According to Einstein's photoelectric equation Maximum kinetic energy of the emitted electrons

\(=\) Incident photon energy - Work function

\(\therefore\)    \({K_{{{\max }_1}}} = 1\,{\text{eV}} - 0.5\,{\text{eV}} = 0.5\,{\text{eV}}\)   ..... \((i)\)

and \({K_{{{\max }_2}}} = 2.5\,{\text{eV}} - 0.5\,{\text{eV}} = 2\,{\text{eV}}\)   ..... \((ii)\)

Divide \((i)\) by \((ii)\), we get

\({\frac{K_{\max }}{K_{\max _{2}}}=\frac{0.5\, \mathrm{eV}}{2\, \mathrm{eV}}=\frac{1}{4}}\)

\(\frac{{\frac{1}{2}m{v_{{{\max }^2}}}}}{{\frac{1}{2}mv_{{{\max }_2}}^2}} = \frac{1}{4}\) or \(\frac{{{v_{\max }}}}{{{v_{{{\max }_2}}}}} = \sqrt {\frac{1}{4}}  = \frac{1}{2}\)