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Every particle in the Standard Model has an associated antiparticle — a partner with the same mass but opposite charge and opposite values for the other additive quantum numbers. When a particle meets its antiparticle, they annihilate, converting their mass entirely into electromagnetic energy in the form of photons. The reverse process, in which a high-energy photon converts into a particle-antiparticle pair, is equally real and routinely observed. This lesson develops the antiparticle concept, introduces the photon as the quantum of electromagnetic energy, and works through the annihilation and pair-production processes that link mass and energy directly via Einstein's relation E = mc².
Specification mapping. This lesson develops AQA A-Level Physics (7408) Particles and Radiation strand, sub-strands 3.2.1.2 (particles and antiparticles, including the positron, antiproton, antineutron and antineutrino) and 3.2.1.3 (photon model of electromagnetic radiation; annihilation and pair production with E = hf and minimum-energy requirements). Refer to the official AQA 7408 specification document for the authoritative wording. The lesson is positioned after the nuclear-stability lesson (order 1, where positrons and antineutrinos were introduced as decay products) so that students already have a working acquaintance with antiparticles before their full classification.
Synoptic links. Three principal synoptic threads run through this content. First, the photoelectric effect (3.2.2.1, our order 7) — the photon concept introduced here is the same E = hf quantum that liberates electrons from metal surfaces in the photoelectric effect; both lessons depend on the discrete, particle-like character of electromagnetic energy. Second, electromagnetic spectrum and wave properties (course 3, waves) — the photon is the quantum of all electromagnetic radiation, from radio waves to gamma rays, and the wave description (frequency, wavelength) is the same continuous quantity that determines photon energy via E = hf = hc/λ. Third, mass-energy equivalence (course 6, nuclear physics) — annihilation and pair production are the cleanest experimental demonstrations of E = mc², converting rest mass directly into photon energy and vice versa.
For every type of particle, an antiparticle exists with the following properties:
A particle and its antiparticle are produced together in pair production and annihilate together in annihilation. Some neutral particles are their own antiparticles (e.g. the photon, the neutral pion π⁰), but most particles have distinct antiparticles.
| Particle | Symbol | Antiparticle | Symbol | Rest energy (MeV) |
|---|---|---|---|---|
| Electron | e⁻ | Positron | e⁺ | 0.511 |
| Proton | p | Antiproton | p̄ | 938.3 |
| Neutron | n | Antineutron | n̄ | 939.6 |
| Electron neutrino | νₑ | Electron antineutrino | ν̄ₑ | ~0 |
| Muon | μ⁻ | Antimuon | μ⁺ | 105.7 |
| Up quark | u | Anti-up quark | ū | ~2 |
| Down quark | d | Anti-down quark | d̄ | ~5 |
| Photon | γ | (its own antiparticle) | γ | 0 |
The rest energy E₀ = m₀c² is a measure of the particle's mass in units of energy. For the electron, m₀ = 9.11 × 10⁻³¹ kg gives E₀ = (9.11 × 10⁻³¹)(3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J = 0.511 MeV. For the proton, the corresponding value is about 938 MeV — nearly 2000 times larger.
The positron — the antiparticle of the electron — was the first antiparticle to be discovered experimentally. Predicted theoretically by Paul Dirac in 1928 from the negative-energy solutions of his relativistic wave equation, it was observed in cloud-chamber tracks by Carl Anderson in 1932 (Nobel Prize 1936). The positron has:
When a positron is produced in β⁺ decay or pair production, it travels a few millimetres in matter before colliding with an electron and annihilating into two 0.511 MeV gamma rays — the back-to-back annihilation signature exploited by PET scanners.
The antineutron has the same mass as the neutron (1.67 × 10⁻²⁷ kg) and zero net charge — but it is not identical to the neutron. Their distinct identities are revealed in three ways:
This is why a neutral particle can still be distinct from its antiparticle — internal structure (quark content) and additive quantum numbers can distinguish them even when charge cannot.
The photon is the quantum of the electromagnetic field — the discrete packet of energy and momentum carried by electromagnetic radiation. Photons:
The energy of a photon depends only on its frequency:
E = hf
where h = 6.63 × 10⁻³⁴ J s is the Planck constant. Equivalently, using c = fλ:
E = hc / λ
A photon carries momentum p = E/c = h/λ. Although it has no rest mass, this non-zero momentum is real and observable — it underlies radiation pressure (the basis of solar sails) and Compton scattering (X-rays bouncing off electrons with a measurable wavelength shift).
| Region | Wavelength | Frequency (Hz) | Photon energy |
|---|---|---|---|
| Radio | 1 m | 3 × 10⁸ | 2 × 10⁻²⁵ J = 1.2 × 10⁻⁶ eV |
| Microwave | 10⁻² m | 3 × 10¹⁰ | 2 × 10⁻²³ J = 1.2 × 10⁻⁴ eV |
| Infrared | 10⁻⁵ m | 3 × 10¹³ | 2 × 10⁻²⁰ J = 0.12 eV |
| Visible (red, 700 nm) | 7 × 10⁻⁷ m | 4.3 × 10¹⁴ | 2.8 × 10⁻¹⁹ J = 1.8 eV |
| Visible (violet, 400 nm) | 4 × 10⁻⁷ m | 7.5 × 10¹⁴ | 5.0 × 10⁻¹⁹ J = 3.1 eV |
| Ultraviolet | 10⁻⁸ m | 3 × 10¹⁶ | 2.0 × 10⁻¹⁷ J = 124 eV |
| X-ray | 10⁻¹⁰ m | 3 × 10¹⁸ | 2.0 × 10⁻¹⁵ J = 12.4 keV |
| Gamma ray | 10⁻¹² m | 3 × 10²⁰ | 2.0 × 10⁻¹³ J = 1.24 MeV |
This table is worth memorising in rough form: visible photons carry single-eV energies; X-ray photons keV; gamma-ray photons MeV.
Calculate the energy of a green photon of wavelength 550 nm.
E = hc/λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) / (550 × 10⁻⁹) E = 3.62 × 10⁻¹⁹ J
Converting to eV: 3.62 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.26 eV
A gamma-ray photon has energy 1.022 MeV. What is its frequency?
E = 1.022 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.635 × 10⁻¹³ J
f = E / h = 1.635 × 10⁻¹³ / 6.63 × 10⁻³⁴ = 2.47 × 10²⁰ Hz
When a particle meets its antiparticle, both are destroyed and their combined energy (rest energy plus kinetic energy) is converted entirely into photons. For electron-positron annihilation:
e⁻ + e⁺ → 2γ
Two photons are produced rather than one, because a single photon cannot conserve both energy and momentum (the centre-of-mass frame has zero net momentum, but a single photon always has momentum p = E/c ≠ 0). Two photons emitted back-to-back in the centre-of-mass frame trivially conserve momentum.
If the electron and positron are at rest before they annihilate, the total available energy is just their combined rest energy:
E_total = 2 m_e c² = 2 × 0.511 MeV = 1.022 MeV
Each of the two photons receives half this energy: 0.511 MeV per photon. Their wavelength is:
λ = hc / E_photon = (6.63 × 10⁻³⁴)(3.00 × 10⁸) / (0.511 × 10⁶ × 1.60 × 10⁻¹⁹) = 2.43 × 10⁻¹² pm = 2.43 pm
This characteristic 0.511 MeV gamma-ray line is the signature used to detect positrons in PET scanners.
An electron and a positron, each travelling at 10% the speed of light, collide head-on and annihilate into two equal-energy photons. Calculate the energy of each photon (treating motion non-relativistically for simplicity).
Kinetic energy of each particle: E_k = ½ m v² = 0.5 × 9.11 × 10⁻³¹ × (0.1 × 3 × 10⁸)² = 4.10 × 10⁻¹⁶ J ≈ 2.6 keV (small compared with rest energy)
Total energy released = 2 × (m_e c² + E_k) = 2 × (0.511 + 0.00256) MeV ≈ 1.027 MeV
Each photon: 0.513 MeV (about 0.4% above the rest-energy-only case).
The same principle applies to all particle-antiparticle pairs:
The reverse of annihilation: a high-energy photon converts into a particle-antiparticle pair. For an electron-positron pair:
γ → e⁻ + e⁺
The photon must carry at least enough energy to provide the combined rest energy of the produced pair:
E_photon ≥ 2 m_e c² = 2 × 0.511 = 1.022 MeV
This corresponds to a minimum photon frequency:
f_min = E_min / h = (1.022 × 10⁶ × 1.60 × 10⁻¹⁹) / (6.63 × 10⁻³⁴) = 2.47 × 10²⁰ Hz (gamma-ray region)
Any photon energy above this threshold is shared as kinetic energy between the two produced particles.
Pair production cannot occur in completely empty space — a free photon converting to an electron-positron pair cannot simultaneously conserve energy and momentum. The total energy of the produced pair is at least 1.022 MeV; the total momentum of the produced pair, if balanced, would be zero. But the incident photon has p = E/c, which is non-zero. So pair production in vacuum violates momentum conservation.
The resolution is that pair production occurs in the Coulomb field of a nearby nucleus (or, more rarely, an electron). The nucleus recoils to absorb the excess momentum:
γ + nucleus → e⁻ + e⁺ + nucleus (slightly faster)
The nucleus, with its very large mass, can absorb the photon's momentum with essentially no change in energy — so the threshold remains 1.022 MeV in practice.
Higher-energy photons can produce heavier pairs:
| Pair | Threshold (MeV) | Photon wavelength (m) |
|---|---|---|
| e⁻ + e⁺ | 1.022 | 1.21 × 10⁻¹² |
| μ⁻ + μ⁺ | 211 | 5.88 × 10⁻¹⁵ |
| p + p̄ | 1876 | 6.61 × 10⁻¹⁶ |
Cosmic-ray gamma rays and accelerator-produced gamma rays can routinely produce these pairs, which is how antimatter is generated for experimental research.
A 2.000 MeV photon undergoes pair production near a nucleus, producing an electron and a positron with equal kinetic energies. Calculate the kinetic energy of each particle.
Total energy of photon: 2.000 MeV Rest energy of pair: 2 × 0.511 = 1.022 MeV Surplus energy (becomes kinetic): 2.000 − 1.022 = 0.978 MeV Each particle: 0.978 / 2 = 0.489 MeV
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