Light quanta with an energy $4.9 \mathrm{eV}$ eject photo electrons from metal with work function $4.5 \mathrm{eV}$. Find the maximum impulse transmitted to the surface of the metal when each electron flies out.

According to Einstein's photo-electric equation

$K_{\max }=\frac{1}{2}m{v}_{\max }^2=h\nu -{\varphi }_0$

$\Rightarrow K_{\max }=4.9-4.5=0.4 \mathrm{eV}$

If  $p$ is the momentum of each ejected photo electron, then

$p=\sqrt{2m{K}_{\max }}$

Also, we know that change in momentum of a body is equal to impulse. Hence the entire momentum of electron is gained when it is ejected out from the metal surface, so impulse $\left(I\right)$ on the surface is given by

$I=\Delta p=\sqrt{2m{K}_{\max }}$

Substituting the values, we get maximum impulse

$I=\sqrt{2\times 9.1\times {10}^{-31}\times 0.4\times 1.6\times {10}^{-19}}$$=3.41\times {10}^{-25} {\mathrm{kgms}}^{-1}$