OCR A Level Physics A: The nuclear atom, nuclear size and fundamental particles (H556)

This guide covers sections 6.4.1 (The nuclear atom) and 6.4.2 (Fundamental particles) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. It is full A level content in Module 6, which the assessment overview places in components 02 and 03. Radioactive decay, half-life and activity, and binding energy, fission and fusion, are separate topics and are not repeated here.

Course pages: OCR A Level Physics hub | checklist | free diagnostics. After this guide, use the revision notes and the practice questions.

Values used: r₀ = 1.2 fm (1 fm = 10⁻¹⁵ m), atomic mass unit u = 1.66 × 10⁻²⁷ kg, e = 1.60 × 10⁻¹⁹ C. In an exam, use the values the question supplies.

What this unit covers

Spec	You must be able to
6.4.1(a)	Describe alpha-particle scattering and the evidence for a small charged nucleus
6.4.1(b)	Describe the simple nuclear model: protons, neutrons, electrons
6.4.1(c)	Compare the sizes of atom and nucleus
6.4.1(d)	Use proton number, nucleon number, isotopes and the notation ᴬ_Z X
6.4.1(e)	Describe the strong nuclear force, its range and where it is attractive or repulsive
6.4.1(f)	Use R = r₀A^(1/3)
6.4.1(g)	Compare mean densities of atoms and nuclei
6.4.2(a)-(b)	Particles and antiparticles; same mass, opposite charge where charged
6.4.2(c)-(d)	Classify hadrons and leptons by the forces they feel
6.4.2(e)-(g)	Quark model with u, d, s and antiquarks; proton uud, neutron udd; quark charges
6.4.2(h)-(k)	Beta-minus and beta-plus decay as quark transformations; balance them by charge
6.4.2(l)	Describe the decay of other particles in the quark model

6.4.1(a)-(b): alpha scattering and the nuclear model

In the alpha-particle scattering experiment, a narrow beam of alpha particles is directed at a thin metal foil (gold is the classic choice) in an evacuated chamber. A detector, such as a screen that flashes where a particle lands, is moved round to count how many particles emerge at each angle. The vacuum matters because alpha particles are stopped by a few centimetres of air.

Observation	Conclusion
Most alpha particles go straight through or are barely deflected	An atom is mostly empty space
A small fraction are deflected through large angles	A concentrated positive charge repels the alpha particles
A very small number rebound by more than 90°	The charge is concentrated in a tiny region and carries most of the atom’s mass

An alpha particle is positive, so only a small, positively charged, massive object can turn it round. A spread-out positive charge, as in the earlier plum pudding picture, would give only small deflections. The experiment therefore provided evidence for a small, charged nucleus, and shows how a model is changed when new evidence contradicts it.

Simple nuclear model. The atom has a tiny central nucleus containing protons (charge +e) and neutrons (no charge). Electrons (charge −e) are found outside it. A neutral atom contains the same number of electrons as protons. Protons and neutrons are called nucleons.

6.4.1(c): relative sizes

Atomic radii are of the order of 10⁻¹⁰ m. A nucleus has a radius of the order of 10⁻¹⁵ to 10⁻¹⁴ m. So the atom is roughly 10⁴ to 10⁵ times wider than its nucleus.

Worked example. A calcium-40 atom has a radius of 1.8 × 10⁻¹⁰ m. Its nuclear radius is R = 1.2 × 10⁻¹⁵ × 40^(1/3) = 4.1 × 10⁻¹⁵ m. The ratio of radii is 1.8 × 10⁻¹⁰ / 4.1 × 10⁻¹⁵ = 4.4 × 10⁴. Volume goes as radius cubed, so the nucleus fills about 1 / (4.4 × 10⁴)³, roughly 10⁻¹⁴, of the atom’s volume.

6.4.1(d): nuclide notation and isotopes

A nuclide is written ᴬ_Z X, where X is the chemical symbol.

• Z is the proton number (number of protons, which fixes the element).
• A is the nucleon number (protons plus neutrons).
• Neutron number N = A − Z.

Isotopes are nuclei of the same element, so the same proton number, with different numbers of neutrons, so different nucleon numbers. Chemical behaviour is set by the electrons, so isotopes behave chemically alike.

Worked example. Chlorine has two common isotopes, ³⁵₁₇Cl and ³⁷₁₇Cl. Both have 17 protons. Neutrons: 35 − 17 = 18 and 37 − 17 = 20. A neutral atom of either has 17 electrons. A lead nucleus ²⁰⁷₈₂Pb has 82 protons and 207 − 82 = 125 neutrons.

6.4.1(e): the strong nuclear force

Protons repel each other electrostatically, yet nuclei exist. The force holding nucleons together is the strong nuclear force. Its features for this specification:

• It acts between nucleons. From a separation of about 0.5 fm out to about 3 fm it is attractive.
• It is repulsive for separations below about 0.5 fm. This stops the nucleus collapsing to a point.
• It is short range: beyond a few femtometres it is negligible, so a nucleon feels only its near neighbours.
• At nuclear separations the attraction outweighs the electrostatic repulsion between protons.

Describe the force against separation graph in words: zero (or tiny) at large separation, a strong attractive region out to about 3 fm, and a steep repulsion inside about 0.5 fm, crossing zero at about 0.5 fm. 1 fm = 10⁻¹⁵ m.

6.4.1(f)-(g): nuclear radius and density

The radius of a nucleus is R = r₀A^(1/3), where r₀ is a constant (about 1.2 fm) and A is the nucleon number. Since volume is (4/3)πR³, volume is proportional to A. Each nucleon takes up about the same volume.

Worked example (radius). Compare helium-4 and lead-208. R(He) = 1.2 × 4^(1/3) = 1.9 fm. R(Pb) = 1.2 × 208^(1/3) = 7.1 fm. A increases 52 times, but the radius only 3.7 times (52^(1/3)). If you plot R against A^(1/3), you get a straight line through the origin with gradient r₀.

Nuclear density. The mass of a nucleus is close to A × u. Then

ρ = mass / volume = A × u / ((4/3)π r₀³A) = 3u / (4π r₀³)

A cancels, so every nucleus has the same density.

Worked example (densities). Aluminium-27 (atom radius taken as 1.4 × 10⁻¹⁰ m).

• Nucleus: R = 1.2 × 27^(1/3) = 3.6 fm, so V = (4/3)π(3.6 × 10⁻¹⁵)³ = 1.95 × 10⁻⁴³ m³. Mass = 27 × 1.66 × 10⁻²⁷ = 4.48 × 10⁻²⁶ kg. ρ = 4.48 × 10⁻²⁶ / 1.95 × 10⁻⁴³ = 2.3 × 10¹⁷ kg m⁻³.
• Atom: V = (4/3)π(1.4 × 10⁻¹⁰)³ = 1.15 × 10⁻²⁹ m³. Electrons contribute almost nothing to the mass, so use the same mass: ρ = 4.48 × 10⁻²⁶ / 1.15 × 10⁻²⁹ = 3.9 × 10³ kg m⁻³.
• The nuclear density is about 6 × 10¹³ times the mean atomic density.

The atom’s mean density is similar to everyday solids and liquids. Nuclear matter is vastly denser: one cubic centimetre of it would have a mass of about 2 × 10¹¹ kg.

6.4.2(a)-(b): particles and antiparticles

Every particle has a corresponding antiparticle with the same mass. A charged particle’s antiparticle has the opposite charge.

Particle	Antiparticle	Notes
electron e⁻	positron e⁺	Same mass; charges −e and +e
proton p	antiproton p̄	Same mass; charges +e and −e
neutron n	antineutron n̄	Both neutral, same mass
neutrino ν	antineutrino ν̄	Both neutral

A bar over a symbol marks an antiparticle. Being neutral does not mean a particle is its own antiparticle; the neutron and antineutron are different particles.

6.4.2(c)-(d): hadrons and leptons

Class	Examples	Forces felt
Hadrons	proton, neutron	Strong nuclear force and weak nuclear force
Leptons	electron, neutrino	Weak nuclear force, but not the strong nuclear force

To classify a particle from a question, ask whether it feels the strong force. Charged particles also feel the electromagnetic force. The weak force is the one responsible for beta decay.

6.4.2(e)-(g): the quark model

Hadrons are made of quarks. This specification uses three: up (u), down (d) and strange (s), and their antiquarks (ū, d̄, s̄).

Quark	Charge / e	Antiquark	Charge / e
up u	+2/3	ū	−2/3
down d	−1/3	d̄	+1/3
strange s	−1/3	s̄	+1/3

• Proton = uud: 2/3 + 2/3 − 1/3 = +1.
• Neutron = udd: 2/3 − 1/3 − 1/3 = 0.
• Antiproton = ūūd̄: −1. Antineutron = ūd̄d̄: 0.

Worked example. The positive kaon is made of an up quark and a strange antiquark, u s̄. Charge = +2/3 + 1/3 = +1 (that is, +e). A hadron with composition d s̄ would have −1/3 + 1/3 = 0. Any hadron’s charge must be a whole number of e, which is a useful check on your arithmetic.

6.4.2(h)-(k): beta decay in the quark model

In beta-minus decay a neutron in the nucleus becomes a proton, emitting an electron and an antineutrino. At quark level a down quark changes to an up quark:

d → u + e⁻ + ν̄

Check charge: left −1/3; right +2/3 − 1 + 0 = −1/3. Balanced.

In beta-plus decay a proton becomes a neutron, emitting a positron and a neutrino. An up quark changes to a down quark:

u → d + e⁺ + ν

Check charge: left +2/3; right −1/3 + 1 + 0 = +2/3. Balanced.

Decay	Nucleon-level	Quark-level	Nucleus Z	Nucleus A
β⁻	n → p + e⁻ + ν̄	d → u + e⁻ + ν̄	+1	unchanged
β⁺	p → n + e⁺ + ν	u → d + e⁺ + ν	−1	unchanged

Only one quark changes; the other two are spectators. Both decays are due to the weak nuclear force. The neutrino is a lepton: a very light neutral particle.

Worked example. Show that neutron decay is balanced in the quark model. Neutron udd → proton uud. One d becomes u. The d quark emits e⁻ and ν̄, so udd → u + u + d + e⁻ + ν̄. Charge on the left: 0. Charge on the right: +1 (proton) − 1 (electron) = 0. Balanced.

6.4.2(l): decay of other particles

The same method applies to any particle whose quark content you are given: identify which quark changes, then balance charge.

Worked example. One decay mode of the lambda particle Λ (uds) produces a proton (uud), an electron and an antineutrino. The strange quark changes to an up quark: s → u + e⁻ + ν̄. Charge: −1/3 on the left; +2/3 − 1 = −1/3 on the right. The u and d quarks are spectators, so uds → uud + e⁻ + ν̄. Total charge is 0 before and +1 − 1 = 0 after.

Common errors

• Writing A and Z the wrong way round. A (top) is the larger number.
• Saying isotopes have different numbers of protons or electrons.
• Giving the strong force as attractive at all separations. It is repulsive below about 0.5 fm.
• Forgetting the cube root, or using R = r₀A.
• Stating that nuclear density depends on A.
• Putting the neutrino in beta-minus decay instead of the antineutrino.
• Quoting charge in coulombs when the question asks for a multiple of e, or the reverse.
• Forgetting that an antiquark has the opposite charge to its quark.

Next steps

Test yourself with the practice questions, then condense with the revision notes. Tick off 6.4.1 and 6.4.2 on the checklist.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 6.4.1 The nuclear atom and 6.4.2 Fundamental particles.