Particles and Antiparticles
In 1932, Carl Anderson at Caltech photographed cosmic ray tracks in a Wilson cloud chamber in a 1.5 T magnetic field. He observed a track with electron-mass curvature but opposite (positive) charge — the antielectron (positron) that Paul Dirac had predicted from his relativistic quantum equation in 1928. Anderson was awarded the Nobel Prize in Physics in 1936. When a positron meets an electron, both annihilate to produce two gamma rays each carrying exactly E = mₑc² = 0.511 MeV — the quantitative signature now used in PET medical scanning.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
When an electron meets a positron (its antiparticle), they disappear and produce gamma rays.
Before reading on, answer:
- Why must the total energy be conserved in this annihilation?
- How many gamma rays are typically produced, and why?
- Could a single proton and antiproton annihilation produce a single photon?
Warm-up: An antiparticle has, compared to its corresponding particle:
Know — Antiparticles
- Same mass, opposite charge
- Annihilation and pair production
- Conservation laws
Understand — Particle Classification
- Hadrons and leptons
- Baryons and mesons
- Quark content
Can Do — Analyse Interactions
- Apply conservation laws
- Calculate annihilation energy
- Identify particle types
Core Content
Matter meets antimatter
In 1932, Carl Anderson at Caltech placed a Wilson cloud chamber in a 1.5 T magnetic field and photographed cosmic ray tracks. One track curved exactly like an electron — same radius, same mass — but in the opposite direction, indicating positive charge. This was the positron (antielectron, symbol $e^+$), the particle Paul Dirac had predicted in 1928 from his relativistic quantum equation. Dirac's theory required that for every particle there exists an antiparticle with the same mass but opposite charge and quantum numbers.
When a particle meets its antiparticle, they annihilate:
$$e^- + e^+ \rightarrow \gamma + \gamma$$The total energy released equals the rest mass energy of both particles plus their kinetic energy. For electron-positron annihilation at rest:
$$E_{total} = 2m_e c^2 = 2 \times 0.511 \text{ MeV} = 1.022 \text{ MeV}$$Two photons are produced (not one) to conserve both energy and momentum. In the centre-of-mass frame the initial momentum is zero, so the two photons travel in exactly opposite directions, each carrying 0.511 MeV. A single photon would have non-zero momentum and would violate conservation of momentum.
The reverse process — pair production — occurs when a high-energy photon near a nucleus converts into a particle-antiparticle pair:
$$\gamma \rightarrow e^- + e^+$$The photon must have energy $E_\gamma \geq 2m_e c^2 = 1.022$ MeV. A nucleus must be present to absorb momentum and make the process possible — pair production cannot occur in empty space.
Similar processes occur for other particle-antiparticle pairs: proton-antiproton, muon-antimuon, quark-antiquark. Each annihilation converts the particles' rest mass energy into photons or other particle pairs.
Figure 1 — Left: electron-positron annihilation producing two 0.511 MeV gamma photons travelling in opposite directions. Right: pair production — a photon with $E \geq 1.022$ MeV creates an $e^-e^+$ pair near a nucleus.
Calculate the minimum photon energy required for proton-antiproton pair production. ($m_p = 938.3$ MeV/c²)
Dirac (1928) predicted antiparticles; Anderson (1932) found the positron. Annihilation: e⁻ + e⁺ → γ + γ, each photon 0.511 MeV (at rest), opposite directions — two photons needed to conserve momentum. Pair production: γ → e⁻ + e⁺, minimum E_γ = 2m_ec² = 1.022 MeV; nucleus required. General: min energy = 2m₀c².
Write the annihilation equation and note why TWO photons — the momentum argument is a standard 2-mark question.
Why are two photons (not one) produced in electron-positron annihilation at rest?
The building blocks of matter
We just saw that particles have antiparticles of equal mass and opposite charge, and that annihilation converts rest mass to photons. That raises a question: how do we categorise all the particles that have been discovered — are they all fundamental? This card answers it → no. Hadrons (baryons and mesons) are made of quarks; leptons are fundamental with no known substructure.
All known matter particles fall into two broad categories based on how they experience the fundamental forces:
Hadrons: Particles that experience the strong nuclear force. They are not fundamental — they are composed of quarks.
- Baryons: Made of three quarks (or three antiquarks for antibaryons). Examples: proton ($uud$, $B=+1$), neutron ($udd$, $B=+1$), lambda ($uds$, $B=+1$). Baryon number $B = +1$ (or $-1$ for antibaryons).
- Mesons: Made of a quark-antiquark pair ($q\bar{q}$). Examples: pion $\pi^+$ ($u\bar{d}$), kaon $K^+$ ($u\bar{s}$). Baryon number $B = 0$.
Leptons: Fundamental particles that do not experience the strong force. They are point-like with no known substructure.
- Charged leptons: electron ($e^-$), muon ($\mu^-$), tau ($\tau^-$) and their antiparticles.
- Neutral leptons (neutrinos): $\nu_e$, $\nu_\mu$, $\nu_\tau$ and their antineutrinos. Neutrinos have very small mass and interact only via the weak force and gravity.
Each lepton has an associated lepton number ($L_e$, $L_\mu$, $L_\tau$) that is conserved in all interactions. For example, in beta decay: $n \rightarrow p + e^- + \bar{\nu}_e$, the electron lepton number is $0 \rightarrow 0 + 1 + (-1) = 0$ (conserved).
Figure 2 — Particle classification hierarchy. Hadrons experience the strong force and are made of quarks; leptons do not experience the strong force and are fundamental. Baryons have baryon number $B = \pm 1$; mesons have $B = 0$.
$e^- + e^+ \rightarrow \gamma + \gamma$ — annihilation (1.022 MeV at rest)
$\gamma \rightarrow e^- + e^+$ — pair production ($E_\gamma \geq 2m_ec^2 = 1.022$ MeV)
$E = m_0c^2$ — rest mass energy
Baryon number $B$ — conserved in all interactions ($B = +1$ for baryons, $-1$ for antibaryons, $0$ for mesons and leptons)
Lepton number $L$ — conserved separately for each family ($L_e$, $L_\mu$, $L_\tau$)
A particle interaction produces a proton, a neutron, and a pion ($\pi^0$). The initial state had baryon number $B = 2$. Is baryon number conserved? What type of particle is the pion?
Hadrons feel the strong force: baryons (3 quarks, B=+1) — proton (uud), neutron (udd); mesons (qq̄, B=0) — pion (ud̄), kaon (us̄). Leptons are fundamental (no strong force): charged (e⁻, μ⁻, τ⁻) and neutral (ν_e, ν_μ, ν_τ). Lepton number conserved per family; baryon number conserved in all interactions.
Draw the hadron-lepton tree in your notes — classifying a given particle is tested almost every year.
Baryons have baryon number $B = +1$, while mesons have $B = 0$.
Leptons experience the strong nuclear force, which is why they are affected by gluons.
The pion $\pi^+$ is a meson because it consists of a quark-antiquark pair ($u\bar{d}$).
The rules that govern every interaction
We just saw that hadrons and leptons carry quantum numbers like baryon number B and lepton number L. That raises a question: how do these numbers help us decide whether a given particle reaction is actually possible? This card answers it → every interaction must conserve Q, B, L_e/L_μ/L_τ, and energy-momentum simultaneously; any violation makes the process forbidden.
Every particle interaction must satisfy a set of conservation laws simultaneously. These laws act as constraints — if any one is violated, the process simply cannot occur. The key conserved quantities are:
- Charge ($Q$) — total charge before = total charge after; conserved in all interactions.
- Baryon number ($B$) — conserved in all interactions. Protons and neutrons: $B = +1$; antiprotons, antineutrons: $B = -1$; mesons and leptons: $B = 0$.
- Lepton number ($L$) — conserved separately for each flavour: $L_e$, $L_\mu$, $L_\tau$. Electrons and electron-neutrinos: $L_e = +1$; their antiparticles: $L_e = -1$.
- Energy and momentum — always conserved; these together explain why a single photon cannot result from annihilation, and why a nucleus is needed for pair production.
- Strangeness ($S$) — conserved in strong and electromagnetic interactions, but can change by $\pm 1$ in weak interactions. Strange quarks ($s$) carry $S = -1$.
To analyse an interaction: sum the conserved quantity for all particles before the interaction, then for all particles after, and check they are equal. Antiparticles carry the opposite quantum numbers to their partners.
When analysing particle interactions, always check conservation of: charge ($Q$), baryon number ($B$), lepton number ($L$), and strangeness (in strong interactions). A common trap: forgetting that antiparticles have opposite quantum numbers. An antiproton has $B = -1$, $Q = -1$. A positron has $L_e = -1$, $Q = +1$. Mesons have $B = 0$ and $L = 0$ but can carry strangeness (e.g., kaon $K^+ = u\bar{s}$ has $S = +1$). Remember that two photons are produced in electron-positron annihilation at rest to conserve momentum — a single photon would have zero momentum in the centre-of-mass frame but the initial system has zero momentum, so two photons moving in opposite directions are required.
Conserved in ALL interactions: Q (charge), B (baryon number), L_e/L_μ/L_τ (lepton number per flavour), energy + momentum. Strangeness S is conserved in strong/EM but can change ±1 in weak interactions. Method: sum each quantity before and after; any mismatch means the process is forbidden.
List the five conserved quantities in your notes — checking them in order is the guaranteed way to score full marks on "is this interaction allowed?" questions.
Three of these statements about conservation laws in particle physics are correct. Pick the odd one out.
In the reaction $\pi^- + p \rightarrow K^0 + \Lambda^0$, which conservation law would be violated if the products were $\pi^0 + n$ instead?
Activities
Practice calculating energies for pair annihilation and pair production
- An electron and positron at rest annihilate. Calculate the energy of each photon produced. ($m_e = 9.11\times10^{-31}$ kg, $c = 3.00\times10^8$ m/s; or use $m_ec^2 = 0.511$ MeV)
- A photon with energy 1.5 MeV passes near a nucleus. Explain whether pair production can occur and calculate the kinetic energy available for the pair.
- Calculate the minimum photon energy needed to produce a proton-antiproton pair. ($m_pc^2 = 938.3$ MeV) How many times greater is this than the electron-positron threshold?
- A muon-antimuon pair is created by a photon. The muon rest mass energy is 105.7 MeV. What is the minimum photon energy? What happens to the muon after creation?
Determine whether particle interactions are allowed
- For each interaction, state whether it is allowed or forbidden. If forbidden, state which conservation law is violated: (a) $p + \bar{p} \rightarrow \pi^+ + \pi^-$; (b) $n \rightarrow p + e^- + \nu_e$; (c) $p \rightarrow n + e^+ + \nu_e$; (d) $e^- \rightarrow \mu^- + \nu_\mu + \bar{\nu}_e$.
- A student claims that the reaction $\pi^+ \rightarrow e^+ + \gamma$ is allowed because charge, energy and baryon number are all conserved. Identify which conservation law the student overlooked.
- Explain why an isolated proton cannot decay into a positron and a photon ($p \rightarrow e^+ + \gamma$).
- The kaon $K^+$ decays as $K^+ \rightarrow \mu^+ + \nu_\mu$. Verify that baryon number, charge, and lepton number are all conserved. ($K^+ = u\bar{s}$, strangeness $S = +1$; which force mediates this decay?)
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(3 marks) 1. An electron and a positron, both at rest, undergo annihilation. (a) Write the equation for this process. (b) Calculate the energy of each photon produced. (c) Explain why two photons must be produced rather than one.
1 mark: correct equation · 1 mark: correct energy (0.511 MeV each) · 1 mark: momentum conservation explanation
AnalyseBand 6(5 marks) 2. (a) Distinguish between hadrons and leptons, and between baryons and mesons, giving an example of each. (b) Explain why electron-positron annihilation at rest produces two photons rather than one. (c) Calculate the energy of each photon produced when a proton and antiproton at rest annihilate. ($m_p c^2 = 938.3$ MeV) (d) A photon with energy 3.0 MeV passes near a nucleus. Explain whether electron-positron pair production can occur and calculate the kinetic energy shared by the pair. (e) State the quark composition of a proton and a neutron, and explain how this accounts for their electric charges.
1 mark: hadron/lepton + baryon/meson distinction · 1 mark: two-photon momentum argument · 1 mark: proton-antiproton energy · 1 mark: pair production check + KE · 1 mark: quark composition and charge
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks): (a) $e^- + e^+ \rightarrow \gamma + \gamma$ (1 mark). (b) Each photon carries half the total rest mass energy: $E = m_ec^2 = 0.511$ MeV (1 mark). (c) In the centre-of-mass frame the total initial momentum is zero. A single photon always has momentum ($p = E/c \neq 0$), which would violate momentum conservation. Two photons of equal energy travelling in exactly opposite directions have zero net momentum, satisfying conservation (1 mark).
Q2 (5 marks): (a) Hadrons experience the strong nuclear force and are composed of quarks; leptons are fundamental particles that do not feel the strong force. Baryons are hadrons made of three quarks ($B = +1$), e.g. proton ($uud$); mesons are hadrons made of a quark-antiquark pair ($B = 0$), e.g. pion $\pi^+$ ($u\bar{d}$). Leptons include charged leptons ($e^-$, $\mu^-$, $\tau^-$) and neutrinos ($\nu_e$, $\nu_\mu$, $\nu_\tau$) (1 mark). (b) In the centre-of-mass frame the system has zero total momentum. A single photon carries momentum $p = E/c \neq 0$, violating momentum conservation. Two photons emitted in opposite directions carry equal and opposite momenta that sum to zero (1 mark). (c) At rest the total energy is $2m_pc^2 = 2 \times 938.3 = 1876.6$ MeV. The two photons share this equally, so each carries 938.3 MeV (1 mark). (d) The threshold for $e^-e^+$ pair production is $2m_ec^2 = 1.022$ MeV. Since 3.0 MeV $> 1.022$ MeV, pair production can occur. The kinetic energy available: $E_{KE} = 3.0 - 1.022 = 1.978$ MeV shared between the pair (1 mark). (e) Proton: $uud$, charges $+\frac{2}{3} + \frac{2}{3} - \frac{1}{3} = +1$. Neutron: $udd$, charges $+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0$. The fractional quark charges sum to give the observed integer charges of the baryon (1 mark).
At the start you were asked about Carl Anderson's 1932 cloud chamber discovery at Caltech — a positive-charge track with electron mass, in a 1.5 T magnetic field, confirming Dirac's 1928 antiparticle prediction. When this positron (e⁺) meets an electron (e⁻), both disappear and two gamma rays are produced, each at exactly 0.511 MeV — the basis of PET scanning. Review your predictions:
- Did you predict energy is conserved because annihilation converts rest mass to photon energy via $E = mc^2$? Correct — Anderson's positron had the same mass as an electron, so annihilation produces 2 × 0.511 MeV = 1.022 MeV of photon energy.
- Did you predict two photons are produced to conserve momentum? Correct — in the centre-of-mass frame, initial momentum is zero, so two equal-energy photons must travel in exactly opposite directions, as in PET scanning.
- Did you predict a single photon cannot result from annihilation because momentum cannot be conserved? Correct — the same argument applies for all particle-antiparticle annihilations; at least two photons are required.
Extend: (a) A kaon $K^0$ ($d\bar{s}$, strangeness $S = -1$) decays via the weak force to $\pi^+\pi^-$. Verify that charge and baryon number are conserved. Show that strangeness is NOT conserved — and explain why this is allowed. (b) Positron emission tomography (PET) scanning uses $\beta^+$ decay. Explain what happens to the positron emitted inside the body, and why this allows medical imaging. (c) Calculate the energy of each annihilation photon in a PET scan, in both MeV and joules.
Five timed questions on particles, antiparticles and conservation laws. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
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