Chapter: The Nucleus
Welcome to one of the most exciting topics in A2 Physics! In this chapter, we zoom in past the electron clouds to explore the very core of matter: the atomic nucleus. Don't worry if nuclear physics sounds intimidating at first—we will break down the experiments, the forces, and the mathematics step-by-step.
By the end of these notes, you will understand how Rutherford discovered the nucleus, how we measure its size, why all nuclei have remarkably similar densities, and what powerful force holds everything together.
---1. Nuclear Constituents and Notation
At the center of every atom lies a nucleus composed of particles called nucleons. There are two types of nucleons:
• Protons (\(p\)): positively charged particles.
• Neutrons (\(n\)): neutral (uncharged) particles.
Standard Nuclear Notation
We represent any nuclide using the standard notation: \(^{A}_{Z}\text{X}\)
• \(\text{X}\) is the chemical symbol of the element.
• \(Z\) is the Atomic Number (or proton number). It equals the total number of protons in the nucleus and defines the positive charge of the nucleus (\(Q = +Ze\)).
• \(A\) is the Nucleon Number (or mass number). It is the total number of protons plus neutrons (\(A = Z + N\)).
• \(N\) is the Neutron Number, calculated simply as \(N = A - Z\).
Isotopes
Isotopes are nuclei that have the same proton number (\(Z\)) but different neutron numbers (\(N\)), and therefore different nucleon numbers (\(A\)). Because they have identical electron arrangements, isotopes share identical chemical properties, but their differing masses lead to different nuclear stabilities.
Key Takeaway: Protons determine what element it is (\(Z\)), while neutrons change the isotope's mass (\(A\)) without altering its chemical identity.
---2. Rutherford Alpha-Particle Scattering & Nuclear Dimensions
How do we know the nucleus exists? In the famous Geiger–Marsden (Rutherford) experiment, energetic alpha particles (\(\alpha\), which are helium nuclei \(^{4}_{2}\text{He}\)) were fired at a very thin gold foil inside an evacuated chamber (vacuum).
Key Observations and Inferences
1. Observation: The vast majority of \(\alpha\)-particles passed straight through the foil with little or no deflection.
Inference: The atom is mostly empty space.
2. Observation: A very small fraction (around 1 in 8000) was deflected through very large angles (greater than \(90^\circ\), known as back-scattering).
Inference: The positive charge and almost all the mass of the atom are concentrated in a tiny, extremely dense central volume called the nucleus.
Distance of Closest Approach (\(d\))
When an \(\alpha\)-particle travels directly towards a nucleus of charge \(+Ze\), it slows down because of the repulsive electrostatic (Coulomb) force. At the point of closest approach (\(d\)), the \(\alpha\)-particle momentarily stops. At this turning point, all of its initial kinetic energy (\(E_k\)) has been converted into electrostatic potential energy (\(E_p\)):
\(E_k = E_p = \frac{1}{4\pi\varepsilon_0} \frac{q_{\alpha} Q_{\text{nucleus}}}{d} = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{d}\)
Rearranging for the distance of closest approach:
\(d = \frac{2Ze^2}{4\pi\varepsilon_0 E_k}\)
Crucial Exam Distinction: The distance \(d\) calculated this way is an upper limit (upper bound) to the nuclear radius, not the exact radius. The \(\alpha\)-particle stops and turns back before it actually touches the nuclear surface due to electrostatic repulsion.
Key Takeaway: Rutherford's scattering experiment proved the nuclear model of the atom, and energy conservation during head-on collisions allows us to calculate an upper bound for the nuclear radius (\(d \approx 10^{-14}\text{ m}\)).
---3. Nuclear Radius and Scaling
Through scattering experiments, physicists established that the nuclear radius \(r\) scales with the nucleon number \(A\) according to a simple power law:
\(r = r_0 A^{1/3}\)
• \(r\) is the radius of the nucleus in metres (\(\text{m}\)).
• \(A\) is the nucleon number (mass number).
• \(r_0\) is a constant representing the radius of a single nucleon / characteristic nuclear radius: \(r_0 \approx 1.2 \times 10^{-15}\text{ m} = 1.2\text{ fm}\).
Unit Reminder: \(1\text{ fm}\) (femtometre) \(= 10^{-15}\text{ m}\). Always convert femtometres to metres before calculating!
Graphical Relationships
Examiners love testing how to linearise the equation \(r = r_0 A^{1/3}\):
Graph 1: Plotting \(r\) against \(A^{1/3}\)
• Comparing \(r = r_0 (A^{1/3})\) to \(y = mx + c\):
• \(y = r\), \(x = A^{1/3}\), \(m = r_0\), and \(c = 0\).
• This yields a straight line through the origin \((0,0)\) with gradient equal to \(r_0\).
Graph 2: Logarithmic Plot (\(\ln r\) against \(\ln A\))
• Taking natural logarithms of both sides: \(\ln(r) = \ln(r_0 A^{1/3}) = \ln(r_0) + \frac{1}{3}\ln(A)\)
• Rearranging into \(y = mx + c\) format: \(\ln(r) = \frac{1}{3}\ln(A) + \ln(r_0)\)
• Plotting \(\ln(r)\) on the y-axis against \(\ln(A)\) on the x-axis gives:
– Gradient: \(\frac{1}{3}\)
– y-intercept: \(\ln(r_0)\) (Note: remember that the intercept is \(\ln(r_0)\), not \(r_0\)).
Key Takeaway: Nuclear radius is directly proportional to the cube root of the nucleon number (\(r \propto A^{1/3}\)). A plot of \(r\) against \(A^{1/3}\) gives a straight line through the origin with gradient \(r_0\).
---4. Nuclear Volume and Nuclear Density
Nuclear Volume (\(V\))
Assuming the nucleus is spherical, its volume is given by:
\(V = \frac{4}{3}\pi r^3\)
Substituting \(r = r_0 A^{1/3}\) into the volume equation:
\(V = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A\)
Since \(\frac{4}{3}\pi r_0^3\) is constant, this shows that nuclear volume is directly proportional to nucleon number (\(V \propto A\)).
Nuclear Density (\(\rho\))
To find the density of nuclear matter, we divide the nuclear mass by the nuclear volume. The mass of a nucleus is approximately \(m_{\text{nucleus}} \approx A \times u\), where \(u = 1.661 \times 10^{-27}\text{ kg}\) (the unified atomic mass unit):
\(\rho = \frac{\text{mass}}{\text{volume}} = \frac{A \cdot u}{\frac{4}{3}\pi r_0^3 A}\)
Notice what happens mathematically: the nucleon number \(A\) cancels out completely!
\(\rho = \frac{3u}{4\pi r_0^3}\)
Crucial Deductions:
1. Because \(A\) cancels out, nuclear density is constant for all nuclei, regardless of the element or mass number.
2. Calculating this value gives \(\rho \approx 10^{17}\text{ to } 10^{18}\text{ kg m}^{-3}\). This is unimaginably dense (a single thimble of nuclear matter would have a mass of hundreds of millions of tonnes!), demonstrating that bulk matter is almost entirely empty space.
Key Takeaway: While volume scales directly with mass number (\(V \propto A\)), the density of nuclear material is independent of \(A\) and is constant across all nuclides (\(\approx 10^{17}\text{ kg m}^{-3}\)).
---5. The Strong Nuclear Force
Inside the nucleus, positively charged protons are packed tightly together. Electrostatic repulsion tries to blow the nucleus apart. Why doesn't it? There must be an attractive force stronger than the electrostatic repulsion holding the nucleons together: the Strong Nuclear Force.
Properties of the Strong Nuclear Force
• Charge-independent: It acts equally between proton–proton, neutron–neutron, and proton–neutron pairs.
• Short-range: It only acts over subatomic distances.
• Variation with distance:
– Below \(\approx 0.5\text{ fm}\): It is repulsive (this prevents the nucleons from collapsing into a single point).
– Between \(\approx 0.5\text{ fm}\) and \(\approx 3.0\text{ fm}\): It is strongly attractive (with maximum attraction around \(1\text{ fm}\)).
– Beyond \(\approx 3.0\text{ fm}\): It falls rapidly to zero.
Key Takeaway: The strong nuclear force is repulsive below \(0.5\text{ fm}\), attractive between \(0.5\text{ fm}\) and \(3.0\text{ fm}\), and negligible beyond \(3.0\text{ fm}\). It acts identically on all nucleons regardless of charge.
---6. Common Mistakes to Avoid in Examinations
1. Confusing Radius and Volume Proportionality:
Remember that \(r \propto A^{1/3}\), but \(V \propto A\). If the nucleon number doubles, the volume doubles, but the radius increases only by a factor of \(2^{1/3} \approx 1.26\).
2. Proving Nuclear Density is Constant:
If an exam question asks you to "show that nuclear density is constant for all nuclei", do not just compute the density of one specific nuclide (e.g. Carbon-12). You must write down \(\rho = \frac{A \cdot u}{\frac{4}{3}\pi r_0^3 A}\) and show algebraically that \(A\) cancels out.
3. Logarithmic Intercept Errors:
In the equation \(\ln(r) = \frac{1}{3}\ln(A) + \ln(r_0)\), the y-intercept is \(\ln(r_0)\). To determine \(r_0\) from a graph, calculate \(e^{\text{intercept}}\), not just the intercept itself.
4. Metric Prefix Conversions:
Remember that \(1\text{ fm} = 10^{-15}\text{ m}\). Failing to convert to metres before calculating volume, density, or electric potential energy will cause power-of-ten errors.
5. Closest Approach Interpretation:
Always state that the distance of closest approach is an upper limit on nuclear size because Coulomb repulsion halts the alpha particle before it touches the nuclear surface.
Quick Summary Checklist
• \(A\) = Nucleon number, \(Z\) = Proton number, \(N = A - Z\) = Neutron number.
• Rutherford scattering showed the atom is mostly empty space with a tiny, dense, positive nucleus.
• Closest approach: \(E_k = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d}\) (gives upper bound on radius).
• Radius scaling: \(r = r_0 A^{1/3}\) where \(r_0 \approx 1.2\text{ fm}\).
• Volume: \(V \propto A\).
• Nuclear density: \(\rho = \frac{3u}{4\pi r_0^3} \approx 10^{17}\text{ kg m}^{-3}\) (constant for all nuclei).
• Strong force: Repulsive \(< 0.5\text{ fm}\), attractive \(0.5\text{ to } 3.0\text{ fm}\), zero \(> 3.0\text{ fm}\).