Question

De-Broglie wavelength is given by
${\lambda }_e=\frac{h}{p}=\frac{h}{\sqrt{2m.E}}$ for electron
De-Broglie wavelength of photon is given by
${\lambda }_p=\frac{h}{p}=\frac{h}{{\displaystyle \frac{E}{p}}}=\frac{hc}{E}$
$\frac{{\lambda }_e}{{\lambda }_p}=\frac{1}{\sqrt{2mE}}.\frac{E}{c}=\frac{1}{c}\sqrt{\frac{E}{2m}}$