Rutherford's experiment, which established the nuclear model of the atom, used a beam of — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Nature of Alpha-Particles An $\alpha$-particle is a fast-moving composite particle consisting of 2 protons and 2 neutrons: $$\ce{^4_2He^{2+}}$$ This is the nucleus of a helium-4 atom stripped of both orbital electrons, carrying a net charge of $+2e$. ### Step 2 - Key Distinction: Helium Atom vs. Helium Nucleus * **Neutral helium atom ($\ce{He}$):** Complete atom with 2 electrons → net charge = 0. Would NOT experience Coulombic repulsion from the gold nucleus. * **Helium nucleus ($\ce{He^{2+}}$, $\alpha$-particle):** Stripped of both electrons → net charge = $+2e$. Experiences strong electrostatic repulsion from the gold nucleus ($Z=79$), producing the characteristic scattering pattern. ### Step 3 - Evaluation of Options * **Option (A) $\beta$-particles:** Negatively charged electrons — not used in Rutherford's experiment. Incorrect. * **Option (B) $\gamma$-rays:** Electromagnetic radiation — massless photons, no Coulombic scattering. Incorrect. * **Option (C) Helium atoms:** Neutral atoms — no charge, no Coulombic repulsion from nucleus. Incorrect. * **Option (D) Helium nuclei:** $\alpha$-particles = positively charged helium nuclei, scattered by gold foil. **Correct.** $$\text{Correct Option: } \boxed{\text{D}}$$