Question 4 of 7: Pair-Production Threshold — Nuclear Field vs. Electron Field

Question 4: Pair-Production Threshold — Nuclear Field vs. Electron Field (6 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Pair-production threshold in a nuclear Coulomb field $E_{\gamma,\text{nucl}}=1.022$ MeV; threshold in an electron's field (triplet production) $E_{\gamma,e}=2.044$ MeV; electron rest energy $m_ec^2=0.511$ MeV.

Find. The physical reason the two thresholds differ by exactly a factor of two.

Approach. Pair production always requires the photon's energy to exceed $2m_ec^2$ (the rest-mass energy of the created $e^-e^+$ pair), but momentum as well as energy must be conserved; the "other body" in the interaction (nucleus or electron) must absorb the photon's recoil momentum, and how much kinetic energy that recoil costs depends entirely on the recoiling body's mass.

1. Nuclear field — negligible recoil energy. A nucleus is thousands of times more massive than an electron, so for a given recoil momentum $p$ its kinetic energy $p^2/2M$ is negligible (it can absorb essentially unlimited momentum for almost no energy cost). The photon's energy is therefore free to go almost entirely into creating the pair's rest mass: $$E_{\gamma,\text{nucl}}^{\text{threshold}} \approx 2m_ec^2 = 2(0.511\ \text{MeV})$$ $$\boxed{E_{\gamma,\text{nucl}}^{\text{threshold}} = 1.022\ \text{MeV}}$$
2. Electron field — comparable-mass recoil (triplet production). When the "target" providing the Coulomb field is itself an electron, its mass is exactly comparable to the particles being created, so it can no longer be treated as an infinitely heavy, energy-free recoil absorber — conservation of momentum forces the target electron itself to recoil with significant kinetic energy. At threshold all three final particles (the created $e^-e^+$ pair and the recoiling target electron) move together with the same velocity, so the minimum photon energy must supply the rest-mass energy of three electron masses, not two. Using the invariant $s=(E_\gamma+m_ec^2)^2-E_\gamma^2$ and requiring $s=(3m_ec^2)^2$ at threshold: $$2E_\gamma m_ec^2 + (m_ec^2)^2 = 9(m_ec^2)^2 \;\Longrightarrow\; E_\gamma = 4m_ec^2$$ $$\boxed{E_{\gamma,e}^{\text{threshold}} = 4m_ec^2 = 4(0.511\ \text{MeV}) = 2.044\ \text{MeV}}$$ exactly double the nuclear-field threshold, because a comparable-mass recoiling body must carry away a full extra rest-mass-energy's worth of kinetic energy that a heavy nucleus would not.