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

💡 Solution & Explanation

### Step 1 - Distance of Closest Approach Formula For an $\alpha$-particle (charge $q_1 = 2e$) projected with kinetic energy $K.E.$ toward a metal nucleus of charge $q_2 = Ze$: $$r_0 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{K.E.} = \frac{1}{4\pi\varepsilon_0} \frac{2e \cdot Ze}{K.E.}$$ ### Step 2 - Determine Proportionality Since $K.E.$, $q_1 = 2e$, and Coulomb's constant are all constants for this experiment: $$r_0 \propto Z$$ ### Step 3 - Compare Atomic Numbers The metal with the **minimum Z** gives the **minimum $r_0$**: | Metal | Z | |-------|---| | Copper (Cu) | 29 | | Silver (Ag) | 47 | | Gold (Au) | 79 | | Calcium (Ca) | 20 | Minimum $Z$ = Calcium (Ca) with $Z = 20$. ### Step 4 - Evaluation of Options * **Option (A) Cu ($Z=29$):** $29 > 20$ → larger $r_0$ than Ca. Incorrect. * **Option (B) Ag ($Z=47$):** Much larger $Z$ → larger $r_0$. Incorrect. * **Option (C) Au ($Z=79$):** Highest $Z$ → maximum $r_0$. Incorrect. * **Option (D) Ca ($Z=20$):** Minimum $Z$ → minimum nuclear charge → minimum repulsion → minimum $r_0$. **Correct.** $$\text{Correct Option: } \boxed{\text{D}}$$