A metal plate of area $1 \times 10^{-4} \mathrm{~m}^{2}$ is illuminated by a radiation of intensity $16 \mathrm{~mW} / \mathrm{m}^{2}$. The work function of the metal is $5 \mathrm{eV}$. The energy of the incident photons is $10 \mathrm{eV}$ and only $10 \%$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum
energy, respectively, will be: $\left[1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right]$
Correct Option: 3
(3) using, intensity $\mathrm{I}=\frac{\mathrm{nE}}{\mathrm{At}}$
$\mathrm{n}=$ no. of photoelectrons
$\Rightarrow 16 \times 10^{-3}=\left(\frac{\mathrm{n}}{\mathrm{t}}\right) \times \frac{10 \times 1.6 \times 10^{-19}}{10^{-4}}$ or, $\frac{\mathrm{n}}{\mathrm{t}}=10^{12}$
So, effective number of photoelectrons ejected per unit time $=10^{12} \times 10 / 100=10^{11}$