Physics • Year 12 • Module 8 • Lesson 16
Particles and Antiparticles
Apply conservation laws, interpret annihilation energy data, and classify particle interactions in real and exam-style scenarios.
1. Interpret particle interaction data — conservation law audit
The table below lists six proposed particle interactions. For each one, state whether it is allowed or forbidden, identify which conservation law (charge, baryon number, or lepton number) would be violated if it is forbidden, and name the particle type of each product. 12 marks (2 each)
| Proposed interaction | Allowed or forbidden? | Conservation law violated (if forbidden) |
|---|---|---|
| e− + e+ → γ + γ | ||
| e− + e+ → γ | ||
| p + p̄ → p + p + p̄ | ||
| n → p + e− + ν̄e | ||
| n → p + e− | ||
| p → e+ + νe |
2. Interpret annihilation photon energy data
Positron Emission Tomography (PET) scans exploit electron–positron annihilation. A hospital physicist records the energy spectrum of photons detected by a PET scanner ring. The graph below shows the number of photon counts vs photon energy (MeV) for coincidence-detected photon pairs. 8 marks
Figure 2. Count rate vs photon energy recorded by a hospital PET scanner. The tall peak corresponds to coincident photon pairs from electron–positron annihilation events inside the patient. Background represents Compton-scattered photons. Illustrative data consistent with clinical PET physics.
2.1 Identify the photon energy at the peak in the spectrum and explain, using E = mec², why photons of exactly this energy are produced by PET. 2 marks
2.2 A PET scanner detects photon pairs in coincidence (two photons arriving at opposite detectors within a few nanoseconds). Explain why electron–positron annihilation produces photons in opposite directions, with reference to conservation of momentum. 3 marks
2.3 A student claims: “The background counts in the spectrum prove that the Law of Conservation of Energy is sometimes violated in particle interactions.” Evaluate this claim. 3 marks
3. Compare baryons, mesons, and leptons across five features
Complete the three-column table. For each feature, write a concise description that contrasts the three particle classes. 10 marks (1 per shaded cell)
| Feature | Baryon | Meson | Lepton |
|---|---|---|---|
| Quark content | |||
| Experiences strong force? | |||
| Baryon number | |||
| Stability / lifetime | |||
| Example particle |
4. Predict and justify — proton–antiproton annihilation at CERN
In the 1980s at CERN’s Super Proton Synchrotron (SPS), Carlo Rubbia’s team collided protons with antiprotons at high energy. Unlike electron–positron annihilation at rest, which always produces two 0.511 MeV photons, proton–antiproton annihilation at high energy can produce pions, kaons, and other hadrons rather than photons alone. 6 marks
4.1 A proton (B = +1) and antiproton (B = −1) annihilate. Predict the total baryon number of the products and explain what types of particle could be produced while conserving all relevant quantum numbers. 3 marks
4.2 The beam energy at the SPS was much greater than the proton rest mass energy of 938.3 MeV. Predict how the total energy of the photons or particles produced would compare with the minimum 2mpc² and justify your prediction. 3 marks
Q1 — Conservation law audit
e− + e+ → γ + γ: Allowed. Charge: −1 + 1 = 0 = 0 + 0. Baryon number: 0 = 0. Lepton number: Le = +1 + (−1) = 0 = 0.
e− + e+ → γ: Forbidden — violates conservation of momentum. In the centre-of-mass frame the total momentum is zero but a single photon carries nonzero momentum; two photons in opposite directions are required.
p + p̄ → p + p + p̄: Allowed. Charge: 1 + (−1) = 0 = 1 + 1 + (−1). Baryon number: 1 + (−1) = 0 = 1 + 1 + (−1) = 1. Wait — check again: left B = 0; right B = 1 + 1 − 1 = +1. Forbidden — violates baryon number (0 ≠ +1). [Note: the reaction p + p̄ → p + p̄ + p + p̄ would be allowed. Accept student analysis if algebra is shown correctly.]
n → p + e− + ν̄e: Allowed (this is beta-minus decay). Charge: 0 = 1 + (−1) + 0. Baryon number: 1 = 1 + 0 + 0. Le: 0 = 0 + 1 + (−1) = 0. All conserved.
n → p + e−: Forbidden — violates lepton number. Le left = 0; right = 0 + 1 = +1. An antineutrino (Le = −1) must also be produced.
p → e+ + νe: Forbidden — violates baryon number. B left = +1; right = 0 + 0 = 0.
Q2.1 — Peak energy and E = mec²
The peak is at 0.511 MeV [1]. When an electron and positron are essentially at rest in tissue, their rest-mass energies each equal mec² = 0.511 MeV. Conservation of energy requires each photon to carry exactly this energy, so both photons emerge at 0.511 MeV [1].
Q2.2 — Opposite directions and momentum conservation
In the reference frame where the electron and positron are approximately at rest, the total initial momentum is zero [1]. Conservation of momentum requires the total momentum of the products to also be zero [1]. With two photons, this is satisfied by them travelling in exactly opposite directions (antiparallel), so their momenta cancel: p1 + p2 = 0 [1].
Q2.3 — Evaluate the claim about background counts
The claim is incorrect [1]. Conservation of energy is not violated; the background counts are Compton-scattered photons that have transferred some of their energy to electrons in the detector or patient tissue, reducing their recorded energy below 0.511 MeV [1]. The total energy in each interaction is still conserved — energy has been transferred to the recoil electron, not destroyed [1].
Q3 — Baryon / meson / lepton comparison table
Quark content: Baryon: three quarks (qqq) or three antiquarks (q̄q̄q̄). Meson: quark–antiquark pair (qq̄). Lepton: no quark content; fundamental.
Experiences strong force? Baryon: Yes. Meson: Yes. Lepton: No.
Baryon number: Baryon: +1 (antibaryon: −1). Meson: 0. Lepton: 0.
Stability / lifetime: Baryon: The proton is stable; other baryons (e.g. neutron, Λ0) decay. Meson: Generally short-lived (10−8 to 10−16 s). Lepton: Electron is stable; muon decays (~2.2 μs); tau decays (~3 × 10−13 s).
Example: Baryon: proton (uud) or neutron (udd). Meson: pion (π+ = ud̄) or kaon. Lepton: electron, muon, electron neutrino.
Q4.1 — Proton–antiproton annihilation products
Total baryon number of products = B(p) + B(p̄) = +1 + (−1) = 0 [1]. The products must therefore carry zero net baryon number. This rules out any net surplus of baryons over antibaryons [1]. The products can include mesons (B = 0, made of quark–antiquark pairs), photons, or baryon–antibaryon pairs (e.g. a proton–antiproton pair), all of which sum to B = 0. Charge must also be conserved (total = 0), so charged products must appear in opposite-charge pairs [1].
Q4.2 — Total energy vs 2mpc²
The total energy of the products would be greater than 2mpc² = 2 × 938.3 MeV = 1876.6 MeV [1]. By conservation of energy, the total energy of the products equals the total energy of the colliding proton and antiproton, which includes their rest-mass energies plus their kinetic energies [1]. At the SPS, beam energies were hundreds of GeV, so the kinetic energy contribution is enormous compared with the rest-mass energy, allowing many additional massive particles (pions, kaons, etc.) to be created in the products [1].