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

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, \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda_2} \therefore \lambda_2=\frac{\mathrm{hc}}{\mathrm{E}}\)
\(
\begin{aligned}
& \therefore \frac{\lambda_1}{\lambda_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \cdot \frac{\mathrm{E}}{\mathrm{hc}}=\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}}} \\
& \therefore \frac{\lambda_1}{\lambda_2} \propto \mathrm{E}^{1 / 2} \quad \therefore \mathrm{n}=\frac{1}{2}
\end{aligned}
\)