The following charged particles accelerated from rest, through the same potential difference, are pr — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Kinetic Energy from Accelerating Potential A charged particle with charge $q$ accelerated from rest through potential difference $V$ gains: $$K.E. = qV$$ ### Step 2 - Energy Conservation at Closest Approach At the distance of closest approach $r_{\text{min}}$, all kinetic energy converts to electrostatic potential energy: $$K.E. = U = \frac{kqQ}{r_{\text{min}}}$$ where $Q = Ze$ is the gold nucleus charge, $k$ is Coulomb's constant. ### Step 3 - Derive $r_{\text{min}}$ Setting $qV = \frac{kqQ}{r_{\text{min}}}$ and solving: $$r_{\text{min}} = \frac{kqQ}{qV} = \frac{kQ}{V}$$ The projectile charge $q$ cancels completely. Therefore $r_{\text{min}}$ depends only on $Q$ (target gold nucleus charge) and $V$ (accelerating potential) — **independent of projectile mass or charge**. ### Step 4 - Evaluation of Options All three particles (alpha-particle, proton, deuteron) are accelerated through the same $V$ toward the same gold nucleus ($Q$ constant). Since $r_{\text{min}} = kQ/V$: * **Options (A), (B), (C) are incorrect:** $r_{\text{min}}$ does not depend on the projectile's charge or mass. * **Option (D) is correct:** Distance of closest approach is the same for all three particles. $$\text{Correct Option: } \boxed{\text{D}}$$