Light from a source of power \( 100\ \mathrm{mW} \) and wavelength \( 500\ \mathrm{nm} \) is incident on a photosensitive metal connected to a circuit. If every photon ejects \( 1 \) electron from the photosensitive metal, then, the photocurrent in the circuit will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 40\ \mathrm{mA} \)
The photocurrent equals the charge of the electrons ejected per second, and with one electron per photon this is number of photons per second × \( e \). Each photon has energy \( E = hc/\lambda \).
\( \displaystyle n = \frac{P}{hc/\lambda} = \frac{P\lambda}{hc} = \frac{(100\times10^{-3}\ \mathrm{W})(500\times10^{-9}\ \mathrm{m})}{(6.63\times10^{-34}\ \mathrm{J\,s})(3\times10^{8}\ \mathrm{m\,s^{-1}})} \)
\( n \approx 2.51\times10^{17}\ \mathrm{s^{-1}} \)
\( I = ne = (2.51\times10^{17})(1.6\times10^{-19}\ \mathrm{C}) \approx 4.0\times10^{-2}\ \mathrm{A} = 40\ \mathrm{mA} \)
Shortcut: Photon energy \( = 1240/500 = 2.48\ \mathrm{eV} \). Then \( I = P/(\text{energy in eV}) = 100\ \mathrm{mW}/2.48\ \mathrm{V} \approx 40\ \mathrm{mA} \), because each electron carries one photon’s worth of energy in eV.