In the different experiments, α-particles, proton, deuteron and neutron are projected towards gold n — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Distance of Closest Approach Formula For a charged projectile with charge $q_1$ and kinetic energy $K.E.$ projected toward a nucleus with charge $q_2 = Ze$: $$r_0 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{K.E.}$$ ### Step 2 - Charges of the Four Particles * $\alpha$-particle ($\ce{He^{2+}}$): $q = +2e$ * Proton ($\ce{H^+}$): $q = +1e$ * Deuteron ($\ce{^2_1H^+}$): $q = +1e$ * Neutron: $q = 0$ ### Step 3 - Apply the Formula All four particles have the same $K.E.$ and target the same gold nucleus. So $r_0 \propto q_1$. For the neutron, $q_1 = 0$: $$r_0 = \frac{1}{4\pi\varepsilon_0} \frac{0 \cdot Ze}{K.E.} = 0$$ The neutron experiences **no electrostatic repulsion** and can approach the nucleus indefinitely — its distance of closest approach (due to electrostatic forces) is effectively zero. ### Step 4 - Evaluation of Options * **Option (A) $\alpha$-particle:** $q = 2e$ → largest repulsion → largest $r_0$. Incorrect. * **Option (B) proton:** $q = e$ → finite $r_0$. Incorrect. * **Option (C) deuteron:** $q = e$ → same $r_0$ as proton. Incorrect. * **Option (D) neutron:** $q = 0$ → no electrostatic barrier → minimum $r_0 \approx 0$. **Correct.** $$\text{Correct Option: } \boxed{\text{D}}$$