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What is antimatter, and how are particles and photons created and destroyed in pair production and annihilation?

Antiparticles and their properties, the photon model of electromagnetic radiation, the photon energy equation, and the processes of annihilation and pair production with their energy calculations.

A focused answer to AQA A-Level Physics 3.2.1.3, covering antiparticles, the photon model, the photon energy equation, and the calculations behind annihilation and pair production using rest energies.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Antiparticles
  3. The photon model
  4. Annihilation
  5. Pair production
  6. Try this

What this dot point is asking

AQA specification point 3.2.1.3 wants you to know that every particle has an antiparticle of equal mass and rest energy but opposite charge, use the photon model and E=hfE = hf, and carry out the energy calculations for annihilation and pair production using rest energies in MeV.

Antiparticles

The rest energy of a particle is the energy equivalent of its mass, E0=mc2E_0 = m c^2, usually quoted in MeV. The electron and positron each have a rest energy of about 0.51 MeV0.51 \text{ MeV}. The neutrino has its own antiparticle, the antineutrino, distinguished in beta decay.

The photon model

The photon model treats light as quantised: energy is delivered in fixed lumps rather than continuously, which is the key to explaining the photoelectric effect and line spectra.

Annihilation

Annihilation is the basis of PET (positron emission tomography) scanning in medicine, where positrons emitted inside the body annihilate with electrons, producing pairs of gamma photons that are detected to build up an image.

Pair production

For electron-positron pair production the photon must have at least 1.02 MeV1.02 \text{ MeV} (twice 0.51 MeV0.51 \text{ MeV}). Any extra photon energy becomes kinetic energy of the two created particles.

Try this

Q1. State two properties an antiparticle shares with its particle and one that differs. [2 marks]

  • Cue. Same mass and rest energy; opposite charge.

Q2. Calculate the minimum frequency of a photon that can cause electron-positron pair production. [3 marks]

  • Cue. Minimum energy =1.02 MeV=1.63×10−13 J= 1.02 \text{ MeV} = 1.63 \times 10^{-13} \text{ J}, then f=Ehf = \dfrac{E}{h}.

Q3. State the name of the antiparticle of the electron. [1 mark]

  • Cue. The positron.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksCalculate the minimum frequency of a photon that can produce an electron-positron pair. The rest energy of an electron is 0.51 MeV0.51 \text{ MeV}, 1 MeV=1.60×10−13 J1 \text{ MeV} = 1.60 \times 10^{-13} \text{ J} and h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}.
Show worked answer →

The photon must supply at least the total rest energy of both particles: Emin⁡=2×0.51=1.02 MeVE_{\min} = 2 \times 0.51 = 1.02 \text{ MeV}.

Convert to joules: E=1.02×1.60×10−13=1.63×10−13 JE = 1.02 \times 1.60 \times 10^{-13} = 1.63 \times 10^{-13} \text{ J}.

Use E=hfE = hf, so f=Eh=1.63×10−136.63×10−34=2.5×1020 Hzf = \dfrac{E}{h} = \dfrac{1.63 \times 10^{-13}}{6.63 \times 10^{-34}} = 2.5 \times 10^{20} \text{ Hz}.

Markers reward doubling the rest energy, converting to joules, and finding the frequency from f=Ehf = \dfrac{E}{h}.

AQA 20214 marksExplain what happens in electron-positron annihilation, and state why two photons rather than one are produced. Calculate the minimum energy of each photon.
Show worked answer →

In annihilation the electron meets its antiparticle, the positron, and both are destroyed, their mass and energy being converted entirely into electromagnetic radiation (photons).

Two photons are produced, travelling in opposite directions, so that momentum is conserved. A single photon could not conserve momentum, because the initial momentum of the pair (at rest) is zero and a single photon must carry momentum.

The total energy available is 2×0.51=1.02 MeV2 \times 0.51 = 1.02 \text{ MeV}, shared equally, so each photon has a minimum energy of 0.51 MeV0.51 \text{ MeV}.

Markers reward the destruction of the pair into photons, two photons for momentum conservation, and the 0.51 MeV0.51 \text{ MeV} minimum per photon.

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