Explanation:

The de-Broglie wavelength of an electron,
\(\lambda = \frac{h}{{mv}} \Rightarrow v = \frac{h}{{m\lambda }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Energy of photon,
\({E_p} = \frac{{hc}}{\lambda }\left( {{\text{since,}}\lambda \,{\text{is}}\,{\text{same}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
Kinetic energy of photon to kinetic
energy of electron
\(\,\,\,\,\,\, = \frac{{{E_p}}}{{{E_e}}} = \frac{{\frac{{hc}}{\lambda }}}{{\left( {\frac{1}{2}} \right)m{v^2}}} = \frac{{2hc}}{{\lambda m{v^2}}}\)
Substituting the value of \(v\) from
Eq. (1), we get
\(\,\,\,\,\,\,\,\,\frac{{{E_p}}}{{{E_e}}} = \frac{{2hc}}{{\lambda m{{\left( {\frac{h}{{m\lambda }}} \right)}^2}}} = \frac{{2\lambda mc}}{h}\)