Alpha-Particle Scattering & Rutherford Nuclear Model of the Atom 🔬

1. The Gold-Foil Adventure 🎯

• A narrow beam of 5.5 MeV α-particles from a ²¹⁴Bi source was aimed at an ultra-thin gold foil only 2.1 × 10^-7 m thick :contentReference[oaicite:0]{index=0}.
• The foil sat inside a vacuum chamber. A rotatable zinc-sulphide screen plus microscope spotted each α-particle through tiny “sparkles” called scintillations :contentReference[oaicite:1]{index=1}.
• Most particles zipped straight through. About 0.14 % deflected more than 1°, and only 1 in 8000 bounced back over 90° – a huge surprise! :contentReference[oaicite:2]{index=2}

2. What the Data Shouted 📣

Such extreme backward kicks demanded a very strong, very tiny repulsive center. The obvious conclusion: almost all positive charge and mass lives in a minuscule nucleus, while electrons swirl far away :contentReference[oaicite:3]{index=3}.

• Nucleus size: ≈ 10^-15 to 10^-14 m.
• Whole atom: ≈ 10^-10 m – that’s 10 000–100 000 × larger than the nucleus! :contentReference[oaicite:4]{index=4}
• Result: the atom is mostly empty space, which is why most α-particles sail through un-touched.

3. The Physics Behind the Curves 📐

3.1 Coulomb Kick

The repulsive force between an α-particle \((2e)\) and a nuclear charge \((Ze)\) is

$$F=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^{2}}$$ :contentReference[oaicite:5]{index=5}

3.2 Impact Parameter & Scattering Angle

The impact parameter \(b\) is the perpendicular offset between the incoming velocity and the nuclear center. Small \(b\) ⇒ big angles (even a U-turn!), large \(b\) ⇒ tiny nudges :contentReference[oaicite:6]{index=6}.

3.3 Distance of Closest Approach 🚧

Energy conservation gives the “slam-on-the-brakes” distance

$$d=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^{2}}{K}$$ :contentReference[oaicite:7]{index=7}

For a 7.7 MeV α-particle hitting gold \((Z=79)\): \(d≈3.0 × 10^{-14}\,\text{m}\) (30 fm). The α-particle reverses before touching the 6 fm nucleus – proof of a tiny core! :contentReference[oaicite:8]{index=8}

4. Electron Orbits in the Rutherford Picture 🛰️

To keep an electron circling the nucleus, the electrostatic pull equals the required centripetal force:

$$\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}}=\frac{mv^{2}}{r} \qquad\text{(Eq 12.2)}$$ :contentReference[oaicite:9]{index=9}

Rearranging: $$r=\frac{4\pi\varepsilon_0\,e^{2}}{mv^{2}} \qquad\text{(Eq 12.3)}$$

The electron’s total energy in an orbit of radius \(r\) is $$E=-\frac{e^{2}}{8\pi\varepsilon_0 r}\qquad\text{(Eq 12.4)}$$ (negative sign = bound state) :contentReference[oaicite:10]{index=10}

5. Why More Work Was Needed 🤔

While the nuclear model nailed the atom’s structure, it could not explain why gases glow at specific colors (discrete wavelengths). That puzzle set the stage for the quantum revolution! :contentReference[oaicite:11]{index=11}

6. Quick NEET Power Points 🚀

1. Back-scattering rarity (1 in 8000) proves the nucleus is tiny and massive.
2. Remember the distance-of-closest-approach formula \(d=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^{2}}{K}\).
3. Impact parameter concept connects beam geometry to scattering angles.
4. Electron orbit condition \(\frac{e^{2}}{4\pi\varepsilon_0 r^{2}}= \frac{mv^{2}}{r}\) – a classic question favorite.
5. The enormous empty space fraction inside atoms (≥ 99.99 %) often appears in conceptual questions.

🧑‍🔬 Keep exploring – tiny distances reveal huge insights! 🚀