-particles are projected towards the nucleus of following metals, with the same kinetic energy. Towa — Atomic Structure Chemistry Question

💡 Solution & Explanation

Distance of closest approach $r_0 = \frac{kQq}{K.E.} = \frac{k (Ze) (2e)}{K.E.} \propto Z$. For the same kinetic energy, the minimum $r_0$ corresponds to the metal with the minimum atomic number ($Z$). Calcium ($Z=20$) has the lowest atomic number among the options. Therefore, correct answer is D.