The wavelength $\lambda_{\mathrm{e}}$ of an electron and $\lambda_{\mathrm{p}}$ of a photon of same

Question

The wavelength $\lambda_{\mathrm{e}}$ of an electron and $\lambda_{\mathrm{p}}$ of a photon of same energy $E$ are related by\

Professor PC Thomas Classes · Answer

Wavelength of an electron of energy $E$ is
$$
\begin{equation*}
\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{E}}} 	ag{i}
\end{equation*}
$$

Wavelength of a photon of same energy E is
$\lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$ or $\mathrm{E}=\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}$
Squaring both sides of Eq (i), we get
$\lambda_{e}^{2}=\frac{h^{2}}{2 m_{e} E}$ or $E=\frac{h^{2}}{2 m_{e} \lambda_{e}^{2}} \ldots$
Equating (ii) and (iii), we get
$$
\frac{h c}{\lambda_{p}}=\frac{h^{2}}{2 m_{e} \lambda_{e}^{2}}
$$
$\lambda_{\mathrm{p}} \propto \lambda_{\mathrm{e}}^{2}$