According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is $K_{\mathrm{m}\mathrm{a}\mathrm{x}}=h\nu -{\varphi }_0$

Where $h\nu$ is the energy of the incident photon and ${\varphi }_0$ is the work function.

But $K_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{2}m{v}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$

$\therefore \frac{1}{2}m{v}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2=h\nu -{\varphi }_0$

As per the question,

$\frac{1}{2}m{v^{\mathit{2}}}_{{\max }_1}=1 \mathrm{eV}-\text{0.5}\mathrm{ }\mathrm{eV}=\text{0.5}\mathrm{ }\mathrm{eV}$ ...(i)

and $\frac{1}{2}m{v^2}_{{\max }_2}=\text{2.5}\mathrm{ }\mathrm{eV}-\text{0.5}\mathrm{ }\mathrm{eV}=2\mathrm{ }\mathrm{eV}$ ...(ii)

Dividing equation (i) by equation (ii), we get

$\frac{{v^2}_{{\max }_1}}{{v^2}_{{\max }_2}}=\frac{\text{0.5} \mathrm{eV}}{2 \mathrm{eV}}=\frac{1}{4}$

$\frac{v_{{\mathit{max}}_1}}{v_{{\mathit{max}}_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$