A proton and an $\alpha$-particle are accelerated from rest to the same energy. The de-Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ are in the ratio

The de-Broglie wavelength associated with a charged particle is given by $$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $$ where, $\mathrm{h}=$ planck's constant and $\mathrm{p}=$ momentum $=\sqrt{2 \mathrm{mKE}}$ (here, KE is the kinetic energy of the charged particle) $$ \Rightarrow \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}} $$ For proton and $\alpha$-particle, the wavelengths are respectively given as, $$ \begin{aligned} \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{KE}_{\mathrm{p}}}} \text { and } \lambda_\alpha=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_\alpha \mathrm{KE}_\alpha}} \\ \therefore \quad \quad \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\sqrt{2 \mathrm{~m}_\alpha \mathrm{KE}_\alpha}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{KE}_{\mathrm{P}}}} \end{aligned} $$ Here, $\mathrm{KE}_\alpha=\mathrm{KE}_{\mathrm{p}}$ and $\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}$ Substituting these above mentioned relations in Eq. (i), we get $$ \Rightarrow \quad \frac{\lambda_{\mathrm{P}}}{\lambda_\alpha}=\sqrt{\frac{4 \mathrm{~m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{p}}}}=\frac{2}{1} \quad \text { or } \quad \lambda_{\mathrm{P}}: \lambda_\alpha=2: 1 $$