# A metal plate of area

Question:

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: $\left[1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right]$

1. (1) $10^{14}$ and $10 \mathrm{eV}$

2. (2) $10^{12}$ and $5 \mathrm{eV}$

3. (3) $10^{11}$ and $5 \mathrm{eV}$

4. (4) $10^{10}$ and $5 \mathrm{eV}$

Correct Option: 3

Solution:

(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 $=10^{12} \times 10 / 100=10^{11}$