A proton and a deuteron are projected towards the stationary gold nucleus, in different experiments, — Atomic Structure Chemistry Question

💡 Solution & Explanation

Distance of closest approach $r_0 = \frac{kQq}{1/2 mv^2} \propto \frac{q}{m}$ (for same speed). For proton, $q/m = e/m_p$. For deuteron, $q/m = e/2m_p$. The proton's $q/m$ is twice the deuteron's, so its distance of closest approach is greater. Therefore, correct answer is B.