Explanation:
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is
\({K_{{\text{max }}}} = hf - {\phi _0}\)
where \(hf\) is the energy of incident photon and \(\phi_{0}\) is the work function
\({\text{ But }}{K_{{\text{max }}}} = \frac{1}{2}\;m{v_{{\text{ma}}{{\text{x}}^{\text{2}}}}}\)
\(\therefore \frac{1}{2}\;mv_{\max {\text{ }}}^2 = hf - {\phi _0}\)
As per question
\(\frac{1}{2}\;mv_{{\text{ma}}{{\text{x}}_{\text{1}}}}^2 = 1\,eV - 0.5\,eV = 0.5\,eV\,\,\,\,\,\,\,\,\,\,(1)\)
\(\frac{1}{2}\;mv_{{\text{ma}}{{\text{x}}_{\text{2}}}}^{\text{2}} = 2.5eV - 0.5eV = 2eV\,\,\,\,\,\,\,\,\,\,(2)\)
Dividing eqn. (1) by eqn. (2), we get
\(\frac{{v_{{{\max }_1}}^2}}{{v_{{{\max }_2}}^2}} = \frac{{0.5eV}}{{2eV}} = \frac{1}{4}\)
\(\frac{{{v_{{{\max }_1}}}}}{{{v_{{{\max }_2}}}}} = \sqrt {\frac{1}{4}} = \frac{1}{2}\)