Question

${10}^{-3}W$ of $5000\stackrel{o}{A}$ 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 photoelectrons, is

• A. 40%
• B. 0.04%
• C. 20%
• D. 10%

Answer 1

Answer: (B)

0.04%

Energy of each photon $E=\frac{hc}{\lambda }=\frac{1.986\times {10}^{-25}}{5000\times {10}^{-10}}=3.936\times {10}^{-19}$

Number of Photons per sec: $N=\frac{{10}^{-3}}{3.936\times {10}^{-19}}=2.54\times {10}^{15}$

Number of electrons per sec: $N^{\prime }=\frac{0.16\times {10}^{-6}}{1.6\times {10}^{-19}}={10}^{12}$

$\therefore \frac{N^{\prime }}{N}=\frac{{10}^{12}}{2.54\times {10}^{15}}\approx 4\times {10}^{-4}$

In percentage $\frac{N^{\prime }}{N}\times 100=0.04$%