Radiations of two frequencies are incident on a metal surface of work function $\displaystyle 2.0$ eV one by one. The energies of their photons are $\displaystyle 2.5$ eV and $\displaystyle 4.5$ eV respectively. Find the ratio of the maximum speed of the electrons emitted in the two cases.

$\displaystyle h \nu=\phi_{0}+K_{\max }$ For radiation having photons of energy $\displaystyle 2.5$ eV: $\displaystyle 2.5 \mathrm{~eV}=2 \mathrm{~eV}+\frac{1}{2} m v_{1}^{2}$ $\displaystyle 0.5 \mathrm{~eV}=\frac{1}{2} m v_{1}^{2}$ .............(i) For radiation having photons of energy $\displaystyle 4.5$ eV: $\displaystyle 4.5 \mathrm{~eV}=2 \mathrm{~eV}+\frac{1}{2} m v_{2}^{2}$ $\displaystyle 2.5 \mathrm{~eV}=\frac{1}{2} m v_{2}^{2}$ ............(ii) Dividing eq. (i) by (ii): $\displaystyle \frac{v_{1}}{v_{2}}=\frac{1}{\sqrt{5}}$$\displaystyle \frac{v_{2}}{v_{1}}=\sqrt{5}$