Question

${10}^{-3}W$ of 5000 A light is directed on a photoelectric cell. If the current in the cell is 0.16 $\mu A$, the percentage of incident photons which produce photo electrons, is

• A. 0.4%
• B. 0.04%
• C. 20%
• D. 10%

Answer 1

The correct option is B 0.04%

Current is $0.16\times {10}^{-6}Amp$ it means $0.16\times {10}^{-6}Coulomb$ charge is flowing $\text{per second}$

So $n=\frac{0.16\times {10}^{-6}C}{1.6\times {10}^{-19}C}={10}^{12}\text{electrons are generated per second}$

Now we notice that $\text{one photon has energy E}$, $E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{5000\times {10}^{-10}m}=3.972\times {10}^{-19}Joule$

So number of photon in ${10}^{-3}W$ will be $N=\frac{{10}^{-3}}{3.972\times {10}^{-19}}=0.25\times {10}^{16}$ this is number of $\text{photons incident per second}$

So required percentage is $\frac{n}{N}\times 100=\frac{{10}^{14}}{0.25\times {10}^{16}}=.04$%

Option B is correct.