A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{ ext{photon}

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A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{	ext{photon}	ext{~}}/\lambda_{	ext{electron}	ext{~}}$ ) of their de Broglie wavelengths is: ( $c$ is the speed of light)

Accepted Answer

For photon, $E = \frac{hc}{\lambda_{Ph}} \Rightarrow \lambda_{Ph} = \frac{hc}{E}$

For electron, $p =$ momentum and $E = \frac{p^{2}}{2m} = \left( \frac{h}{\lambda_{e}} \right)^{2} \times \frac{1}{2m}$

$$\begin{matrix} & \ \Rightarrow \lambda_{e} = \frac{h}{\sqrt{2mE}} \\ & \ \therefore\frac{\lambda_{Ph}}{\lambda_{e}} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} = c\sqrt{\frac{2m}{E}} \end{matrix}$$

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