🔬 Chapter 3.12.1: The Radius of the Nucleus ⚛️

Welcome to the fascinating world of Nuclear Physics! In this chapter, we are going to shrink down to the smallest scale imaginable—the size of the atomic nucleus.
Understanding the size of the nucleus is crucial because it tells us about the power and range of the Strong Nuclear Force, the force that holds the universe together.
Don't worry if measuring something so small seems impossible; physicists have developed two clever methods to estimate this tiny dimension. Let's dive in!

1. Setting the Scale: Typical Nuclear Radius Values

Before we measure the nucleus, let's appreciate how small it is:

  • A typical atom has a radius of about \(10^{-10}\) m (this is 0.1 nanometres).
  • The nucleus, however, is significantly smaller—on the order of \(10^{-15}\) m. This distance is called a femtometre (fm) or a fermi.

To put this into perspective: If an atom were the size of a football stadium, the nucleus would be like a tiny fly sitting in the very centre!

Quick Review: The Nucleus Components

Remember, the nucleus is made up of nucleons (protons and neutrons). The number of nucleons is defined by the Nucleon Number (A).

2. Method 1: Closest Approach of Alpha Particles

This method is an extension of the famous Rutherford Scattering experiment, and it provides an estimate of the nuclear radius.

How the Experiment Works (The Concept)

1. We fire high-speed alpha particles (which are positively charged, \(Q_\alpha = +2e\)) towards a target nucleus (which is also positively charged, \(Q_{nucleus} = +Ze\)).
2. Because both particles are positive, they experience a strong Coulomb electrostatic repulsion.
3. We assume the alpha particle travels head-on towards the nucleus. It slows down due to the repulsion, stops momentarily, and then reverses direction.
4. At the point where it stops, all its initial Kinetic Energy (\(E_k\)) has been converted entirely into Electric Potential Energy (\(E_{PE}\)). This distance is the distance of closest approach (\(r_{min}\)).

Calculating the Distance of Closest Approach (\(r_{min}\))

We apply the Principle of Conservation of Energy:
$$E_k \text{ (initial)} = E_{PE} \text{ (at closest approach)}$$

The Coulomb equation for electric potential energy (or the work done to bring the charges together) is:

$$E_{PE} = \frac{1}{4\pi\epsilon_0} \frac{Q_{nucleus} Q_{alpha}}{r_{min}}$$

Therefore, we set:
$$\text{Initial } E_k \text{ of alpha} = \frac{1}{4\pi\epsilon_0} \frac{(Ze) (2e)}{r_{min}}$$

If you know the initial kinetic energy of the alpha particle, you can rearrange this equation to solve for \(r_{min}\), which provides an upper limit for the nuclear radius \(R\):

$$\text{Estimated Radius } r_{min} = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{E_k}$$

⚠️ Limitation of the Closest Approach Method

This calculation relies purely on the electromagnetic interaction (Coulomb repulsion). However, the nucleus is held together by the Strong Nuclear Force (SNF).

  • If the alpha particle gets close enough (\(R \approx 10^{-15}\) m), the powerful SNF starts to pull the alpha particle in.
  • If this happens, our assumption (that all KE converts only to EPE) breaks down.
  • Therefore, the distance of closest approach, \(r_{min}\), is only an estimate and is generally larger than the true nuclear radius \(R\).

Key Takeaway: Alpha particles are good for estimating size, but because they are made of nucleons, they interact via the SNF, meaning we can't measure the exact boundary of the nucleus.

3. Method 2: Electron Diffraction (The Precise Method)

To get a more accurate measurement of the nuclear radius, we use high-energy electrons.

Why Electrons?

1. Not Subject to SNF: Electrons are leptons (fundamental particles) and do not interact via the Strong Nuclear Force. They only interact via the electromagnetic force and the weak force. This means they can probe the charge distribution of the nucleus without being affected by the binding force, giving a more accurate measurement of the nuclear boundary.
2. Wave-Particle Duality: To see a small object, the probing wave must have a wavelength (\(\lambda\)) comparable to the size of the object. Since the nucleus is tiny (\(10^{-15}\) m), we need very high energy electrons to produce a small enough de Broglie wavelength (\(\lambda = h/p\)).

The Diffraction Process

When a beam of high-energy electrons hits a nucleus, the electrons scatter, and a diffraction pattern is observed (similar to light passing through a slit).

Interpreting the Diffraction Pattern

The pattern involves measuring the intensity of the scattered electrons against the angle of scattering (\(\theta\)).

  • The resulting graph shows clear maxima and minima, characteristic of diffraction.
  • The angle at which the first minimum occurs (\(\theta_{min}\)) is crucial.

The relationship between the nuclear radius \(R\), the de Broglie wavelength \(\lambda\), and the angle of the first minimum \(\theta_{min}\) is given by:

$$\sin \theta_{min} \approx \frac{1.22 \lambda}{2R}$$


(Note: While you must be familiar with the graph of intensity against angle, you are generally not required to recall this exact formula for calculations, but you must understand that \(R\) is determined from the position of the first minimum.)

By measuring \(\theta_{min}\) and knowing the electron's wavelength \(\lambda\), we can determine the nuclear radius \(R\).

Key Takeaway: Electron diffraction is more precise because electrons don't feel the Strong Nuclear Force, allowing them to map the charge distribution and size of the nucleus more accurately.

4. The Universal Nuclear Radius Relationship

When physicists measured the radii of many different nuclei using electron diffraction, a remarkable pattern emerged: the radius \(R\) is not linearly dependent on the nucleon number \(A\), but rather on the cube root of \(A\).

The Experimental Relationship

The radius \(R\) of a nucleus is directly proportional to the cube root of its nucleon number \(A\):

$$R \propto A^{1/3}$$

This proportionality is written as the defining equation for nuclear radius:

$$\mathbf{R = R_0 A^{1/3}}$$

  • \(R\) is the radius of the nucleus (in fm).
  • \(A\) is the Nucleon Number (or mass number).
  • \(R_0\) is the Fermi Constant (or nuclear radius constant).

The typical value for \(R_0\) derived from experimental data is approximately:

$$R_0 \approx 1.2 \times 10^{-15} \text{ m} \quad (\text{or } 1.2 \text{ fm})$$

Interpreting the Equation: Constant Nuclear Density

The relationship \(R = R_0 A^{1/3}\) provides powerful evidence that the density of nuclear material is constant, regardless of the size of the nucleus.

Don't worry if this sounds complicated—we can prove it with a simple derivation!

Step-by-Step Proof of Constant Density

We define density \(\rho\) as mass \(M\) divided by volume \(V\).

Step 1: Relate Mass to \(A\)
The mass \(M\) of the nucleus is proportional to the number of nucleons \(A\) (since all nucleons have roughly the same mass, \(m_n\)). $$M \propto A$$

Step 2: Relate Volume to \(R\)
Assuming the nucleus is spherical, its volume is: $$V = \frac{4}{3} \pi R^3$$

Step 3: Substitute \(R\) using the empirical formula
We know \(R = R_0 A^{1/3}\). Substitute this into the volume equation: $$V = \frac{4}{3} \pi (R_0 A^{1/3})^3$$ $$V = \frac{4}{3} \pi R_0^3 A$$

Notice that \(\frac{4}{3} \pi R_0^3\) is a constant value. This means the Volume \(V\) is directly proportional to the Nucleon Number \(A\): $$V \propto A$$

Step 4: Calculate Density (\(\rho\))
Density is Mass/Volume. $$\rho = \frac{M}{V}$$ Since \(M \propto A\) and \(V \propto A\), the factors of \(A\) cancel out! $$\rho = \frac{A \cdot m_{nucleon}}{A \cdot (\frac{4}{3} \pi R_0^3)} = \frac{m_{nucleon}}{(\frac{4}{3} \pi R_0^3)}$$

Since \(m_{nucleon}\) and \(R_0\) are both constants, the Nuclear Density \(\rho\) is constant for all nuclei.

Calculation of Nuclear Density

This density is incredibly high. We can calculate its typical value:

  • Approximate mass of one nucleon \(m_{nucleon} \approx 1.67 \times 10^{-27}\) kg
  • \(R_0 \approx 1.2 \times 10^{-15}\) m

$$\rho = \frac{1.67 \times 10^{-27} \text{ kg}}{\frac{4}{3} \pi (1.2 \times 10^{-15} \text{ m})^3} \approx 2.3 \times 10^{17} \text{ kg m}^{-3}$$

Did you know? This density is so immense that a teaspoon of nuclear matter would weigh billions of tonnes! This constant density confirms that nucleons are packed tightly together, filling the nuclear volume like marbles in a box.

✅ Quick Review and Key Takeaways

| Method | Probing Particle | Interaction Used | Result | Limitation | |---|---|---|---|---| | Closest Approach | Alpha Particle | Electromagnetic (Coulomb) | Provides an Estimate (\(r_{min}\)) | Alpha particles are affected by the Strong Nuclear Force, invalidating the result if they touch the nucleus. | | Electron Diffraction | High-Energy Electron | Electromagnetic (Diffraction) | Provides a Precise Radius (\(R\)) | Electrons are not affected by the SNF, allowing accurate measurement of the charge distribution boundary. |

Key Formulae:
1. Closest Approach Energy Balance: \(E_k = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min}}\)
2. Radius Dependence: \(\mathbf{R = R_0 A^{1/3}}\)
3. The interpretation of the dependence is that Nuclear Density is constant for all nuclei.