Annihilation and Pair Production

Spec mapping: OCR H556 Module 6.4 — Annihilation and pair production. Particle-antiparticle annihilation $e^- + e^+ \to 2\gamma$e−+e+→2γ producing back-to-back $511$511 keV photons; the requirement of at least two photons from momentum conservation; pair production $\gamma \to e^- + e^+$γ→e−+e+ near a nucleus with photon-energy threshold $E_\gamma \geq 2 m_e c^2 = 1.022$Eγ​≥2me​c2=1.022 MeV; the role of conservation of energy, momentum, charge, lepton number and baryon number in both processes; the link to PET scanning later in the module. Refer to the official OCR H556 specification document for exact wording.

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 later in this course.

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 Module 6.4 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:

$E_{\gamma} = m_{e} c^{2} = 0.511\,\text{MeV} = 511\,\text{keV}$Eγ​=me​c2=0.511MeV=511keV

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:

$E_{\gamma} = \dfrac{1.222}{2} = 0.611\,\text{MeV}$Eγ​=21.222​=0.611MeV

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:

$E_{\text{min}} = 2 m_{e} c^{2} = 2 \times 0.511 = 1.022\,\text{MeV}$Emin​=2me​c2=2×0.511=1.022MeV

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:

$\text{KE}_{\text{total}} = E_{\gamma} - 2 m_{e} c^{2}$KEtotal​=Eγ​−2me​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:

$$\begin{aligned} E_{\text{min}} &= 2 m_{e} c^{2} = 2 \times 0.511\,\text{MeV} = 1.022\,\text{MeV} = 1.022 \times 10^{6} \times 1.60 \times 10^{-19}\,\text{J} \\ &\approx 1.635 \times 10^{-13}\,\text{J} \end{aligned}$$Emin​​=2me​c2=2×0.511MeV=1.022MeV=1.022×106×1.60×10−19J≈1.635×10−13J​

Minimum frequency from E = hf:

$$\begin{aligned} f_{\text{min}} = E / h = 1.635 \times 10^{-13} / 6.63 \times 10^{-34} \\ \approx 2.47 \times 10^{20} Hz \end{aligned}$$fmin​=E/h=1.635×10−13/6.63×10−34≈2.47×1020Hz​

That's a gamma ray — specifically, a high-energy gamma well above medical imaging energies (which are tens to hundreds of keV). Pair production happens inside cosmic ray showers and in particle accelerators, but not in a hospital X-ray room.

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Worked Example 3: Kinetic energy of an electron-positron pair

A 2.5 MeV photon undergoes pair production into an e^- e^+ pair. Assuming the pair shares the available kinetic energy equally, what is the kinetic energy of each particle?

Solution. Available kinetic energy:

$\text{KE}_{\text{total}} = E_{\gamma} - 2 m_{e} c^{2} = 2.5 - 1.022 = 1.478\,\text{MeV}$KEtotal​=Eγ​−2me​c2=2.5−1.022=1.478MeV

Shared equally:

$\text{KE}_{\text{each}} = 0.739\,\text{MeV}$KEeach​=0.739MeV

In reality the sharing is not always equal, but energy conservation tells us the sum of the kinetic energies is 1.478 MeV regardless.

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Annihilation and Pair Production: Schematic

flowchart TB
    subgraph Annihilation
        em["e⁻"]
        ep["e⁺"]
        g1["γ (511 keV)"]
        g2["γ (511 keV)"]
        em --> g1
        em --> g2
        ep --> g1
        ep --> g2
    end
    subgraph PairProduction
        g["γ (E > 1.022 MeV)"]
        N["Nucleus<br/>(provides momentum)"]
        em2["e⁻"]
        ep2["e⁺"]
        g --> em2
        g --> ep2
        N --> em2
        N --> ep2
    end

In annihilation, momentum conservation forces the two photons to emerge in exactly opposite directions when the particles are at rest. In pair production, the nucleus absorbs the small amount of recoil momentum that the e^- e^+ pair cannot balance by themselves.

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Applications and Consequences

Annihilation is the physical basis of PET scanning (Lesson 14). Positron-emitting tracers are injected into a patient; positrons travel at most a millimetre or so in tissue before annihilating with a nearby electron; the resulting 511 keV photons are detected by a ring of detectors surrounding the patient. Because the photons fly back-to-back, a coincident pair of detections defines a line along which the annihilation took place, and enough such lines can be combined by computer to reconstruct a three-dimensional image.

Pair production occurs naturally in cosmic ray air showers (a high-energy gamma from a supernova or active galaxy enters the atmosphere, produces an e^- e^+ pair, which emits bremsstrahlung radiation, which pair-produces again, and so on). It is one of the processes that attenuates high-energy X-rays and gamma-rays in matter, especially above 1.022 MeV. At medical imaging energies (keV to hundreds of keV), photoelectric absorption and Compton scattering dominate, and pair production is negligible.

Historically, the positron was discovered via pair production in cloud chambers in 1932. Before then, Dirac's prediction (1928) that every fermion should have an antiparticle was regarded with scepticism; after Anderson's cloud-chamber photographs, it became one of the central pillars of modern physics.

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Conservation Laws

In both annihilation and pair production, all of the following are conserved:

• Total energy (including rest energy).
• Total momentum (vector sum).
• Electric charge.
• Lepton number.
• Baryon number.
• Angular momentum.

Let's check annihilation:

e⁻ (charge -1, L = +1) + e⁺ (charge +1, L = -1) → 2γ (charge 0, L = 0)

Charge: -1 + 1 = 0 ✓. Lepton number: +1 + (-1) = 0 ✓. Energy and momentum: conserved by emitting two photons in opposite directions with total energy 2 m_e c^2.

And pair production:

γ (L = 0) → e⁻ (L = +1) + e⁺ (L = -1)

Charge: 0 = -1 + 1 ✓. Lepton number: 0 = +1 + (-1) ✓. Energy: photon must carry at least 2 m_e c^2. Momentum: balanced with the help of a nearby nucleus.

It is not possible, for example, to have \gamma \to e^- + e^- (charge not conserved) or e^- + e^- \to \gamma (charge not conserved, lepton number not conserved). Conservation laws are strict.

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Annihilation and Pair Production: Diagrams

The two processes are inverses of each other. The geometry of each is dictated by momentum conservation:

In annihilation, the two photons must emerge in opposite directions (back-to-back) when the initial particles are at rest, because the initial momentum is zero. A single photon would carry a non-zero momentum and so violate the conservation law. In pair production, the photon's forward momentum must be balanced by the products: the $e^-$e− and $e^+$e+ alone carry too little, so a heavy "spectator" — a nearby nucleus — absorbs the recoil. Without the nucleus the process is kinematically forbidden, which is why pair production never occurs in pure vacuum.

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Synoptic Links

• Mass-energy equivalence (earlier in Module 6.4): $E = mc^2$E=mc2 is invoked literally in both processes. Annihilation converts the entire rest energy of the $e^- e^+$e−e+ pair into photon energy; pair production converts photon energy back into rest energy plus kinetic energy. The two are the cleanest experimental demonstration of Einstein's relation at the particle scale.
• Conservation laws (next lesson): Energy, momentum, charge, baryon number and lepton number must all balance. Annihilation and pair production are textbook applications of all five rules at once, and exam questions routinely test the conservation reasoning rather than just numerical thresholds.
• Photon energy (Module 4.5): $E = hf = hc/\lambda$E=hf=hc/λ from the photoelectric effect is the energy formula you substitute into both the $511$511 keV annihilation line and the $1.022$1.022 MeV pair-production threshold. The same photon-energy machinery as for visible-light experiments is used at $\sim 10^5$∼105 higher energies.
• PET scanning (later in this module): The $511$511 keV back-to-back photon pair is the entire physical basis of positron emission tomography. The "line of response" defined by a coincident detection event is exactly the back-to-back direction predicted here. Annihilation is the PET signature.
• Quarks, leptons and hadrons (last lesson): Lepton-number conservation is the reason that $\gamma \to e^- + e^+$γ→e−+e+ is allowed ($L = 0 \to L = +1 + (-1) = 0$L=0→L=+1+(−1)=0) but $\gamma \to e^- + \mu^+$γ→e−+μ+ would be forbidden (electron lepton number not conserved). The conservation reasoning links directly to the lepton families introduced in the last lesson.

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Specimen question modelled on the OCR H556 paper format