An electron of mass ' \(m\) ' and a photon have same energy ' \(E\) '. The ratio of de-Broglie wavelength of electron to the wavelength of photon is \((\mathrm{c}=\) velocity of light \()\)

If \(\mathrm{E}\) is the kinetic energy of the electron, then
\(
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(\mathrm{p}=\text { momentum }) \\
& \therefore \mathrm{p}=\sqrt{2 \mathrm{mE}} \\
& \therefore \text { de Broglie wavelength of electron } \\
& \lambda_{\mathrm{e}}=\frac{1}{\sqrt{2 \mathrm{mE}}} \\
& \text { Energy of photon, } \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \therefore \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}} \\
& \therefore \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{\mathrm{E}}{\mathrm{hc}}=\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}}}
\end{aligned}
\)