10^{-3} , W of 5000 , mathring{A} light is di | vedclass

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(B) The number of photoelectrons emitted per second is given by $\frac{n}{\Delta t} = \frac{I}{e}$.Given $I = 0.16 \, \mu A = 0.16 	imes 10^{-6} \, A

Vedclass · Accepted Answer

$0.04$

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(B) The number of photoelectrons emitted per second is given by $\frac{n}{\Delta t} = \frac{I}{e}$.Given $I = 0.16 \, \mu A = 0.16 	imes 10^{-6} \, A$ and $e = 1.6 	imes 10^{-19} \, C$.$\frac{n}{\Delta t} = \frac{0.16 	imes 10^{-6}}{1.6 	imes 10^{-19}} = 10^{12} \, 	ext{photons/s}$.The number of incident photons per second is given by $\frac{N}{\Delta t} = \frac{W \lambda}{hc}$.Given $W = 10^{-3} \, W$,$\lambda = 5000 \, \mathring{A} = 5 	imes 10^{-7} \, m$,$h = 6.63 	imes 10^{-34} \, J \cdot s$,and $c = 3 	imes 10^8 \, m/s$.$\frac{N}{\Delta t} = \frac{10^{-3} 	imes 5 	imes 10^{-7}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx \frac{5 	imes 10^{-10}}{19.89 	imes 10^{-26}} \approx 0.251 	imes 10^{16} = 2.51 	imes 10^{15} \, 	ext{photons/s}$.Percentage efficiency = $\frac{n/\Delta t}{N/\Delta t} 	imes 100 = \frac{10^{12}}{2.51 	imes 10^{15}} 	imes 100 \approx 0.04 \%$.

$10^{-3} \, W$ of $5000 \, \mathring{A}$ 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 .......$\%$

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