Physics/Lessons/Particles & Radiation/Particles, Antiparticles & Photons
Physics · Particles & Radiation · AS Level

Particles, Antiparticles & Photons

Every particle of matter has a mirror-image counterpart made of the same stuff but opposite in charge. This lesson sets out what an antiparticle is, how photons fit into the particle picture of light, and what happens when matter and antimatter meet.

9 min read

Antiparticles

For every type of particle, there exists a corresponding antiparticle. An antiparticle has exactly the same mass as its particle, and therefore the same rest energy, but has properties like charge that are exactly reversed. The four antiparticles required at this level, alongside the particles they correspond to, are:

ParticleAntiparticleCharge (particle → antiparticle)
ElectronPositron$-1 \rightarrow +1$
ProtonAntiproton$+1 \rightarrow -1$
NeutronAntineutron$0 \rightarrow 0$
NeutrinoAntineutrino$0 \rightarrow 0$

Even the two neutral particles — the neutron and neutrino — have genuinely distinct antiparticles, despite no charge to reverse. What flips instead is lepton number (for the neutrino) or baryon number (for the neutron): "no charge" doesn't mean "no antiparticle."

Particle / antiparticle pairMassRest energy
Electron / positron$9.11\times10^{-31}\,\text{kg}$ (equal)$0.511\,\text{MeV}$ (equal)
Proton / antiproton$1.673\times10^{-27}\,\text{kg}$ (equal)$938\,\text{MeV}$ (equal)
What's the same, what's opposite

A particle and its antiparticle always have identical mass and therefore identical rest energy. What's opposite is charge (where the particle is charged at all) and certain other quantum properties. They are not "the same particle with a minus sign" in every respect — only in the properties that can meaningfully be reversed.

A note on terminology across exam boards

Every major specification requires the same four antiparticle pairs above. Charge-to-mass ratio may be called "specific charge" (the AQA term) — same idea, different framing.

Rest energy and mass-energy equivalence

Mass itself is a form of energy — a particle at rest still represents a store of energy simply by having mass. This is why particle masses above are quoted in MeV as well as kilograms: a particle's rest energy is the energy equivalent of its mass, the natural unit for comparing particles once you start considering particles being created or destroyed. You don't need to calculate with the mass-energy relationship directly at this level — what matters is the concept: every particle's mass corresponds to a definite quantity of energy, allowing mass to convert into other forms of energy and back, in the processes described later.

The photon model of electromagnetic radiation

Electromagnetic radiation can be modelled as a stream of discrete packets of energy called photons, rather than only as a continuous wave. Each photon carries energy fixed by the frequency of the radiation:

$E = hf$

where $h$ is the Planck constant, $h = 6.63\times10^{-34}\,\text{J s}$, and $f$ is the frequency in hertz. Higher-frequency radiation — gamma rays, X-rays — is made of individually more energetic photons than lower-frequency radiation such as radio waves, even though both travel at the same speed.

DerivationE = hf = hc/λ
1
Start from the photon energy equation, $E = hf$.
2
All electromagnetic radiation travels at the speed of light $c$ in a vacuum, and for any wave, wave speed is related to frequency and wavelength by $c = f\lambda$.
This is the standard wave equation, applied here to light specifically.
3
Rearranging the wave equation for frequency: $f = \dfrac{c}{\lambda}$.
4
Substituting into $E = hf$ gives $E = \dfrac{hc}{\lambda}$.
This form is often more convenient when a wavelength is given directly rather than a frequency.
▪ derived

Both forms describe the same quantity — the energy of a single photon — expressed using whichever of frequency or wavelength is given. Higher frequency means shorter wavelength (since $c$ is fixed), and both point toward higher photon energy.

Worked exampleEnergy of a photon of red light
1
Find the energy of a single photon of red light with wavelength $\lambda = 650\,\text{nm} = 650\times10^{-9}\,\text{m}$.
2
Use $E = \dfrac{hc}{\lambda}$, with $h = 6.63\times10^{-34}\,\text{J s}$ and $c = 3.00\times10^{8}\,\text{m}\,\text{s}^{-1}$.
3
$E = \dfrac{6.63\times10^{-34} \times 3.00\times10^{8}}{650\times10^{-9}}$
4
$E = 3.06\times10^{-19}\,\text{J}$
Converting to electronvolts, using $1\,\text{eV} = 1.60\times10^{-19}\,\text{J}$, gives $E \approx 1.91\,\text{eV}$ — a useful sanity check, since visible-light photon energies should always come out in the low-eV range.
▪ answer

Annihilation

When a particle meets its own antiparticle, the two can undergo annihilation: both particles disappear entirely, and their combined mass is converted into energy, released as electromagnetic radiation.

ReasoningWhy annihilation produces two photons, not one
1
Consider an electron and a positron annihilating while both are essentially at rest (or moving slowly), so their total momentum before the interaction is very close to zero.
This is the typical situation for annihilation at low energies.
2
Momentum must be conserved: the total momentum of whatever is produced must also be (very close to) zero.
3
A photon always carries momentum in the direction it travels — it cannot have zero momentum while carrying energy. If only a single photon were produced, it would have to travel off in some one direction, carrying non-zero momentum away with it.
A lone outgoing photon can't simultaneously carry away energy (which it must, since the rest energy of the original particles has to go somewhere) and carry zero net momentum.
4
A single photon is therefore incompatible with starting from (approximately) zero total momentum, so annihilation from rest cannot produce just one photon.
5
Producing two photons resolves this: if they travel in exactly opposite directions with equal energy (and therefore equal momentum magnitude), their momenta cancel, giving zero total momentum, while their combined energy still equals the total rest energy released.
Two oppositely directed, equal-energy photons satisfy both conservation of energy and conservation of momentum simultaneously.
▪ derived

This is why electron–positron annihilation is written as $e^- + e^+ \rightarrow \gamma + \gamma$, with two photons rather than one — forced by momentum conservation whenever the annihilating particles have negligible momentum beforehand.

Drawing annihilation as an interaction diagram

Annihilation is drawn differently from the exchange-particle diagrams met in the next lesson, which show a force acting between two separate particles at two separate vertices. Annihilation is a direct conversion, not a force: the incoming electron and positron lines run into a single shared vertex, and the two outgoing photon lines emerge from that same point — one vertex, not two, since there's no go-between particle here, just matter turning directly into radiation.

A single-vertex Feynman-style interaction diagram with time running upward, on a cream background with flat black lines. One straight line enters from the bottom left with an arrowhead pointing toward a single central vertex, labelled electron near its lower end. A second straight line enters from the bottom right with an arrowhead pointing toward the same central vertex, labelled positron near its lower end. From that same central vertex, two wavy lines diverge outward toward the top of the frame: one wavy line running up and to the left, labelled photon, and one wavy line running up and to the right, also labelled photon, the two wavy lines mirroring each other symmetrically about a vertical centre line through the vertex. A single small solid dot marks the shared vertex where all four lines meet.
Annihilation drawn as an interaction diagram: incoming electron and positron lines meet at a single vertex, converting directly into two outgoing photons — one vertex, not two, since this is a direct matter-to-radiation conversion rather than a force exchanged between separate particles.

Pair production

Pair production is the reverse process: a single photon, if it carries enough energy, converts into a particle–antiparticle pair, e.g. $\gamma \rightarrow e^- + e^+$. Since a particle and its antiparticle have equal mass and thus equal rest energy, the photon must supply at least enough energy for the combined rest energy of both particles — the minimum, called the threshold energy. A photon below threshold cannot produce the pair; above threshold, the surplus appears as kinetic energy shared between the two new particles.

Pair production needs a nearby nucleus

A lone photon in empty space cannot convert into a particle–antiparticle pair, even with enough energy, because — as with annihilation — momentum conservation gets in the way: the pair would carry away momentum, but the single incoming photon's momentum can't match it while also satisfying energy conservation. In practice, pair production happens near a nucleus, which absorbs a small amount of momentum, letting both conservation laws be satisfied together.

A single-vertex Feynman-style interaction diagram with time running upward, on a cream background with flat black lines. One wavy line enters from the bottom centre with an arrowhead pointing toward a single central vertex, labelled photon near its lower end. A separate nearly-straight line runs vertically up the right-hand side of the frame, passing close beside the vertex without quite touching it, with a very slight kink at the point of closest approach to show a small deflection, labelled nucleus (recoils to conserve momentum) beside it. From the central vertex, two straight lines diverge outward toward the top of the frame: one running up and to the left, labelled electron, and one running up and to the right, labelled positron, mirroring each other symmetrically about a vertical centre line through the vertex. A single small solid dot marks the vertex where the photon line and the two outgoing straight lines meet.
Pair production drawn as an interaction diagram: an incoming photon converts at a single vertex into an outgoing electron–positron pair. The nucleus track passing nearby is included to show where the small recoil momentum goes — without it, the diagram would violate momentum conservation.

Key takeaways

  • Every particle has an antiparticle with identical mass (and rest energy) but opposite charge and other reversed properties; the four required pairs are electron/positron, proton/antiproton, neutron/antineutron, neutrino/antineutrino.
  • Mass and energy are equivalent — every particle's mass corresponds to a rest energy, usually quoted in MeV, though $E=mc^2$ itself isn't required for calculation at this level.
  • The photon model treats EM radiation as discrete energy packets: $E = hf = \dfrac{hc}{\lambda}$, using $c = f\lambda$ to connect the two forms. E.g. a 650 nm red-light photon carries $E = 3.06\times10^{-19}\,\text{J} \approx 1.91\,\text{eV}$.
  • Annihilation: particle + antiparticle → (usually two) photons; two photons are needed (not one) so that momentum, as well as energy, can be conserved when the initial particles have negligible momentum.
  • Pair production: a sufficiently energetic photon → particle + antiparticle, requiring at least the threshold energy equal to the pair's combined rest energy, and typically needs a nearby nucleus to conserve momentum.
  • Both are drawn as single-vertex interaction diagrams (matter converting directly to radiation, or radiation to matter) — a different shape from the two-vertex exchange-particle diagrams met in the next lesson, since no force-carrying go-between particle is involved.