\(K=\frac{p^{2}}{2 m}=\frac{(h / \lambda)^{2}}{2 m}=\frac{h^{2}}{2 m \lambda^{2}}\)
So, maximum energy of photon \(=K\)
\(\frac{{hc}}{{{\lambda _0}}} = \frac{{{h^2}}}{{2m{\lambda ^2}}}\) \(\therefore {\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}\)
$( h=6.6 \times 10^{-34} \;J\cdot sec)$