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A photon and an electron (mass $m$) have the same energy $E$. The ratio $(\lambda_{\text{photon}} / \lambda_{\text{electron}})$ of their de Broglie wavelengths is ($c$ is the speed of light):
Explanation
Photon wavelength is $\lambda_p = \frac{hc}{E}$ and electron wavelength is $\lambda_e = \frac{h}{\sqrt{2mE}}$. Their ratio is $\frac{\lambda_p}{\lambda_e} = c\sqrt{\frac{2m}{E}}$.
Detailed Solution
For a photon: $E = \frac{h c}{\lambda_p} \implies \lambda_p = \frac{h c}{E}$. For a non-relativistic electron of energy $E$: $p = \sqrt{2mE} \implies \lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$. Taking the ratio: $\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{h c / E}{h / \sqrt{2mE}} = c \frac{\sqrt{2mE}}{E} = c \sqrt{\frac{2m}{E}}$.