Free Pearson Edexcel International A Level Physics lesson

Matter, antimatter and conservation

Use particle families and conserved numbers to test proposed interactions.

What you’ll learn

  • Classify quarks, leptons and composite particles.
  • Convert particle energies and masses.
  • Check particle equations against conservation laws.

Energy can create rest mass

Mass and energy are linked by ΔE = c²Δm. In particle creation, available energy can become particle rest energy and kinetic energy; in annihilation, particle and antiparticle rest energy can become radiation or other allowed products. Total energy and momentum must both be conserved. Rest mass by itself is not a separately conserved total in these processes.

An electronvolt is the energy gained by one elementary charge across one volt: 1 eV = 1.60 × 10⁻¹⁹ J. MeV and GeV mean 10⁶ and 10⁹ eV. A mass quoted in MeV/c² is converted by first converting the numerator energy to joules, then dividing by c². MeV is an energy unit; MeV/c² is a mass unit.

A particle–antiparticle pair must have at least the combined rest energy available, with additional constraints from momentum. An electron and positron at rest can annihilate into two photons travelling in opposite directions, so momentum cancels. A single photon cannot carry zero total momentum while retaining nonzero energy, and pair production usually needs a nearby nucleus or another interaction partner to conserve momentum.

ΔE = c²Δm; 1 MeV = 1.60 × 10⁻¹³ J; 1 GeV = 1.60 × 10⁻¹⁰ J; m = Erest/c²

Symbols and units

  • E is energy in J
  • m is rest mass in kg
  • c is light speed in m s⁻¹
  • eV is an energy unit, not a voltage.

How the equation explains the idea

  1. Rest energy is mc². For an allowed energy conversion, divide available rest-energy change by c² to obtain the corresponding mass change, while also respecting charge and momentum conservation.
  2. Using the rounded conversion, 1 MeV = 10⁶ × 1.60 × 10⁻¹⁹ J = 1.60 × 10⁻¹³ J. MeV/c² is therefore a mass unit; MeV by itself is energy.

Worked example

Convert a hypothetical rest mass

  1. A particle has rest mass 500 MeV/c².
  2. Rest energy = 500 × 1.60 × 10⁻¹³ = 8.00 × 10⁻¹¹ J.
  3. Mass = 8.00 × 10⁻¹¹/(3.00 × 10⁸)² = 8.89 × 10⁻²⁸ kg.

The c² in the unit is essential to distinguish mass from energy.

Keep this idea. Convert the energy numerator first, and keep rest energy distinct from kinetic energy.

Organise particles by their constituents

The quark–lepton model separates fundamental leptons from hadrons made of quarks. Baryons contain three quarks, such as proton uud and neutron udd. Mesons contain a quark and an antiquark. Electrons and neutrinos are leptons rather than tiny hadrons. Photons are also fundamental particles but are neither baryons nor leptons.

Up-type quarks u, c and t have charge +2e/3; down-type d, s and b have −e/3. Antiquarks have opposite charge and opposite baryon number. Adding constituent charges explains a proton’s +e and neutron’s zero charge. A neutral composite particle can still have charged constituents, so neutral is not synonymous with structureless.

Every particle has an antiparticle with equal mass and reversed additive quantum numbers. An antineutron is neutral electrically but has baryon number −1 and antiquark constituents. Some particles, including the photon, are their own antiparticles. The organised pattern of quark generations supported prediction of the top quark before its observation, illustrating how a model can guide a testable search.

u,c,t: +2e/3; d,s,b: −e/3; proton uud: +e; neutron udd: 0; quark B = +⅓

Symbols and units

  • Q is signed electric charge
  • e is positive elementary charge
  • B is dimensionless baryon number
  • a quark has B = 1/3 and an antiquark has B = −1/3.

How the equation explains the idea

  1. Add constituent charges algebraically: uud gives (2/3 + 2/3 − 1/3)e = +e; udd gives (2/3 − 1/3 − 1/3)e = 0.
  2. Three quarks give B = 1, while a quark–antiquark meson gives B = 0. An antiparticle reverses its additive charge and baryon/lepton numbers, not its positive rest mass.

Worked example

A charged meson

  1. A meson contains u and anti-d.
  2. Its charge is +2e/3 + e/3 = +e.
  3. Its baryon number is +1/3 − 1/3 = 0.

A quark–antiquark particle can be charged while having zero baryon number.

Keep this idea. Identify constituents and add their charges and conserved numbers separately.

A proposed reaction must pass several tests

Test a particle equation by listing electric charge, baryon number and lepton number on each side. A baryon contributes +1 and an antibaryon −1; a meson contributes zero. A lepton contributes +1 and an antilepton −1 under the relevant lepton-number accounting. A neutral particle can still be essential for one of these balances.

For beta-minus decay, n → p + e⁻ + anti-νe conserves electric charge: 0 = +1 −1 +0. Baryon number stays one and electron lepton number stays zero because electron +1 is balanced by antineutrino −1. Write the supplied symbols carefully: replacing an antineutrino with a neutrino leaves charge unchanged but breaks lepton-number balance.

Passing these number tests is necessary, not a proof that a reaction will occur. Energy, momentum, available interactions and kinematics must also permit it. At a detector vertex, combine charge bookkeeping with momentum vectors and energy estimates. Do not infer a decay merely by rearranging symbols into a numerically balanced equation.

ΣQ, ΣB and ΣL conserved; n → p + e⁻ + anti-νe; also conserve total energy and momentum

Symbols and units

  • Q is electric charge in units of e
  • B is baryon number and L is total lepton number, both dimensionless additive counts.

How the equation explains the idea

  1. For neutron beta decay, n → p + e⁻ + anti-νe, charge is 0 = +1 − 1 + 0 and baryon number is 1 = 1 + 0 + 0.
  2. Lepton number is 0 = 0 + 1 − 1, requiring the antineutrino rather than a neutrino in this process. Passing these tests is necessary but does not replace energy, momentum and other applicable conservation checks.

Worked example

Reject a tempting equation

  1. Proposed equation: n → p + e⁻ + νe.
  2. Charge: 0 = +1 −1 +0; baryon number: 1 = 1.
  3. Lepton number: initial 0, final +1 +1 = +2, so the proposal fails.

Use an electron antineutrino, not a neutrino, for beta-minus decay.

Keep this idea. Audit each conserved quantity independently before evaluating energy and momentum.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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