The de-Broglie wavelength of an electron is the same as that of a...

For photon,

${E}_{P} = \dfrac{hc}{{\lambda}_{p}} ⇒ {\lambda}_{p} = \dfrac{hc}{{E}_{P}}$

For electron

$\begin{cases}{\lambda}_{e} & = & \dfrac{h}{{m}_{e}{v}_{e}} \\ & = & \dfrac{h{v}_{e}}{2{K}_{e}}\end{cases}$

Given ${v}_{e} = 0 . 25c$. So,

$\begin{cases}{\lambda}_{e} & = & \dfrac{h \times 0 . 25c}{2{K}_{e}} \\ & = & \dfrac{hc}{8{K}_{e}}\end{cases}$

Also ${\lambda}_{p} = {\lambda}_{e}$. Thus,

$\dfrac{hc}{{E}_{p}} = \dfrac{hc}{8{K}_{e}} ⇒ \dfrac{{K}_{e}}{{E}_{p}} = \dfrac{1}{8}$