Estimating Nuclear Radius Revision Notes | A Level AQA Physics
Nuclear radius is estimated from the closest approach of alpha particles or from the diffraction of high-energy electrons by a nucleus. 10 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Estimating Nuclear Radius is part of Radioactivity (A-level only) in A Level AQA Physics.
6 min
Reading time
8
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A Level AQA
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What this lesson covers
Nuclear radius is estimated from the closest approach of alpha particles or from the diffraction of high-energy electrons by a nucleus. 10 questions answered.
Explain how the closest approach of alpha particles to a nucleus allows an estimate of nuclear radius, using the Coulomb equation
Describe how electron diffraction experiments determine nuclear radius and interpret the resulting intensity-angle graph
Calculate the closest approach distance using conservation of energy and the Coulomb equation
Compare the two methods for estimating nuclear radius and evaluate their assumptions and limitations
Key ideas to keep in view
Use this lesson to stay inside Radioactivity (A-level only) while connecting the detail back to the wider Nuclear physics (A-level only) topic.
Closest approach distance: The minimum distance between an alpha particle and a nucleus when all the alpha particle's kinetic energy has been converted to electrostatic potential energy. Used to estimate an upper bound for nuclear radius.
Electron diffraction: A technique in which high-energy electrons are fired at a nucleus; the resulting diffraction pattern (intensity vs angle) reveals the nuclear radius from the position of the first diffraction minimum.
Nuclear radius formula: The empirical relationship $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}$ m and $A$ is the nucleon number, showing that nuclear radius scales with the cube root of mass number.
Coulomb potential energy: The electrostatic potential energy stored between two charged particles separated by distance $r$, given by $E = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r}$. Central to the closest approach calculation.
Revision notes
Read the core explanations from Estimating Nuclear Radius before testing yourself from memory.
1. Why Measuring the Nucleus Is So Hard
The nucleus is roughly $10^{-15}$ m across — about 100,000 times smaller than the atom itself. No optical microscope can resolve it. So physicists had to get creative: fire particles at nuclei and read the clues left behind.
Two distinct experimental approaches give us nuclear radius estimates: the closest approach of alpha particles (using electrostatics) and electron diffraction (using wave behaviour). Each method has its own assumptions and its own level of precision — and the AQA exam tests both.
2. Method 1 — Closest Approach of Alpha Particles
Picture an alpha particle ($\alpha$) fired directly at a gold nucleus. As it approaches, the repulsive Coulomb force slows it down. At some minimum distance $d$, the particle momentarily stops before being repelled back. At that instant, all of its initial kinetic energy has been converted into electrostatic potential energy.
This distance $d$ is called the closest approach distance, and it gives an upper bound for the nuclear radius — the nucleus must be at least as small as $d$, because the alpha particle never actually touches it in this model.
The key equation comes from conservation of energy. At the point of closest approach, kinetic energy equals electrostatic potential energy:
Rearranging for $d$ gives $d = \frac{Q_\alpha Q_{\text{nucleus}}}{4\pi\varepsilon_0 E_k}$. The larger the initial kinetic energy, the closer the alpha particle gets — so higher-energy alpha particles give a tighter (better) upper bound on nuclear radius.
Worked example: Calculating Closest Approach Distance for Gold
An alpha particle with kinetic energy $5.0 \times 10^{-13}$ J is fired directly at a gold-197 nucleus ($Z = 79$). Calculate the closest approach distance $d$. (Use $\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9$ N m$^2$ C$^{-2}$, $e = 1.6 \times 10^{-19}$ C)
$Q_\alpha = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$ C
$Q_{\text{Au}} = 79e = 79 \times 1.6 \times 10^{-19} = 1.264 \times 10^{-17}$ C — The alpha particle has charge +2e (2 protons). Gold has Z = 79, so its nuclear charge is +79e. [M1]
$d = \frac{Q_\alpha \times Q_{\text{Au}}}{4\pi\varepsilon_0 \times E_k}$ — At closest approach, all kinetic energy equals electrostatic potential energy. Rearrange for d. [M1]
$Q_\alpha \times Q_{\text{Au}} = 3.2 \times 10^{-19} \times 1.264 \times 10^{-17}$
$= 3.2 \times 1.264 \times 10^{-19-17}$
$= 4.0448 \times 10^{-36}$ C$^2$ — Multiply the two charges together, keeping track of powers of ten separately.
$8.99 \times 10^9 \times 4.0448 \times 10^{-36}$
$= 8.99 \times 4.0448 \times 10^{9-36}$
$= 36.36 \times 10^{-27}$
$= 3.636 \times 10^{-26}$ J m — This gives the numerator of the full expression $\frac{Q_\alpha Q_{\text{Au}}}{4\pi\varepsilon_0}$.
$d \approx 7.3 \times 10^{-14}$ m — This is an upper bound for the gold nuclear radius. The actual radius (from electron diffraction) is closer to $7 \times 10^{-15}$ m — about ten times smaller, confirming this method overestimates.
Answer: $d \approx 7.3 \times 10^{-14}$ m (upper bound for nuclear radius of gold)
Notice that the closest approach distance is about ten times larger than the true nuclear radius. This is expected: the alpha particle is repelled before it reaches the nuclear surface, so $d$ is always an overestimate. The method gives an order-of-magnitude estimate, not a precise measurement.
The closest approach method gives us a useful first estimate, but to pin down nuclear radius more precisely, physicists turned to a completely different tool: the wave nature of electrons.
3. Method 2 — Electron Diffraction
When high-energy electrons are fired at a thin target, they behave as waves with a de Broglie wavelength comparable to nuclear dimensions ($\sim 10^{-15}$ m). The nucleus acts as a diffracting object, and the electrons scatter at various angles, producing a characteristic diffraction pattern.
The pattern is recorded as a graph of intensity against angle. You will see a large central maximum at $\theta = 0°$, followed by a sharp minimum, then a much weaker secondary maximum, and so on — very similar in shape to single-slit diffraction of light, but on a nuclear scale.
The angle of the first diffraction minimum ($\theta_{\text{min}}$) is the key measurement. A smaller nucleus diffracts electrons through a larger angle — the same relationship as in single-slit diffraction, where a narrower slit spreads the pattern more widely.
From the angle of the first minimum and the known de Broglie wavelength of the electrons, the nuclear radius $R$ can be calculated. This method gives a direct, precise measurement of nuclear radius — far more reliable than the closest approach estimate.
Electron diffraction experiments across many different nuclei reveal a consistent pattern: nuclear radius scales with the cube root of the nucleon number $A$. This leads directly to the nuclear radius formula.
4. The Nuclear Radius Formula and What It Tells Us
The $A^{1/3}$ dependence is not arbitrary. If nuclear density is constant (the same for all nuclei), then volume is proportional to $A$, and since volume $\propto R^3$, it follows that $R \propto A^{1/3}$. Electron diffraction confirms this: nuclear density is approximately constant across all nuclei.
This is a profound result. It means nucleons pack together at a fixed density — roughly $10^{17}$ kg m$^{-3}$ — regardless of which element you are looking at. Adding more nucleons increases the volume proportionally, not the density.
Worked example: Using the Nuclear Radius Formula
Calculate the radius of a carbon-12 nucleus ($A = 12$). Use $R_0 = 1.2 \times 10^{-15}$ m.
$R = R_0 A^{1/3}$ — State the formula first — this earns the method mark. [M1]
$R = 1.2 \times 10^{-15} \times 12^{1/3}$ — A = 12 for carbon-12. [M1]
$R \approx 2.7 \times 10^{-15}$ m (to 2 s.f.) — Always include units and round to match the precision of the data given.
Answer: $R \approx 2.7 \times 10^{-15}$ m
5. Comparing the Two Methods
Both methods probe nuclear size, but they differ fundamentally in what they measure and how reliable the result is. The table below sets out the key comparison — the kind of analysis an examiner might ask you to produce.
Feature
Closest Approach (Alpha Scattering)
Electron Diffraction
What is measured
Distance at which alpha KE = electrostatic PE
Angle of first diffraction minimum
Physics used
Conservation of energy + Coulomb's law
Wave diffraction (de Broglie wavelength)
Result type
Upper bound (overestimate)
Direct, precise measurement
Why it overestimates / is precise
Alpha stops before reaching nuclear surface
Diffraction minimum directly encodes nuclear size
Assumption
No nuclear force; nucleus is a point charge
Nucleus acts as a hard sphere diffracting object
Typical accuracy
Order of magnitude only
Consistent with $R = R_0 A^{1/3}$
A key point for evaluation questions: the closest approach method assumes the alpha particle does not penetrate the nucleus and that only the Coulomb force acts. The primary reason $d$ overestimates the nuclear radius is that the alpha particle is brought to rest by electrostatic repulsion before it reaches the nuclear surface — so $d$ is always greater than the true radius. Note that the strong nuclear force (which is attractive at short range) is ignored in this model; if it were included, it would pull the alpha particle slightly closer, meaning the Coulomb-only model actually gives a value of $d$ that is marginally larger than it would otherwise be. The dominant limitation, however, remains that the alpha never reaches the nuclear surface.
6. Exam Technique and Common Mistakes
AQA questions on this topic tend to fall into three categories: calculation of closest approach distance, description or interpretation of the electron diffraction graph, and evaluation of one or both methods. Each has its own mark-winning habits and traps.
Bringing both methods together: the closest approach experiment (historically linked to Rutherford-style scattering) established that the nucleus is tiny and positively charged, while electron diffraction quantified its radius precisely and revealed the constant-density structure of nuclear matter. Both spec requirements connect to the same underlying story — probing the nucleus with projectiles and reading the physics from what bounces back.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
How does electron diffraction measure nuclear radius?
High-energy electrons have a de Broglie wavelength comparable to nuclear dimensions, about 10⁻¹⁵ m, so a nucleus diffracts them. A graph of intensity against angle shows a central maximum then a first minimum; a smaller nucleus gives a larger angle for the first minimum, so that angle and the electron wavelength give the nuclear radius precisely.
How do you estimate nuclear radius using closest approach?
Nuclear radius is estimated by closest approach: an alpha particle fired directly at a nucleus is slowed by Coulomb repulsion, and at the closest approach distance d all its kinetic energy has become electrostatic potential energy. Rearranging that energy equation gives d, an upper bound for the nuclear radius, because the alpha particle never reaches the nuclear surface.
How do you calculate the distance of closest approach?
The distance of closest approach is d = (charge of alpha × charge of nucleus) ÷ (4πε₀ × kinetic energy). For a 5.0 × 10⁻¹³ J alpha particle, charge 3.2 × 10⁻¹⁹ C, aimed at a gold nucleus of charge 1.264 × 10⁻¹⁷ C, d = 8.99 × 10⁹ × 4.0448 × 10⁻³⁶ ÷ 5.0 × 10⁻¹³ ≈ 7.3 × 10⁻¹⁴ m.
Why does the closest approach method overestimate nuclear radius?
The closest approach method overestimates nuclear radius because the alpha particle is brought to rest by electrostatic repulsion before it reaches the nuclear surface, so d is always greater than the true radius. It also treats the nucleus as a point charge and ignores the strong nuclear force, so it gives only an order-of-magnitude estimate.
What is the difference between electron diffraction and alpha scattering for nuclear radius?
Alpha scattering uses conservation of energy and Coulomb's law to give an upper bound on nuclear radius, only an order-of-magnitude estimate. Electron diffraction uses the wave behaviour of electrons and the angle of the first diffraction minimum to give a direct, precise measurement, consistent with R = R₀∛A.
What does an electron diffraction graph for a nucleus look like?
An electron diffraction graph of intensity against angle for a nucleus shows a large central maximum at θ = 0°, a sharp minimum, then a much weaker secondary maximum, very similar in shape to single-slit diffraction of light. The angle of the first diffraction minimum is the key measurement.
How does nuclear size affect the electron diffraction angle?
A smaller nucleus diffracts electrons through a larger angle, the same relationship as in single-slit diffraction, where a narrower slit spreads the pattern more widely. From the angle of the first minimum and the known de Broglie wavelength of the electrons, the nuclear radius can be calculated.
Why are high-energy electrons used to measure nuclear radius?
High-energy electrons are used because their de Broglie wavelength is then comparable to nuclear dimensions, around 10⁻¹⁵ m, so the nucleus acts as a diffracting object. The electrons scatter at various angles, producing a characteristic diffraction pattern.
How does alpha particle energy affect the closest approach distance?
Higher-energy alpha particles get closer to the nucleus, because at closest approach all the kinetic energy has become electrostatic potential energy, which increases as the distance falls. A larger kinetic energy therefore gives a smaller closest approach distance, and a tighter, better upper bound on nuclear radius.
What assumptions does the closest approach method make?
The closest approach method assumes only the Coulomb force acts, ignoring the strong nuclear force, and treats the nucleus as a point charge. Its main limitation is that the alpha particle is brought to rest by electrostatic repulsion before reaching the nuclear surface, so d is always greater than the true radius.