Nuclear Size, Density, and the A Relation Revision Notes | A Level AQA Physics

1. How Small Is a Nucleus, Really?

You already know that atoms are mostly empty space — the nucleus sits at the centre like a fly in a cathedral. But how do physicists actually measure something so vanishingly small? And once they have those measurements, what pattern emerges?

The answer turns out to be one of the most elegant results in nuclear physics: nuclear radii follow a beautifully simple rule that connects directly to the composition of the nucleus. That rule also reveals something profound — all nuclear material has the same density, regardless of which element you are looking at.

2. Typical Nuclear Radii

Nuclear radii are measured in femtometres (fm), where $1 \text{ fm} = 1 \times 10^{-15}$ m. For context, a typical atom has a radius of around $1 \times 10^{-10}$ m — so the nucleus is roughly 100,000 times smaller than the atom itself.

Across the periodic table, nuclear radii range from about $1 \times 10^{-15}$ m for a proton (hydrogen nucleus) up to roughly $7 \times 10^{-15}$ m for the heaviest naturally occurring nuclei such as uranium. These values come from high-energy electron diffraction experiments, where electrons are fired at nuclei and the diffraction pattern reveals the nuclear size.

Knowing the scale is the first step. The next question is whether there is a pattern — and there is a very clear one once you plot nuclear radius against nucleon number $A$.

3. The Nuclear Radius Equation: $R = R_0 A^{1/3}$

When experimental data for nuclear radii are plotted against nucleon number $A$, the relationship is not linear. Instead, plotting $R$ against $A^{1/3}$ gives a straight line through the origin — telling us that $R$ is directly proportional to $A^{1/3}$. This is where the equation comes from: it is derived from experimental data, not from first principles alone.

Notice that $A^{1/3}$ means the cube root of $A$. So doubling the nucleon number does not double the radius — it only increases it by a factor of $2^{1/3} \approx 1.26$. The nucleus grows, but much more slowly than you might expect.

Worked example: Calculating Nuclear Radius Using $R = R_0 A^{1/3}$

Calculate the radius of a carbon-12 nucleus ($A = 12$). Use $R_0 = 1.2 \times 10^{-15}$ m. Give your answer to 2 significant figures.

1. $R = R_0 A^{1/3}$ — [M1] Stating the correct formula earns the method mark. Make sure the exponent is $1/3$, not $3$.
2. $R_0 = 1.2 \times 10^{-15}$ m, $A = 12$ — Carbon-12 has 6 protons and 6 neutrons, giving $A = 12$. Always state your known values before substituting.
3. $12^{1/3} = \sqrt[3]{12} \approx 2.289$ — The cube root of 12 is not a whole number. Use your calculator: $12^{1/3} \approx 2.289$. Do not round here — carry the full value forward.
4. $R = 1.2 \times 10^{-15} \times 2.289 = 2.747 \times 10^{-15}$ m — [M1] Correct substitution. $1.2 \times 2.289 = 2.7468$, which rounds to $2.747$.
5. $R \approx 2.7 \times 10^{-15}$ m — [A1] Final answer with correct units. The question asked for 2 s.f., so $2.747$ rounds to $2.7$. Always include the unit — metres here, not fm unless the question specifies.

Answer: $R \approx 2.7 \times 10^{-15}$ m

4. Why $A^{1/3}$? The Evidence for Constant Nuclear Density

The exponent $1/3$ is not arbitrary — it carries a deep physical meaning. To see why, think about what determines the volume of a sphere: $V = \frac{4}{3}\pi R^3$. If $R = R_0 A^{1/3}$, then substituting gives $V = \frac{4}{3}\pi R_0^3 A^{1/3 \times 3} = \frac{4}{3}\pi R_0^3 A$.

So the volume of the nucleus is directly proportional to $A$ — the nucleon number. Since $A$ is also proportional to the nuclear mass (each nucleon has approximately the same mass $m_p \approx 1.67 \times 10^{-27}$ kg), the mass of the nucleus is also proportional to $A$. Density is mass divided by volume, and both scale with $A$, so the density is constant.

This is the key interpretive point that examiners test: the $R \propto A^{1/3}$ relationship is direct experimental evidence that all nuclei have the same density of nuclear material, regardless of the element. A carbon nucleus and a uranium nucleus are made of the same stuff, packed at the same density — uranium is bigger.

5. Calculating Nuclear Density

Now that you can see why density must be constant, the next step is to calculate its actual value. The approach uses the definition of density ($\rho = m/V$), the nuclear radius equation, and the mass of a nucleon.

Each nucleon (proton or neutron) has a mass of approximately $1.67 \times 10^{-27}$ kg. A nucleus with nucleon number $A$ therefore has a total mass of approximately $A \times 1.67 \times 10^{-27}$ kg. The volume is $\frac{4}{3}\pi R_0^3 A$ as derived above. Dividing mass by volume, the factor of $A$ cancels — confirming that density is independent of $A$.

Worked example: Deriving and Calculating Nuclear Density

Show that nuclear density is approximately $2.3 \times 10^{17}$ kg m$^{-3}$, using $R_0 = 1.2 \times 10^{-15}$ m and nucleon mass $m_p = 1.67 \times 10^{-27}$ kg.

1. $m = A \times m_p = A \times 1.67 \times 10^{-27}$ kg\n$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$ — The volume simplifies because $(A^{1/3})^3 = A$. This is the algebraic step that makes the $A$ dependence visible.
2. $\rho = \frac{m}{V} = \frac{A \times m_p}{\frac{4}{3}\pi R_0^3 A}$ — [M1] Setting up the ratio correctly. Notice that $A$ appears in both numerator and denominator.
3. $\rho = \frac{m_p}{\frac{4}{3}\pi R_0^3}$ — The factor of $A$ cancels completely. This is the algebraic proof that nuclear density does not depend on which nucleus you choose — it is the same for all nuclei.
4. $R_0^3 = (1.2 \times 10^{-15})^3$\n$= 1.2^3 \times (10^{-15})^3$\n$= 1.728 \times 10^{-45}$ m$^3$ — $1.2^3 = 1.2 \times 1.2 \times 1.2 = 1.44 \times 1.2 = 1.728$. And $(10^{-15})^3 = 10^{-45}$. Expand powers carefully — do not skip this step.
5. $\frac{4}{3}\pi R_0^3 = \frac{4}{3} \times 3.1416 \times 1.728 \times 10^{-45}$\n$= 4.189 \times 1.728 \times 10^{-45}$\n$= 7.238 \times 10^{-45}$ m$^3$ — $\frac{4}{3} \times \pi = 4.189$ (to 4 s.f.). Then $4.189 \times 1.728 = 7.238$. Keep enough decimal places to avoid rounding errors in the final step.
6. $\rho = \frac{1.67 \times 10^{-27}}{7.238 \times 10^{-45}}$\n$= \frac{1.67}{7.238} \times 10^{-27-(-45)}$\n$= 0.2308 \times 10^{18}$\n$= 2.308 \times 10^{17}$ kg m$^{-3}$ — [A1] The exponent: $-27 - (-45) = -27 + 45 = +18$. Then $1.67 \div 7.238 = 0.2308$. So $\rho \approx 2.3 \times 10^{17}$ kg m$^{-3}$ (using $R_0 = 1.2 \times 10^{-15}$ m).

Answer: $\rho \approx 2.3 \times 10^{17}$ kg m$^{-3}$ (using $R_0 = 1.2 \times 10^{-15}$ m). This is approximately $10^{17}$ kg m$^{-3}$, many orders of magnitude greater than everyday materials.

To put that number in perspective: water has a density of $1000$ kg m$^{-3}$. Nuclear material is roughly $2 \times 10^{14}$ times denser. A teaspoon of nuclear material would have a mass of around $10^{11}$ kg — comparable to a mountain.

6. Exam Technique and Common Pitfalls

Questions on this topic tend to fall into three categories: recall of typical values, calculation using $R = R_0 A^{1/3}$, and explanation of what the equation implies about nuclear density. Each requires a slightly different approach.

For calculation questions, the most common source of lost marks is handling the cube root incorrectly. Always evaluate $A^{1/3}$ as a separate step before multiplying by $R_0$. For explanation questions, the phrase to reach for is: the $A^{1/3}$ dependence means volume is proportional to $A$, and since mass is also proportional to $A$, density is constant. Examiners want to see that chain of reasoning, not a bare assertion.

Bringing this together: the $R = R_0 A^{1/3}$ equation is not merely a formula to plug numbers into. It encodes the physical reality that nuclear matter is incompressible — every nucleus, from hydrogen to uranium, is packed at the same extraordinary density of around $10^{17}$ kg m$^{-3}$. That single insight connects the experimental measurement of nuclear size to the fundamental structure of matter.