Question

$3.0mW$ of $400nm$ light is incident on a photoelectric cell, if $0.1\%$ of the photons are contributing in ejection of electrons, then the current in the cell is

• A. $0.48\mu A$
• B. Resistance value not given
• C. Zero
• D. $0.96\mu A$

Answer 1

The correct option is D $0.96\mu A$

For wavelength 400 nm , energy of each photon be $E=h\frac{c}{\lambda }$
If n be the number of photons. total energy = $nE$.
Given $nE=3mW$
$n=$ $\frac{3}{E}=6.04\times {10}^{15}$
0.1% of the photons are contributing in ejection of electrons =$6.04\times {10}^{12}$
Current , I= $\frac{dQ}{dt}$=$6.04\times {10}^{12}e$
$e$ is charge of coulomb.
$e=1.6\times {10}^{-19}$ Coulomb
$\therefore I=0.96\mu A$