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FractionRush AQA A-Level Physics 2

🌌 Particles & Antiparticles

Discover antimatter, pair production, annihilation and conservation laws

AQA A-Level Physics 2
🔁State the properties of antiparticles for each particle
Describe pair production and the conditions required
💥Describe pair annihilation and calculate photon energies
⚖️Apply conservation of mass-energy (E = mc²)
🛡️State the conservation laws obeyed in particle interactions
📐Use E = hf to find photon energies in annihilation

Antiparticles

Every particle has a corresponding antiparticle with the same mass but opposite charge (and opposite quantum numbers). Antiparticles were predicted by Paul Dirac in 1928 from his relativistic quantum equation.

ParticleSymbolAntiparticleSymbolCharge
Electrone⁻Positrone⁺+e
ProtonpAntiproton−e
NeutronnAntineutron0
Neutrinoν_eAntineutrinoν̄_e0
The photon is its own antiparticle (γ = γ̄). The antineutron has the same mass and zero charge as the neutron, but opposite baryon number and quark content.
Antimatter is matter composed of antiparticles. When matter and antimatter meet, they annihilate — converting all their rest mass energy into photons.

Pair Production

Pair production occurs when a high-energy photon spontaneously converts into a particle-antiparticle pair (most commonly an electron-positron pair) in the presence of a nucleus (which absorbs recoil momentum).
γ → e⁻ + e⁺
Minimum photon energy required: E_min = 2m_e c²
m_e = 9.11 × 10⁻³¹ kg; c = 3.00 × 10⁸ m s⁻¹
E_min = 2 × 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 1.64 × 10⁻¹³ J = 1.02 MeV
Any energy above 1.02 MeV becomes kinetic energy of the produced particles. The photon must have energy ≥ 2m_e c² for pair production to occur.
Pair production conserves: charge (0 = −e + e), lepton number, baryon number, and mass-energy. A nucleus must be nearby to conserve momentum.

Pair Annihilation

Annihilation occurs when a particle meets its antiparticle. Their combined rest mass energy is converted entirely into two photons emitted in opposite directions (to conserve momentum).
e⁻ + e⁺ → γ + γ
Each photon has minimum energy: E = m_e c²
E = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.19 × 10⁻¹⁴ J = 0.511 MeV

If the particles are at rest, both photons have equal energy (0.511 MeV each) and travel in exactly opposite directions. If the particles have kinetic energy, the photons have more energy than the minimum.

Two photons are produced (not one) to satisfy conservation of momentum. One photon alone could not have zero total momentum as required when a particle and antiparticle at rest annihilate.

Conservation Laws in Particle Physics

Every particle interaction must obey these conservation laws. If any is violated, the interaction cannot occur:

Conservation LawExplanation
Conservation of chargeTotal electric charge before = after
Conservation of mass-energyTotal energy (rest mass + kinetic) is conserved
Conservation of momentumTotal momentum is conserved
Conservation of baryon numberBaryon number: proton = +1, antiproton = −1, others = 0
Conservation of lepton numberElectron lepton number: e⁻, ν_e = +1; e⁺, ν̄_e = −1
Baryon number B: Each baryon (proton, neutron) has B = +1; each antibaryon has B = −1; all other particles have B = 0.
Lepton number L_e: Electron and electron-neutrino have L_e = +1; their antiparticles have L_e = −1.
Calculate the minimum frequency of a photon required for pair production (electron-positron pair). m_e = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s.
1Minimum photon energy = 2m_e c² (to create both particles at rest)
2E = 2 × 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 1.64 × 10⁻¹³ J
3E = hf → f = E/h = 1.64 × 10⁻¹³ / 6.63 × 10⁻³⁴ = 2.47 × 10²⁰ Hz
f_min = 2.47 × 10²⁰ Hz (gamma-ray region)
An electron and positron, both at rest, annihilate. Calculate the wavelength of each photon produced. h = 6.63 × 10⁻³⁴ J s.
1Each photon energy = m_e c² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.199 × 10⁻¹⁴ J
2E = hc/λ → λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 8.199 × 10⁻¹⁴
3λ = 1.989 × 10⁻²⁵ / 8.199 × 10⁻¹⁴ = 2.43 × 10⁻¹² m
λ = 2.43 × 10⁻¹² m = 2.43 pm (gamma ray)
Check whether the following interaction is possible by applying all conservation laws: p + p̄ → π⁺ + π⁻ + γ. (Pions π have baryon number 0 and lepton number 0.)
1Charge: left = +1 + (−1) = 0; right = +1 + (−1) + 0 = 0 ✓
2Baryon number: left = 1 + (−1) = 0; right = 0 + 0 + 0 = 0 ✓
3Lepton number: all particles are non-leptons → L = 0 on both sides ✓
All conservation laws satisfied → this interaction is allowed
A photon of energy 2.50 MeV undergoes pair production. What is the kinetic energy available for the produced electron and positron? (1 MeV = 1.60 × 10⁻¹³ J)
1Rest mass energy of one electron = 0.511 MeV. Two particles = 2 × 0.511 = 1.022 MeV
2KE available = E_photon − 2m_e c² = 2.50 − 1.022 = 1.478 MeV
3Shared between electron and positron (symmetrically if identical): each gets ~0.739 MeV
Total KE available = 1.48 MeV (to 3 s.f.)

1. What is the antiparticle of the neutron?

2. Why are two photons produced in electron-positron annihilation rather than one?

3. What is the minimum photon energy (in MeV) required for electron-positron pair production?

4. The baryon number of an antiproton is:

5. Two photons are produced when a proton and antiproton annihilate at rest. Each photon has energy equal to m_p c². Calculate this energy in MeV. (m_p = 1.673 × 10⁻²⁷ kg; 1 MeV = 1.60 × 10⁻¹³ J)

1. A photon of wavelength 1.80 × 10⁻¹² m undergoes pair production. Calculate (a) the energy of the photon in joules and eV, and (b) the kinetic energy shared by the produced pair.

2. Determine whether the following interaction conserves all laws: n → p + e⁻ + ν_e. Check charge, baryon number and electron lepton number.

3. PET (Positron Emission Tomography) scanners rely on annihilation photons. Explain why the two photons are detected at exactly 180° to each other, and suggest why this is useful in medical imaging.