Pair-Production Threshold in a Nuclear vs. an Electron Field: 17-Phys-B1 Radiation Physics, May 2016, Q4

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

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Given. Pair production requires simultaneous conservation of total energy and momentum; a third body (nucleus or electron) must absorb recoil momentum since a photon alone cannot decay into an $e^+e^-$ pair in vacuum.

Find. Why the threshold is $2m_ec^2=1.022$ MeV for a (heavy) nuclear field but $4m_ec^2=2.044$ MeV for an electron field.

Approach. Write the threshold condition as the minimum invariant mass of the final state (all products moving together at threshold) and evaluate it for a target of mass $M$, then take the two limits $M\to\infty$ (nucleus) and $M=m_e$ (electron, "triplet production").

General threshold formula. At threshold, the photon (energy $E_\gamma$, momentum $E_\gamma/c$) strikes a stationary target of mass $M$; conservation of energy and momentum together require the final-state invariant mass to be minimized, which occurs when every final particle moves with the same velocity (nothing left over as extra kinetic energy). Working in natural units ($c=1$), the invariant $s=(E_\gamma+M)^2-E_\gamma^2=2E_\gamma M+M^2$ must equal the square of the total final rest mass. Setting that equal to $(M+2m_e)^2$ (target $M$ plus the created pair, $2m_e$) and solving for $E_\gamma$: $$E_{\gamma,\text{th}} = 2m_e\left(1+\frac{m_e}{M}\right)$$
Pair production in a nuclear field. A nucleus has $M\gg m_e$ (by a factor of thousands), so $m_e/M\to0$ and $$\boxed{E_{\gamma,\text{th}}(\text{nucleus}) = 2m_ec^2 = 2(0.511\ \text{MeV}) = 1.022\ \text{MeV}}$$ Physically, the heavy nucleus can absorb whatever recoil momentum is needed while carrying away negligible kinetic energy ($p^2/2M\to0$), so essentially all of $E_\gamma$ converts directly into the two electron rest masses.
Pair production in an electron field (triplet production). Here the target has the same mass as the created particles, $M=m_e$, so $m_e/M=1$ and $$\boxed{E_{\gamma,\text{th}}(\text{electron}) = 2m_ec^2(1+1) = 4m_ec^2 = 4(0.511\ \text{MeV}) = 2.044\ \text{MeV} = 2\times E_{\gamma,\text{th}}(\text{nucleus})}$$ Physically, the target electron is no longer a passive, near-stationary recoil absorber: because it has the same mass as the created pair, it necessarily carries away a non-negligible share of the momentum and kinetic energy at threshold, so more photon energy must be supplied before the process can occur at all — the final state now comprises three free electron-mass particles (the recoiling target plus the created $e^+e^-$ pair) instead of two, exactly doubling the threshold.

Question 4 — results
Field	Threshold energy
Nuclear field ($M\gg m_e$)	$2m_ec^2=1.022$ MeV
Electron field ($M=m_e$, triplet production)	$4m_ec^2=2.044$ MeV

Topic. Pair production threshold and the role of the recoiling third body.

Key relations. $E_{\gamma,\text{th}}=2m_ec^2(1+m_e/M)$; nuclear limit $M\to\infty\Rightarrow2m_ec^2$; electron ("triplet") limit $M=m_e\Rightarrow4m_ec^2$.

Why this works. A single free photon can never pair-produce in vacuum (a two-body final state cannot simultaneously conserve energy and momentum with a massless initial particle); a third body of finite mass is required precisely to carry away recoil momentum, and how much extra energy that costs depends on how heavy the recoiling body is.

Common pitfall. Assuming the threshold is always $2m_ec^2$ regardless of what the pair forms near; it only reaches that minimum in the heavy-nucleus limit.

Source. F. H. Attix, Introduction to Radiological Physics and Radiation Dosimetry, Ch. 7 (pair and triplet production).

Question 4 of 7: Pair-Production Threshold in a Nuclear vs. an Electron Field

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