AQA A-level Physics (7408) · 3.2 Particles and radiation
Mini-Lesson
Particles and radiation
This mini-lesson covers AQA 3.2 — Particles and radiation: the quark and lepton building blocks, baryons and mesons, the conservation laws that decide which reactions can happen, annihilation and pair production, and the quantum evidence — the photoelectric effect, energy levels, and de Broglie matter waves.
Section 3.2 is compulsory and appears on Paper 1.
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.2.1 · constituents of the atom
Nucleons, isotopes and specific charge
An atom is a nucleus of protons and neutrons (together: nucleons) surrounded by electrons. We write a nuclide as AZX:
Z = proton number (atomic number) — defines the element.
A = nucleon number (mass number) = protons + neutrons.
Isotopes have the same Z but different A (same chemistry, different nuclear stability).
specific charge = charge ÷ mass (C kg−1)proton: 1.60×10−19 ÷ 1.67×10−27 = 9.58×107 C kg−1 · electron: 1.76×1011 C kg−1
Order-of-magnitude check: the electron has the largest specific charge of any particle you meet — about 1800× that of the proton, because it carries the same charge with a mass 1836 times smaller.
3.2.1 · quarks & leptons
Quarks, leptons and the particle zoo
AQA needs three quarks and their antiquarks (u, d, s) plus the leptons.
u: Q = +⅔e d: Q = −⅓e s: Q = −⅓eevery quark has baryon number B = +⅓; the strange quark alone has strangeness S = −1
Hadrons feel the strong interaction and are made of quarks. Two families:
Mesons = quark + antiquark (B = 0). π+ = u̅d (u and anti-d); K+ = u̅s (u and anti-s).
Leptons are fundamental — not made of quarks: electron e−, muon µ− and their neutrinos, plus antiparticles. Lepton number L = +1 for particles, −1 for antiparticles.
Proton stability: the proton is the only stable baryon — every other baryon eventually decays into a proton. A free neutron is unstable (half-life about 10 minutes) and undergoes β− decay.
Sort it
Baryon, meson or lepton?
Tap a particle, then tap its family.
🧲 Baryon (3 quarks)
🎯 Meson (quark + antiquark)
⚡ Lepton (fundamental)
Match it
Match the quark structure to its particle
Tap a card on the left, then its partner on the right.
Statement
Answer
3.2.1 · conservation laws
Conservation laws and the four interactions
A reaction can only happen if every conserved quantity balances:
Charge (Q) — always conserved.
Baryon number (B) — always conserved.
Lepton number — conserved separately for electron-type and muon-type leptons.
Energy and momentum — always conserved.
Strangeness (S) — conserved in the strong interaction, but can change by 0 or ±1 in the weak interaction.
β−: n → p + e− + ν̅eβ+: p → n + e+ + νe (inside a nucleus) — both are weak interactions, exchange particles W− / W+
Why the antineutrino? In β− decay the electron has lepton number +1, so the other product must have L = −1 — an electron antineutrino. Its existence was postulated precisely because energy and lepton number would otherwise not balance.
Quick check
Think it through
?Which conservation law is violated in a proposed strong interaction if the strangeness changes from 0 to −1?
Quick check
Think it through
?Which exchange particle mediates β− decay?
3.2.2 · antimatter
Antiparticles, annihilation and pair production
Every particle has an antiparticle with the same mass and rest energy but opposite charge, baryon number, lepton number and strangeness.
E = mc²rest energy of an electron (or positron) = 0.511 MeV = 8.19 × 10−14 J
Annihilation: a particle meets its antiparticle and both vanish, producing two photons (two, so that momentum is conserved). Minimum energy of each photon = the rest energy of one particle.
Pair production: a single photon of energy ≥ 2 × rest energy creates a particle–antiparticle pair. It must happen near a nucleus so that momentum is conserved.
Worked example — minimum photon energy for pair production
To create an electron–positron pair: Emin = 2 × 0.511 MeV = 1.02 MeV.
In joules: 1.02 × 106 × 1.60 × 10−19 = 1.64 × 10−13 J.
Real-world link: a PET scanner detects the two back-to-back 0.511 MeV gamma photons made when a positron from a tracer annihilates with an electron in the patient.
Calculate
Your turn — calculation 1
1An electron and a positron, both effectively at rest, annihilate and produce two identical photons. The rest energy of an electron is 0.511 MeV. Calculate the frequency of each photon. Give the number in front of ×1020 Hz. (h = 6.63×10−34 J s, e = 1.60×10−19 C)
× 10²⁰ Hz
Hint: Each photon takes 0.511 MeV = 0.511×106 × 1.60×10−19 = 8.18×10−14 J. Then f = E ÷ h.
3.2.3 · photoelectric effect
The photoelectric effect
Shine light on a metal and electrons can be emitted — but only if the frequency is high enough. This is impossible to explain with waves, and is the strongest evidence for the photon model.
hf = φ + Ek(max)φ = work function (minimum energy to free an electron) · threshold frequency f0 = φ ÷ h
Below f0, no electrons are emitted — however intense the light and however long you wait.
Above f0, emission is instantaneous. Increasing intensity increases the number of electrons per second, but not their maximum kinetic energy.
Increasing frequency increases Ek(max) — one photon transfers all its energy to one electron.
Worked example
Light of wavelength 300 nm hits a metal of work function 2.00 eV.
Wave model fails: waves would let any frequency work if you waited long enough for energy to build up, and would make Ek depend on intensity. Neither is observed. Energy must arrive in discrete photons of energy hf.
Calculate
Your turn — calculation 2
2A metal has a work function of 3.20 × 10−19 J. Calculate its threshold frequency. Give the number in front of ×1014 Hz. (h = 6.63×10−34 J s)
× 10¹⁴ Hz
Hint: f0 = φ ÷ h = 3.20×10−19 ÷ 6.63×10−34.
Quick check
Think it through
?A metal is illuminated with light above its threshold frequency. The intensity of the light is then doubled, with the frequency unchanged. What happens?
3.2.3 · energy levels
Energy levels, excitation and photon emission
Electrons in an atom occupy discrete energy levels (measured from the ground state upwards, all negative because the electron is bound).
hf = E1 − E2a photon is emitted when an electron drops from a higher level E1 to a lower level E2
Excitation: an electron absorbs exactly the right energy (from a photon, or a colliding electron) and jumps up a level.
Ionisation: the electron gains enough energy to leave the atom completely (n = ∞, E = 0).
Discrete levels → discrete photon energies → a line spectrum that fingerprints the element.
Worked example — hydrogen
An electron falls from −3.40 eV to −13.6 eV: ΔE = 10.2 eV = 10.2 × 1.60×10−19 = 1.63×10−18 J.
Fluorescent tube: mercury atoms are excited by electron collisions, emit UV photons on de-excitation, and the phosphor coating absorbs the UV and re-emits several lower-energy visible photons.
Calculate
Your turn — calculation 3
3In a hydrogen atom an electron falls from the −3.40 eV level to the −13.6 eV ground state. Calculate the wavelength of the emitted photon, in nm. (h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)
nm
Hint: ΔE = 10.2 eV = 1.63×10−18 J; then λ = hc ÷ ΔE, and convert m → nm.
3.2.4 · wave-particle duality
Wave–particle duality and de Broglie
The photoelectric effect shows light behaves as particles; diffraction and interference show it behaves as waves. De Broglie proposed that matter does the same:
λ = h ÷ p = h ÷ mvevery moving particle has a wavelength — big momentum means a tiny wavelength
Electron diffraction through a thin graphite film gives concentric rings — a wave effect — and confirms de Broglie.
Speed up the electrons (higher accelerating pd) → larger momentum → smaller λ → the diffraction rings shrink.
We never see people diffract: with m ≈ 70 kg and v ≈ 1 m s−1, λ ≈ 10−35 m — unmeasurably small.
Why it matters: the electron microscope works because fast electrons have wavelengths far shorter than visible light, so they resolve far finer detail.
Calculate
Your turn — calculation 4
4An electron has momentum 3.30 × 10−24 kg m s−1. Calculate its de Broglie wavelength in nm. (h = 6.63×10−34 J s)
nm
Hint: λ = h ÷ p = 6.63×10−34 ÷ 3.30×10−24 = 2.0×10−10 m. Now convert to nm.
Calculate
Your turn — calculation 5
5Light of wavelength 300 nm falls on a metal with work function 2.00 eV. Calculate the maximum kinetic energy of the emitted photoelectrons, in eV. (h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C)
eV
Hint: Photon energy = hc/λ = 6.63×10−19 J = 4.14 eV. Then Ek(max) = 4.14 − 2.00.
Quick check
Think it through
?In an electron diffraction tube, the accelerating potential difference is increased. What happens to the diffraction rings on the screen?
Recap
The big ideas to know
Building blocks: quarks (u, d, s) make hadrons · leptons (e, µ, neutrinos) are fundamental