AQA A-Level Physics: Particles and radiation (7408)

This guide teaches section 3.2 Particles and radiation of the AQA AS and A-level Physics (7407/7408) specification, version 1.4 (AS and A-level exams June 2016 onwards). It covers 3.2.1 Particles and 3.2.2 Electromagnetic radiation and quantum phenomena, every numbered sub-section. Section 3.2 is AS and A-level content: at A-level it is examined in Paper 1 (sections 1-5 and 6.1) and assumed in Paper 2. Sections 3.9-3.13 are options – you study one, assessed in Paper 3 Section B – but this topic is compulsory for everyone.

Course links: AQA A-level Physics hub and the printable checklist. Then try the free diagnostics, the revision notes and the practice questions. For J to eV conversion see SI units and their prefixes.

What this topic covers

Spec	What you must be able to do
3.2.1.1	Describe the atom; charge and mass of p, n, e in SI and relative units; specific charge; Z, A, nuclide notation; isotopes
3.2.1.2	Strong nuclear force and its ranges; alpha and β⁻ decay equations; why the neutrino was proposed
3.2.1.3	Antiparticles; photon model, E = hf = hc/λ; annihilation and pair production
3.2.1.4	Four fundamental interactions; exchange particles; weak interaction diagrams
3.2.1.5	Hadrons (baryons, mesons), leptons, baryon and lepton number, strangeness
3.2.1.6	Quark properties and quark make-up of p, n, their antiparticles, pions and kaons; neutron decay
3.2.1.7	Quark change in β⁻ and β⁺ decay; apply conservation laws
3.2.2.1	Photoelectric effect: threshold frequency, work function, stopping potential, hf = ϕ + Ek(max)
3.2.2.2	Ionisation, excitation, fluorescent tube, the electronvolt
3.2.2.3	Line spectra and hf = E1 − E2
3.2.2.4	Wave-particle duality and λ = h/mv

This section has no required practical. Constants used below: h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, proton or neutron mass 1.67 × 10⁻²⁷ kg.

3.2.1.1 Constituents of the atom

The atom is a small nucleus of protons and neutrons (nucleons) with electrons around it.

Particle	Charge / C	Relative charge	Mass / kg	Relative mass
Proton	+1.60 × 10⁻¹⁹	+1	1.67 × 10⁻²⁷	1
Neutron	0	0	1.67 × 10⁻²⁷	1
Electron	−1.60 × 10⁻¹⁹	−1	9.11 × 10⁻³¹	about 0.0005

Proton number Z is the number of protons; nucleon number A is protons plus neutrons. Nuclide notation writes the element X with A above and Z below on the left: ᴬ_Z X. The neutron number is A − Z. Isotopes have the same Z but different A, so different numbers of neutrons. Use isotopic data to count each particle.

Specific charge = charge ÷ mass, in C kg⁻¹. Electron: 1.76 × 10¹¹ C kg⁻¹. Proton: 9.58 × 10⁷ C kg⁻¹. For a nucleus or ion, use the net charge and take the mass as A × 1.67 × 10⁻²⁷ kg (electron mass is negligible).

Worked example. Find the specific charge of the ion ²⁷₁₃Al³⁺.

charge = 3 × 1.60 × 10⁻¹⁹ = 4.80 × 10⁻¹⁹ C
mass   = 27 × 1.67 × 10⁻²⁷ = 4.509 × 10⁻²⁶ kg
specific charge = 4.80 × 10⁻¹⁹ / 4.509 × 10⁻²⁶ = 1.06 × 10⁷ C kg⁻¹

3.2.1.2 Stable and unstable nuclei

The strong nuclear force holds nucleons together against the electrostatic repulsion between protons. It is attractive up to about 3 fm, and is repulsive closer than about 0.5 fm, which stops the nucleus collapsing. Beyond about 3 fm it has no significant effect.

Alpha decay: a large nucleus emits ⁴₂α (two protons, two neutrons). A falls by 4, Z by 2. ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α

β⁻ decay: a neutron changes to a proton, emitting an electron and an electron antineutrino. A stays the same, Z rises by 1. ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁β + ν̄ₑ

Beta particles leave with a range of kinetic energies, up to a maximum. Without another particle, energy would not be conserved. The neutrino (here an antineutrino) was hypothesised to carry the missing energy.

3.2.1.3 Particles, antiparticles and photons

Every particle has an antiparticle with the same mass and rest energy and opposite charge. Learn: positron (e⁺) for the electron, antiproton for the proton, antineutron for the neutron, antineutrino for the neutrino. Rest energies: electron and positron 0.511 MeV; proton and antiproton 938.3 MeV; neutron and antineutron 939.6 MeV.

Electromagnetic radiation travels as photons, each with energy E = hf = hc/λ, where h is the Planck constant.

Annihilation: a particle meets its antiparticle; both are converted into two photons, travelling in opposite directions so that momentum is conserved. Each photon has at least the rest energy of one particle. PET scanners detect the photon pairs from positron annihilation.

Pair production: a photon becomes a particle and its antiparticle, usually near a nucleus. The minimum photon energy is twice the rest energy of the particle. You do not need E = mc² in calculations; work from rest energies in MeV.

Worked example. A slow electron and slow positron annihilate. Find the frequency and wavelength of each photon.

E = 0.511 MeV = 0.511 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.18 × 10⁻¹⁴ J
f = E/h = 8.18 × 10⁻¹⁴ / 6.63 × 10⁻³⁴ = 1.23 × 10²⁰ Hz
λ = c/f = 2.43 × 10⁻¹² m

3.2.1.4 Particle interactions

There are four fundamental interactions: gravity, electromagnetic, weak nuclear and strong nuclear (the strong interaction). Forces between particles are carried by exchange particles.

• Electromagnetic force: virtual photon (γ).
• Weak interaction: W⁺ and W⁻. You need β⁻ decay, β⁺ decay, electron capture and electron-proton collisions.

Diagrams show incoming particles at the bottom, outgoing at the top, and the exchange particle as a wavy line. Charge balances at each vertex.

Interaction	Equation	Exchange
β⁻ decay	n → p + e⁻ + ν̄ₑ	W⁻ (n → p + W⁻, then W⁻ → e⁻ + ν̄ₑ)
β⁺ decay	p → n + e⁺ + νₑ	W⁺ (p → n + W⁺, then W⁺ → e⁺ + νₑ)
Electron capture	p + e⁻ → n + νₑ	W⁺ from p to e⁻ (or W⁻ from e⁻ to p)
Electron-proton collision	e⁻ + p → n + νₑ	W⁻ from e⁻ to p

Knowledge of the gluon, Z⁰ and graviton is not tested.

3.2.1.5 Classification of particles

Hadrons feel the strong interaction. They are:

• baryons (proton, neutron) and antibaryons (antiproton, antineutron), baryon number B = +1 and −1;
• mesons (pion π, kaon K), B = 0.

Baryon number is a quantum number and is always conserved. The proton is the only stable baryon; other baryons eventually decay into it. The pion is the exchange particle of the strong nuclear force. The kaon can decay into pions.

Leptons do not feel the strong interaction: electron, muon, electron neutrino, muon neutrino and their antiparticles. Lepton numbers Lₑ and L_μ are each conserved separately. Particles have +1, antiparticles −1. The muon decays into an electron: μ⁻ → e⁻ + ν̄ₑ + ν_μ.

Strange particles (for example kaons) are produced by the strong interaction and decay by the weak interaction. Strangeness S reflects that strange particles are always created in pairs. S is conserved in strong interactions; in weak interactions it can change by 0, +1 or −1. New particle results are validated by large teams of scientists and engineers working together.

3.2.1.6 Quarks and antiquarks

Quark	Charge	B	S
u	+⅔	+⅓	0
d	−⅓	+⅓	0
s	−⅓	+⅓	−1

Antiquarks (ū, d̄, s̄) have all three signs reversed, so s̄ has S = +1.

Particle	Quarks	Particle	Quarks
p	uud	p̄	ūūd̄
n	udd	n̄	ūd̄d̄
π⁺	ud̄	π⁻	dū
π⁰	uū or dd̄	K⁺	us̄
K⁻	sū	K⁰	ds̄
K̄⁰	sd̄

Baryons are three quarks, antibaryons three antiquarks, mesons a quark-antiquark pair. Neutron decay: n → p + e⁻ + ν̄ₑ, with udd → uud.

3.2.1.7 Applications of conservation laws

In β⁻ decay a d quark changes to a u quark; in β⁺ decay a u quark changes to a d quark. In any interaction check: charge, baryon number and lepton numbers (always conserved), and strangeness (conserved in strong interactions only). Energy and momentum are also always conserved.

Worked example. Can π⁻ + p → K⁰ + n happen by the strong interaction?

Charge: −1 + 1 = 0     0 + 0 = 0      conserved
B:       0 + 1 = 1     0 + 1 = 1      conserved
S:       0 + 0 = 0    +1 + 0 = +1     NOT conserved

Strangeness changes, so it cannot be a strong interaction.

3.2.2.1 The photoelectric effect

Electromagnetic radiation above a threshold frequency f₀ releases electrons from a metal surface. Each electron absorbs one photon. The work function ϕ is the minimum energy needed to free an electron from the surface, so f₀ = ϕ/h. Below f₀ no electrons are emitted however bright the light: one photon does not carry enough energy. Increasing intensity above f₀ releases more electrons per second but does not raise Ek(max).

hf = ϕ + Ek(max). The stopping potential Vs is the potential difference that just stops the fastest photoelectrons: eVs = Ek(max). You do not need the experimental method for Vs.

Worked example. Light of wavelength 250 nm falls on a metal with ϕ = 4.30 eV. Find Ek(max), Vs and f₀.

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 250 × 10⁻⁹ = 7.96 × 10⁻¹⁹ J = 4.97 eV
Ek(max) = 4.97 − 4.30 = 0.67 eV = 1.08 × 10⁻¹⁹ J
Vs = 0.67 V
f₀ = ϕ/h = (4.30 × 1.60 × 10⁻¹⁹) / 6.63 × 10⁻³⁴ = 1.04 × 10¹⁵ Hz

3.2.2.2 Collisions of electrons with atoms

One electronvolt is the energy transferred when an electron moves through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J. Divide joules by 1.60 × 10⁻¹⁹ to get eV.

Excitation: an electron in an atom moves to a higher energy level, absorbing exactly the energy difference. Ionisation: an electron gains enough energy to leave the atom. A free electron colliding with an atom can give any part of its kinetic energy, as long as it matches a level gap; a photon must be absorbed whole.

Fluorescent tube: a high voltage accelerates free electrons through mercury vapour. They collide with mercury atoms, exciting (and some ionising) them. The atoms de-excite and emit ultraviolet photons. The phosphor coating absorbs the UV, its atoms are excited, and they de-excite in smaller steps, emitting visible light.

3.2.2.3 Energy levels and photon emission

Line spectra, for example from atomic hydrogen, show only certain wavelengths. This is evidence that electrons occupy discrete energy levels. A transition down emits a photon of energy hf = E1 − E2. Levels may be given in J or eV. Hydrogen’s lowest levels: −13.6, −3.40, −1.51, −0.85 eV.

Worked example. Find the wavelength emitted when hydrogen drops from −1.51 eV to −3.40 eV.

ΔE = 3.40 − 1.51 = 1.89 eV = 1.89 × 1.60 × 10⁻¹⁹ = 3.024 × 10⁻¹⁹ J
λ = hc/ΔE = 1.989 × 10⁻²⁵ / 3.024 × 10⁻¹⁹ = 6.58 × 10⁻⁷ m (658 nm, red)

3.2.2.4 Wave-particle duality

The photoelectric effect shows electromagnetic waves have a particle nature. Electron diffraction shows particles have wave properties. The de Broglie wavelength is λ = h/mv, where mv is the momentum. Diffraction is greatest when λ is similar to the gap or atomic spacing. Increasing momentum (for example a higher accelerating voltage) shortens λ, so there is less diffraction and the rings get smaller. Our model of matter has changed as such evidence appeared.

Worked example. An electron is accelerated from rest through 150 V. Find its de Broglie wavelength.

Ek = 150 eV = 2.40 × 10⁻¹⁷ J
v = √(2Ek/m) = √(2 × 2.40 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 7.26 × 10⁶ m s⁻¹
λ = h/mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 7.26 × 10⁶) = 1.00 × 10⁻¹⁰ m

Common errors

• Forgetting the antineutrino in β⁻ decay, or writing a neutrino.
• Applying strangeness conservation to a weak decay.
• Giving one photon in annihilation: there are two, to conserve momentum.
• Leaving energy in eV inside hc/λ without converting to joules.

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.2 Particles and radiation.