Alpha Scattering and the Rutherford Model

In the early 20th century, the prevailing view of the atom was the 'plum pudding' model, which suggested that positive charge and mass were spread uniformly throughout the atom. This changed forever in 1911 when Ernest Rutherford, along with Hans Geiger and Ernest Marsden, conducted the famous gold foil experiment. By firing alpha particles at a thin sheet of gold, they observed behaviour that could not be explained by the existing model, leading to the discovery of the atomic nucleus.

For your A-Level Physics exams, understanding this experiment is crucial. It is not just a historical milestone; it provides the foundation for nuclear physics and our understanding of electrostatic potential energy. In this guide, we will break down the experimental setup, the observations, and the mathematical principles used to determine the size of the nucleus.

The Experimental Setup

The experiment involved a radioactive source emitting a collimated beam of alpha particles (helium nuclei, $^4_2\text{He}^{2+}$) directed at a very thin sheet of gold foil. The foil was surrounded by a fluorescent zinc sulphide screen, which produced a tiny flash of light whenever an alpha particle struck it. This allowed researchers to detect the scattering angle of the particles.

If the plum pudding model were correct, the alpha particles should have passed through the thin foil with only minor deflections. Instead, the team observed that while most particles passed straight through, some were deflected at large angles, and a very small fraction rebounded by nearly 180 degrees.

Observations and Conclusions

The results of the experiment led to three key conclusions that define the Rutherford model:

1. Most of the atom is empty space: Since the vast majority of alpha particles passed through undeflected, the atom cannot be a solid sphere of charge.
2. The nucleus is positively charged: The large-angle deflections of the positively charged alpha particles indicate a strong electrostatic repulsion from a concentrated positive centre.
3. The nucleus contains most of the mass: The rare 180-degree rebounds suggest that the alpha particles were hitting something extremely dense and massive.

The Distance of Closest Approach

One of the most important calculations in A-Level Physics is determining the 'distance of closest approach' ($D$). This is the minimum distance an alpha particle can get to the nucleus before the electrostatic repulsion brings it to a momentary stop and forces it to reverse direction.

At this point of closest approach, the initial kinetic energy ($E_k$) of the alpha particle is entirely converted into electric potential energy ($E_p$). We use the formula for electric potential energy between two point charges:

$$E_k = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_1 Q_2}{D}$$

Where:

• $Q_1 = 2e$ (charge of the alpha particle)
• $Q_2 = Ze$ (charge of the target nucleus, where $Z$ is the atomic number)
• $\epsilon_0$ is the permittivity of free space
• $D$ is the distance of closest approach

Worked Example 1: Calculating Closest Approach

An alpha particle with a kinetic energy of $5.0 \text{ MeV}$ is fired at a gold nucleus ($Z=79$). Calculate the distance of closest approach.

Step 1: Convert units. $E_k = 5.0 \times 10^6 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 8.0 \times 10^{-13} \text{ J}$.

Step 2: Rearrange the formula for $D$. $D = \frac{1}{4\pi\epsilon_0} \cdot \frac{2e \cdot Ze}{E_k}$

Step 3: Substitute values. Using $\frac{1}{4\pi\epsilon_0} \approx 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$ and $e = 1.60 \times 10^{-19} \text{ C}$: $D = (8.99 \times 10^9) \cdot \frac{2 \cdot 79 \cdot (1.60 \times 10^{-19})^2}{8.0 \times 10^{-13}}$ $D \approx 4.55 \times 10^{-14} \text{ m}$.

Assumptions in the Model

To perform these calculations, we make several simplifying assumptions:

• The nucleus is a point charge.
• The nucleus is stationary (recoil is neglected due to its large mass).
• The collision is perfectly elastic (kinetic energy is conserved).
• Only electrostatic forces act between the alpha particle and the nucleus.

Worked Example 2: Comparing Energies

If the kinetic energy of the alpha particle in the previous example were doubled, what would happen to the distance of closest approach?

Solution: Looking at the formula $D = \frac{k \cdot Q_1 Q_2}{E_k}$, we can see that $D$ is inversely proportional to $E_k$ ($D \propto 1/E_k$). If $E_k$ is doubled, $D$ will be halved. New $D = 4.55 \times 10^{-14} / 2 = 2.275 \times 10^{-14} \text{ m}$.

Common Mistakes

1. Forgetting to convert MeV to Joules: Always convert energy to Joules before using the electrostatic potential energy formula.
2. Using the wrong charge: Remember that the alpha particle has a charge of $2e$, not $e$.
3. Confusing $D$ with nuclear radius: The distance of closest approach is an upper limit for the nuclear radius; the actual radius is usually smaller than $D$ because the alpha particle does not touch the nucleus.

Frequently Asked Questions

Why was gold used in the experiment? Gold is highly malleable, allowing it to be hammered into an extremely thin foil (a few atoms thick), which ensures alpha particles only interact with a few nuclei.

What is the impact parameter? The impact parameter ($b$) is the perpendicular distance by which the alpha particle would miss the nucleus if no scattering occurred. It determines the scattering angle.

Does the alpha particle touch the nucleus? No. The electrostatic repulsion stops the alpha particle before it reaches the surface of the nucleus, which is why $D$ is only an estimate of the upper bound of the nuclear radius.

Conclusion

The Rutherford scattering experiment remains a cornerstone of A-Level Physics, illustrating how experimental data can fundamentally shift our understanding of the subatomic world. To see these concepts in motion, head over to MathInstructor AI to generate a free, narrated animated lesson on alpha scattering and the nuclear model.