Nuclear Size
The nucleus sits at the heart of every atom, yet it occupies only about one ten-trillionth of the atom's total volume. Despite this incredible smallness — nuclear radii are typically measured in femtometres (1 fm = 10⁻¹⁵ m) — physicists have been able to measure nuclear sizes precisely by firing high-energy electrons or alpha particles at nuclei and observing how they scatter. What emerged from decades of such experiments is a beautifully simple pattern: nuclear radius scales with the cube root of the mass number A. This relationship tells us something profound about nuclear matter. Unlike electrons in an atom, which are spread across a vast cloud with no definite edge, protons and neutrons (collectively called nucleons) pack together like marbles in a bag. Each nucleon takes up roughly the same volume regardless of which nucleus it lives in. As you add more nucleons, the nucleus grows in volume proportionally — much like adding identical building blocks to a pile keeps the average density of the pile constant. This property is called nuclear saturation. Understanding nuclear size is not merely an academic exercise. It underpins our models of radioactive decay, nuclear reactions in stars, and the design of particle accelerators and medical imaging devices. The empirical formula R = R₀ A^(1/3) is one of the most useful and elegant results in all of nuclear physics, connecting a measurable geometric property to a single integer: the number of nucleons inside.
- What is Mass Number A?
- Modelling Nucleons as Hard Spheres
- The Empirical Formula
- Nuclear Density is Constant
- Why Nuclei Don't Just Collapse
What is Mass Number A?
The mass number A is the total number of nucleons (protons + neutrons) in a nucleus. For example, carbon-12 has A = 12, while uranium-238 has A = 238. A is always a positive integer and is written as a superscript before the element symbol: ¹²C or ²³⁸U.
Modelling Nucleons as Hard Spheres
Experimental scattering data shows that nuclear density is nearly constant across all known nuclei — roughly 2.3 × 10¹⁷ kg/m³, which is about 200 million tonnes per cubic centimetre. If every nucleus has the same density ρ, then the nuclear mass M ∝ A and the nuclear volume V = M/ρ ∝ A. Since V = (4/3)πR³, we get R³ ∝ A, leading immediately to R ∝ A^(1/3).
The Empirical Formula
Fitting the formula to electron-scattering experiments gives R = R₀ A^(1/3), where the constant R₀ ≈ 1.2 fm (some textbooks quote 1.1–1.3 fm depending on the measurement method). This single constant encodes the size of one nucleon's 'territory' inside a nucleus. For hydrogen (A = 1), R = R₀ ≈ 1.2 fm — essentially the proton radius. For lead-208 (A = 208), R = 1.2 × 208^(1/3) ≈ 1.2 × 5.93 ≈ 7.1 fm.
Nuclear Density is Constant
Because R ∝ A^(1/3), the volume V = (4/3)πR³ = (4/3)π R₀³ A. The nuclear density is ρ = (A × mₙ) / V = mₙ / [(4/3)πR₀³], which is independent of A. Here mₙ ≈ 1.67 × 10⁻²⁷ kg is the nucleon mass. Substituting R₀ = 1.2 × 10⁻¹⁵ m gives ρ ≈ 2.3 × 10¹⁷ kg/m³ — the same for every nucleus, from hydrogen to oganesson.
Why Nuclei Don't Just Collapse
Nuclear saturation — the fact that each nucleon only 'feels' its nearest neighbours — arises because the strong nuclear force has very short range (about 1–2 fm). It is strongly repulsive at distances below ~0.5 fm (the hard core) and attractive between ~0.5–2 fm. Beyond that it falls essentially to zero. This short-range nature prevents a cascade of attraction that would pull all nucleons into one point, much like a bag of marbles sits at a stable size rather than collapsing.
The Key Equation
- R — Nuclear radius (fm (femtometres)): The approximate radius of the nucleus, treated as a uniformly dense sphere.
- R_0 — Empirical nuclear radius constant (fm): Approximately 1.2 fm (1.2 × 10⁻¹⁵ m), determined from electron-scattering experiments. Represents the effective radius of a single nucleon in nuclear matter.
- A — Mass number (dimensionless): Total number of protons and neutrons in the nucleus. Determines how many nucleons must be packed into the sphere.
Worked Example
Problem. Estimate the radius of a gold-197 nucleus (¹⁹⁷Au, A = 197) using R₀ = 1.2 fm. Then calculate the nuclear density and compare it to the density of gold metal (19,300 kg/m³).
Answer. Nuclear radius ≈ 7.0 fm; nuclear density ≈ 2.3 × 10¹⁷ kg/m³, roughly 10 trillion times denser than solid gold.
Origins · 1 minRobert Hofstadter and how this was discovered
Robert Hofstadter was awarded the 1961 Nobel Prize in Physics for his pioneering electron-scattering experiments at Stanford that precisely mapped the charge distribution inside atomic nuclei. By firing electrons at energies above 100 MeV at various nuclei and measuring how the scattered electrons emerged at different angles, he confirmed the nuclear radius formula R = R₀A^(1/3) and showed that nuclear density is roughly uniform inside the nucleus with a diffuse surface of thickness ~2.4 fm. His work transformed nuclear physics from a purely theoretical field into one grounded in detailed experimental data. Hofstadter's same scattering technique revealed that the proton itself has internal structure — it is not a point charge but a sphere of finite size, presaging the later discovery of quarks inside protons and neutrons.
Related Concepts
- Nuclear Binding Energy — The semi-empirical mass formula (Bethe–Weizsäcker) uses nuclear radius and volume to calculate how tightly nucleons are bound, directly linking nuclear size to stability.
- The Strong Nuclear Force — Nuclear saturation — the reason density is constant and R ∝ A^(1/3) holds — arises from the short-range repulsive and attractive behaviour of the strong force between nucleons.
- Alpha Decay — Quantum tunnelling calculations for alpha decay depend on the nuclear radius as the starting point of the Coulomb barrier, connecting nuclear size directly to decay rates.
- Rutherford Scattering — The earliest estimates of nuclear size came from Rutherford's alpha-scattering experiments, which established that the nucleus is a concentrated, dense core — the conceptual foundation of nuclear radius measurements.