Annihilation and Pair Production

Two of the most striking phenomena in particle physics are the direct conversion of matter into radiation and of radiation into matter. When a particle and its antiparticle meet, they annihilate, vanishing and leaving behind a pair of high-energy photons carrying all the mass-energy. Conversely, when an energetic photon passes near a nucleus, it can disappear and leave behind a particle-antiparticle pair: pair production. Both processes are dramatic demonstrations of Einstein's mass-energy equivalence, and the annihilation process is the physical basis of PET scanning — the medical imaging technique we shall meet in Lesson 14.

This lesson treats annihilation and pair production quantitatively, introduces the minimum-energy thresholds for each process, and works through the calculations you will see in OCR exam questions. The material forms part of section 6.4.4 of the OCR A-Level Physics A specification (H556).

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Annihilation

Annihilation is what happens when a particle meets its antiparticle: the two particles vanish and their combined mass-energy appears as electromagnetic radiation (usually two photons). The most common example, and the one relevant to medical imaging, is electron-positron annihilation:

e⁻ + e⁺ → 2γ

If the electron and positron are at rest (or nearly so) when they meet, the two photons emerge in exactly opposite directions (back-to-back) to conserve momentum, and each carries an energy equal to the rest energy of the electron:

E_γ = m_e c² = 0.511 MeV = 511 keV

This is the characteristic 511 keV annihilation line, seen in any experiment with positron sources and in every PET scanner in the world.

Why two photons and not one? Because of momentum conservation. In the rest frame of the electron-positron pair, the total momentum is zero. A single photon always has non-zero momentum (p = E/c), so you cannot produce just one photon from a zero-momentum initial state. You need at least two, and they must emerge in opposite directions with equal energy.

Why not three? You can get three photons (about 0.3% of the time), and four and five are also allowed. But two is overwhelmingly the most common outcome for an electron-positron pair with no net angular momentum.

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Energy Conservation in Annihilation

Let us make the energetics explicit. Before annihilation:

Total energy = 2 m_e c² + KE

where KE is any kinetic energy the particles have (often negligible in the laboratory). After annihilation, the two photons carry away this total energy:

2 E_γ = 2 m_e c² + KE
E_γ = m_e c² + KE/2

If the initial kinetic energy is small compared to m_e c^2, each photon simply carries 0.511 MeV. If the kinetic energy is significant, the photons carry slightly more — but in medical physics the positrons emitted by PET tracers lose their kinetic energy very quickly (within a millimetre of tissue) and annihilate essentially at rest, so the 511 keV value is exact to a very good approximation.

Worked Example 1: Photon energy from annihilation of a fast positron

A positron with kinetic energy 0.2 MeV annihilates with an electron at rest. Assuming the two photons emerge in opposite directions with equal energy, what is the energy of each photon?

Solution. Total energy before annihilation:

E = 2 m_e c² + KE
  = 2(0.511) + 0.2
  = 1.222 MeV

Each photon carries:

E_γ = 1.222/2 = 0.611 MeV

Each photon has 0.611 MeV, which is more than 511 keV — the excess coming from the initial kinetic energy.

In reality, if the positron has significant kinetic energy, the two photons need not be collinear or equal-energy; but the total energy and total momentum are conserved. For PET imaging, the positron is treated as annihilating at rest.

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Pair Production

Pair production is the inverse of annihilation. A single high-energy photon, in the presence of a nucleus (which is needed to conserve momentum), can disappear and create a particle-antiparticle pair:

γ → e⁻ + e⁺  (near a nucleus)

The nucleus is essential. A photon alone cannot create an e^- e^+ pair, because in the photon's rest frame (there isn't one, but take a frame where the two final particles are at rest momentarily) momentum would not be conserved. A nearby nucleus can absorb a tiny amount of recoil momentum — almost without gaining any energy, because it is so much heavier than the electron — and make the process kinematically allowed.

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Threshold Energy

To create an electron-positron pair at rest, the photon must supply at least the total rest energy of the pair:

E_min = 2 m_e c² = 2 × 0.511 = 1.022 MeV

Photons with less energy than this cannot produce an e^- e^+ pair, no matter what. Photons with more energy can produce the pair and give the extra energy as kinetic energy of the electron and positron. In general, any kinetic energy comes out as:

KE(total) = E_γ - 2 m_e c²

for pair production of an electron-positron pair.

More massive pairs need correspondingly higher threshold energies:

• μ⁻ + μ⁺: E_min = 2 × 105.7 = 211.4 MeV
• p + p̄: E_min = 2 × 938.3 = 1876.6 MeV

To produce heavy particle pairs you need accelerators — the 3.8 GeV Bevatron was the first machine powerful enough to produce antiprotons (1955). In electron-positron colliders (SLC, LEP), e^+ e^- collide at high energy and annihilate, and the energy can then re-materialise as any particle-antiparticle pair allowed by conservation laws, including quark-antiquark pairs that immediately hadronise.

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Worked Example 2: Minimum frequency of a pair-producing photon

What is the minimum frequency of a photon that can produce an electron-positron pair?

Solution. Minimum photon energy:

E_min = 2 m_e c² = 2 × 0.511 MeV = 1.022 MeV = 1.022 × 10⁶ × 1.60 × 10⁻¹⁹ J
     ≈ 1.635 × 10⁻¹³ J

Minimum frequency from E = hf: