Q. A metal plate of area $1 \times 10^{-4} \; m^2$ is illuminated by a radiation of intensity 16 mW/m$^2$. The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : $[1 \; eV = 1.6 \times 10^{-19} J]$


$I =\frac{nE}{At} $
$ 16\times10^{-3} =\left(\frac{n}{t}\right)_{\text{Photon}} \frac{10\times1.6 \times10^{-19}}{10^{-4}} =10^{12} $

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions