$h\nu  – {W_0} = \frac{1}{2}mv_{\max }^2$

$\Rightarrow \frac{{hc}}{\lambda } – \frac{{hc}}{{{\lambda _0}}} = \frac{1}{2}mv_{\max }^2$

$ \Rightarrow hc\left( {\frac{{{\lambda _0} – \lambda }}{{\lambda {\lambda _0}}}} \right) = \frac{1}{2}mv_{\max }^2$

$ \Rightarrow {v_{\max }} = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} – \lambda }}{{\lambda {\lambda _0}}}} \right)} $

जब तरंगदैध्र्य $\lambda $ एवं वेग $v$ है, तब

$v = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} – \lambda }}{{\lambda {\lambda _0}}}} \right)} $                                               …. $(i)$

 जब तरंगदैध्र्य $\frac{{3\lambda }}{4}$ एवं वेग $v$' है तब

$v' = \sqrt {\frac{{2hc}}{m}\left[ {\frac{{{\lambda _0} – (3\lambda /4)}}{{(3\lambda /4) \times {\lambda _0}}}} \right]} $                           ….$(ii)$

 समीकरण $(ii)$ को $(i)$ से भाग देने पर

 $\frac{{v'}}{v} = \sqrt {\frac{{[{\lambda _0} – (3\lambda /4)]}}{{\frac{3}{4}\lambda {\lambda _0}}} \times \frac{{\lambda {\lambda _0}}}{{{\lambda _0} – \lambda }}} $

 $v' = v{\left( {\frac{4}{3}} \right)^{1/2}}\sqrt {\frac{{[{\lambda _0} – (3\lambda /4)]}}{{{\lambda _0} – \lambda }}} $ अर्थात् $v' > v{\left( {\frac{4}{3}} \right)^{1/2}}$