Kinetic energy of a proton is equal to energy \(E\) of a photon. Let ' \(\lambda_1\) ' be the de-Broglie wavelength of proton and ' \(\lambda_2\) ' be the wavelength of photon. If \(\left(\frac{\lambda_1}{\lambda_2}\right) \propto \mathrm{E}^{\mathrm{n}}\) 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
\(\therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}\)
\(\therefore \quad \lambda_1=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\)
Similarly, for a photon, \(E=\frac{h c}{\lambda_2}\)
\(\begin{aligned} & \therefore \quad \dot{\lambda}_2=\frac{h c}{E} \\ & \therefore \quad \frac{\lambda_1}{\lambda_2}=\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c}=\frac{1}{c} \sqrt{\frac{E}{2 m}} \\ & \therefore \quad \\ & \therefore \quad \frac{\lambda_1}{\lambda_2} \propto E^{1 / 2} \Rightarrow n=0.5\end{aligned}\)