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Kinetic energy of a proton is equal to energy ' $E$ ' of a photon. Let ' $\lambda_1$ ' be the de-Broglie wavelength of proton and ' $\lambda_2$ ' is the wavelength of photon. If $\frac{\lambda_1}{\lambda_2} \propto \mathrm{E}^{\mathrm{n}}$, then the value of ' $n$ ' is
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The correct answer is:
$\frac{1}{2}$
If $E$ is the kinetic energy of the proton, then $E=\frac{p^2}{2 m}$ where $p$ is the momentum and $m$ is the mass of proton
$$
\therefore \mathrm{p}=\sqrt{2 \mathrm{mE}} \quad \therefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}
$$
For photon, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda_2} \therefore \lambda_2=\frac{\mathrm{hc}}{\mathrm{E}}$
$$
\begin{aligned}
& \therefore \frac{\lambda_1}{\lambda_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \cdot \frac{\mathrm{E}}{\mathrm{hc}}=\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}}} \\
& \therefore \frac{\lambda_1}{\lambda_2} \propto \mathrm{E}^{1 / 2} \quad \therefore \mathrm{n}=\frac{1}{2}
\end{aligned}
$$
$$
\therefore \mathrm{p}=\sqrt{2 \mathrm{mE}} \quad \therefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}
$$
For photon, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda_2} \therefore \lambda_2=\frac{\mathrm{hc}}{\mathrm{E}}$
$$
\begin{aligned}
& \therefore \frac{\lambda_1}{\lambda_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \cdot \frac{\mathrm{E}}{\mathrm{hc}}=\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}}} \\
& \therefore \frac{\lambda_1}{\lambda_2} \propto \mathrm{E}^{1 / 2} \quad \therefore \mathrm{n}=\frac{1}{2}
\end{aligned}
$$
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