Alpha-Particle Scattering Experiment

Scientists: H. Geiger and E. Marsden, under Rutherford
Year: 1911
Source: Radioactive $^{214}_{83}\mathrm{Bi}$
Particles: $\alpha$-particles of energy 5.5 MeV
Target: Thin gold foil
Gold foil thickness: $2.1\times10^{-7}\,\text{m}$

Why gold foil was used?

Gold can be beaten into an extremely thin foil, so α-particles can pass through it.

Observation

1. Most α-particles passed straight through the foil.
   • Therefore, most of the atom is empty space.
2. Very few α-particles were deflected through small angles.
   • Only about 0.14% scattered by more than $1^\circ$.
3. About 1 in 8000 α-particles was deflected through an angle greater than $90^\circ$.
4. A very small number of α-particles were scattered backwards.

---

Rutherford’s Conclusions

1. Most of the atom is empty space

Since most α-particles passed through without deflection, the atom must contain a large amount of empty space.

2. Positive charge is concentrated

Large deflections require a strong repulsive force.

Therefore, the positive charge of the atom must be concentrated in a very small region at the centre.

This region is called the nucleus.

3. Most of the mass is in the nucleus

The nucleus contains:

• Entire positive charge
• Almost entire mass of the atom

4. Nucleus is extremely small

Size of nucleus:$$\boxed{10^{-15}\text{ m to }10^{-14}\text{ m}}$$

Size of atom:$$\boxed{10^{-10}\text{ m}}$$

Therefore, the nucleus is extremely small compared with the atom.

5. Electrons revolve around nucleus

Electrons are present at a considerable distance from the nucleus and move around it, similar to planets revolving around the Sun.

---

Rutherford Nuclear Model

According to Rutherford:$$\boxed{\text{Atom}=\text{tiny positive nucleus}+\text{electrons around it}}$$

Main points

• Nucleus is at the centre of the atom.
• Nucleus is positively charged.
• Almost the entire mass of the atom is concentrated in the nucleus.
• Electrons revolve around the nucleus.
• Most of the atom is empty space.
• Atomic electrons are very light and therefore do not appreciably affect α-particle scattering.

---

Coulomb Force in α-Particle Scattering

An α-particle has charge:$$q_1=+2e$$

Gold nucleus has charge:$$q_2=+Ze$$

For gold:$$Z=79$$

Using Coulomb’s law:$$F=\frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r^2}$$

Therefore,$$\boxed{F=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r^2}}$$

This force is repulsive because both α-particle and gold nucleus are positively charged.

Important assumption

The gold nucleus is about 50 times heavier than an α-particle, so the nucleus can be considered stationary during scattering.

Also, because the gold foil is very thin, an α-particle is assumed to undergo not more than one scattering.

Most important exam line

$$\boxed{\text{Most particles pass straight} \Rightarrow \text{atom mostly empty space}}$$$$\boxed{\text{Few particles deflect strongly} \Rightarrow \text{small, dense, positively charged nucleus}}$$