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Conservation laws in particle interactionsAQA A-Level Physics: Revision notes

Section 1

Quark character in beta decay

In beta-minus decay a neutron changes into a proton, so a down quark changes into an up quark:

n → p + e⁻ + νˉe\bar{\nu}_e (udd → uud)

In beta-plus decay a proton changes into a neutron, so an up quark changes into a down quark:

p → n + e⁺ + νe\nu_e (uud → udd)

Quark charges are +⅔ (up) and −⅓ (down). The change in quark charge is carried away by the emitted electron or positron.

Key termsbeta-minus decaybeta-plus decayquark character
Exam tip

Beta-minus emits an electron and an antineutrino. Beta-plus emits a positron and a neutrino. Check lepton number to remember which.

Section 2

Conserved quantities

In every interaction in these questions the following are conserved:

  • Charge: the total charge before equals the total after
  • Baryon number: baryons +1, antibaryons −1, mesons and leptons 0
  • Lepton number: leptons +1, antileptons −1. The electron family (e⁻, νe\nu_e) and the muon family (μ⁻, νμ\nu_\mu) are counted separately
  • Strangeness: strange quark −1, anti-strange +1. The data you need for other particles is given in the question

The energy and momentum of the whole system are also conserved.

Key termschargebaryon numberlepton numberstrangeness

Section 3

Applying the conservation laws

To test whether an interaction can occur, add up each quantity on each side and compare.

Worked example: π⁻ + p → K⁰ + Λ⁰

  • Charge: (−1) + (+1) = 0 before; 0 + 0 = 0 after ✓
  • Baryon number: 0 + 1 = 1 before; 0 + 1 = 1 after ✓
  • Strangeness: 0 + 0 = 0 before; (+1) + (−1) = 0 after ✓

All quantities balance, so the interaction is allowed. If any one quantity does not balance, the interaction cannot occur.

Key termsallowed interaction
Common mistake

Do not just say a quantity is conserved. Write the totals before and after, with numbers, for every quantity you check.

Section 4

Finding an unknown particle

If one particle in an interaction is unknown, use the conservation laws to work out its properties.

Example: π⁻ → μ⁻ + ?

  • Charge: −1 → −1 + ?, so the unknown is neutral
  • Muon lepton number: 0 → +1 + ?, so the unknown has −1

The unknown particle is a muon antineutrino.

In beta-minus decay the electron has electron lepton number +1, so the second particle must be an electron antineutrino to keep the total at 0.

Key termsunknown particle

Section 5

Energy and momentum

Energy is conserved in every interaction: the total energy, including rest energy and kinetic energy, is the same before and after. A decay can only happen if the products have no more rest energy than the particle that decays. A collision that creates new particles needs enough kinetic energy to supply their rest energy.

Momentum is also conserved. In beta decay the electron is emitted with a range of kinetic energies, so a third particle (the antineutrino in beta-minus decay) must share the energy and momentum.

Key termsconservation of energyconservation of momentum

Must Know

  • β⁻: d → u + e⁻ + νˉe\bar{\nu}_e; β⁺: u → d + e⁺ + νe\nu_e
  • Charge, baryon number, lepton number (by family) and strangeness are conserved in these interactions
  • Show totals before and after for each quantity
  • Energy and momentum are conserved in all interactions
  • Strangeness must balance in the collisions studied; use the data given

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Conservation laws in particle interactions

  1. Caesium-137 in stored nuclear waste decays by beta-minus emission. Inside each decaying nucleus a neutron changes into a proton, and an electron is emitted at high speed together with one other, very weakly interacting, particle.
    Show that charge and baryon number are both conserved in the decay n → p + e⁻ + νˉe\bar{\nu}_e.2 marks
  2. A positron emission tomography (PET) scanner uses a tracer containing fluorine-18. Each fluorine-18 nucleus decays by beta-plus emission: a proton in the nucleus changes into a neutron and a positron is emitted together with a neutrino.
    A free proton does not decay in this way, although a proton inside some nuclei can. Using conservation of energy, explain why.2 marks
  3. At a particle accelerator a beam of negative pions is fired into a target of liquid hydrogen, so that pions collide with protons. Data: a pion π⁻ has charge −1, baryon number 0 and strangeness 0; a proton has charge +1, baryon number +1 and strangeness 0. The Λ⁰ has charge 0, baryon number +1 and strangeness −1. The K⁰ has charge 0, baryon number 0 and strangeness +1. The K⁻ has charge −1, baryon number 0 and strangeness −1. The Σ⁺ has charge +1, baryon number +1 and strangeness −1. Strangeness is conserved in these collisions.
    Show that the interaction π⁻ + p → K⁰ + Λ⁰ obeys the conservation of charge, baryon number and strangeness.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).