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