Match the following ratio of De-Broglie wavelengths of a proton and...

Here's the detailed matching: - A. Have the same speed (List-II: r) For the same speed, the de Broglie wavelength (\lambda) is inversely proportional to mass ($\left(\lambda \propto \frac{1}{m}\right). Since the$mass of a proton is about half that of an a-particle (which has twice the mass of a proton), the ratio of de Broglie wavelengths will be (\lambda_p: \lambda_$\alpha=$2: 1). - B. Have the same potential difference (List-II: s) For the same potential difference, the kinetic energy gained by the particles is proportional to their charge. Since the a-particle has twice the charge of a proton, the energy will differ, leading to a wavelength ratio of (2 $\sqrt{2}$: 1). - C. Have the same kinetic energy (List-II: p) For the same kinetic energy, the de Broglie wavelength is inversely proportional to the square root of the mass ($\left(\lambda \propto \frac{1}{\sqrt{m}}\right). Since the a-particle is 4 times more massive than the proton, the$ratio will be (\lambda_p: \lambda_$\alpha=2: 1). D. Have the same linear momentum (List-II: q) For$the same momentum, the de Broglie wavelength is the same for both particles ((\lambda=\frac{h}{p}), so the ratio will be $1: 1$. Thus, the correct match is r s p q.