A metal plate of area $1\times10^{-4}\ \text{m}^2$ is illuminated by a radiation of intensity $16\ \text{mW/m}^2$. The work function of the metal is $5\ \text{eV}$. The energy of the incident photons is $10\ \text{eV}$ and only $10\%$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: $[1\ \text{eV} = 1.6\times10^{-19}\ \text{J}]$
Answer: (C) $10^{11}$ and $5\ \text{eV}$
Power on the plate $= 16\times10^{-3}\times10^{-4} = 1.6\times10^{-6}\ \text{W}$.
Photons per second $= \dfrac{1.6\times10^{-6}}{10\times1.6\times10^{-19}} = 10^{12}$; $10\%$ of them eject electrons, so $10^{11}$ electrons per second.
Maximum kinetic energy $= 10 - 5 = 5\ \text{eV}$.
Solution by Sreeraj P, M.Sc Physics