E.1.3 Properties of nuclei and high-energy scattering (HL only) Notes

Nuclear Radius, Density, High-Energy Scattering, and Distance of Closest Approach

1. Consider the case when you are trying to measure the size of something so small that it’s 1/100,000th the size of the atom itself.
2. How would you even begin?
3. The atomic nucleus is incredibly tiny and dense, yet its properties are fundamental to understanding atomic structure and the forces that govern matter.

Nuclear Radius and its Relationship to Nucleon Number

1. The size of a nucleus is determined by its radius, which depends on the number of nucleons (protons and neutrons) it contains.
2. This relationship is described by the formula: $$R = R_0 A^{\frac{1}{3}}$$ where:
   1. $R$ is the nuclear radius,
   2. $R_0$ is a constant, approximately $1.2 \times 10^{-15} \, \text{m}$ (or 1.2 femtometers),
   3. $A$ is the nucleon number (mass number) of the nucleus.

Why this formula works

1. The formula $R = R_0 A^{\frac{1}{3}}$ reflects the fact that nuclei are approximately spherical, and their volume is proportional to the number of nucleons.
2. Since the volume of a sphere is proportional to the cube of its radius ($V \propto R^3$), the radius must scale with the cube root of the nucleon number.

Nuclear Density: Why Nuclei Are So Dense

1. Despite their small size, nuclei contain nearly all the mass of an atom.
2. This results in extraordinarily high densities.
3. The density of a nucleus can be calculated using the formula for the density of a sphere: $$\rho = \frac{\text{Mass}}{\text{Volume}}$$
4. For a nucleus:
   1. The mass is approximately $A \cdot m_{\text{nucleon}}$, where $m_{\text{nucleon}} \approx 1.67 \times 10^{-27}, \text{kg}$.
   2. The volume is given by $\frac{4}{3} \pi R^3$.
5. Substituting $R = R_0 A^{\frac{1}{3}}$, the nuclear density simplifies to: $$\rho \approx \frac{3}{4 \pi R_0^3} \cdot m_{\text{nucleon}}$$

Key Insight

1. The density of the nucleus is independent of $A$ because both the volume and mass scale with $A$.
2. This means that all nuclei, regardless of size, have roughly the same density, approximately: $$\rho \approx 2.3 \times 10^{17} \, \text{kg/m}^3$$

High-Energy Scattering: Deviations from Rutherford’s Predictions

1. Rutherford’s gold foil experiment demonstrated that atoms have a small, dense nucleus.
2. However, as particle accelerators enabled higher-energy scattering experiments, deviations from Rutherford’s predictions were observed.

What Rutherford Predicted

Rutherford’s model, based on Coulomb’s law, predicted that alpha particles would scatter elastically off the nucleus, with their deflection angles determined by the electric force between the positively charged alpha particle and the nucleus.

What Was Observed

At very high energies, alpha particles began to penetrate the nucleus, revealing new phenomena:

1. Inelastic Scattering:
   1. Some of the kinetic energy of the alpha particle was absorbed by the nucleus, exciting it to a higher energy state.
2. Nuclear Forces:
   1. The strong nuclear force, which operates at very short ranges, began to influence the interactions, deviating from the purely electrostatic predictions of Rutherford’s model.

Distance of Closest Approach

Energy Conservation

1. As an alpha particle approaches a nucleus, its initial kinetic energy ($E_k$) is converted into electrostatic potential energy ($E_p$) at the point of closest approach.
2. At this point, the particle momentarily comes to rest before being repelled.
3. The electrostatic potential energy is given by: $$E_p = \frac{k q_1 q_2}{r}$$ where:
   1. $k = 8.99 \times 10^9 \, \text{N·m}^2 \ \text{C}^{-2}$ is Coulomb’s constant,
   2. $q_1$ and $q_2$ are the charges of the alpha particle and the nucleus,
   3. $r$ is the distance of closest approach.
4. Setting $E_k = E_p$: $$\frac{1}{2} m v^2 = \frac{k q_1 q_2}{r}$$
5. Rearranging for $r$: $$r = \frac{2 k q_1 q_2}{m v^2}$$