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The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by
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Verified Answer
The correct answer is:
$\lambda_p \propto \lambda_e^2$
Wavelength of electron,
$$
\begin{aligned}
& \lambda_e=\frac{h}{\sqrt{2 m E}} \\
& \text { and proton } \lambda_p=\frac{h c}{E} \\
& \Rightarrow \quad \lambda_e^2=\frac{h^2}{2 m E} \text { or } E=\frac{h c}{\lambda_p} \\
& \therefore \quad \lambda_e^2=\frac{h^2}{2 m \cdot \frac{h c}{\lambda_p}} \Rightarrow \lambda_e^2=\frac{h^2}{2 m h C} \lambda_p \\
& \Rightarrow \quad \lambda_e^2 \propto \lambda_p
\end{aligned}
$$
$$
\begin{aligned}
& \lambda_e=\frac{h}{\sqrt{2 m E}} \\
& \text { and proton } \lambda_p=\frac{h c}{E} \\
& \Rightarrow \quad \lambda_e^2=\frac{h^2}{2 m E} \text { or } E=\frac{h c}{\lambda_p} \\
& \therefore \quad \lambda_e^2=\frac{h^2}{2 m \cdot \frac{h c}{\lambda_p}} \Rightarrow \lambda_e^2=\frac{h^2}{2 m h C} \lambda_p \\
& \Rightarrow \quad \lambda_e^2 \propto \lambda_p
\end{aligned}
$$
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