ 48\mu A $
C) $ 1.8\mu A $
D) $ 0.48\mu A $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The number of photons per second directed at the cell is $ n=\frac{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 $
BETA
