Question

$1.5mW$ of $400nm$ light is directed at a photoelectric cell. If $0.10$ percent of the incident photons produce photoelectrons, then the current in the cell is

• A. $4.8\mu A$
• B. $48\mu A$
• C. $1.8\mu A$
• D. $0.48\mu A$

Answer 1

The correct option is D $0.48\mu A$

$P=1.5mW=1.5\times {10}^{-3}W$

Energy $=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}=4.96\times {10}^{-19}J$

Number of photons incident per second

$=\frac{P}{Energyofeachphoton}=\frac{1.5\times {10}^{-3}}{4.96\times {10}^{-19}}\simeq 3\times {10}^{15}$

Number of photo $-$ electrons produce $=0.1$% $\times 3\times {10}^{15}=3\times {10}^{12}$

Current $i=3\times {10}^{12}\times 1.6\times {10}^{-19}A$

$=4.8\times {10}^{-7}A=0.48\mu A$