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We know that a gamma photon with energy that is higher than 1.02 MeV produce electron and positron when its close to the nucleus.

So, my question is if the photon has a very high energy, lets say 70 MeV or something. Why it does not produce 2 electron and 2 positrons (2 pair or 4 body decay) instead of only 1 pair?

Is that suppressed by the coupling constant of QED or its dynamically impossible?

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  • $\begingroup$ It's certainly not dynamically impossible. @anna v 's answer covers that. $\endgroup$ – Cosmas Zachos Feb 18 '20 at 14:32
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In this article of the Wikipedia the full photon nucleus interaction for pair production is given

In pair production, a photon creates an electron positron pair. In the process of photons scattering in air (e.g. in lightning discharges), the most important interaction is the scattering of photons at the nuclei of atoms or molecules. The full quantum mechanical process of pair production can be described by the quadruply differential cross section given here

It is not a simple matter.

So, my question is if the photon has a very high energy, lets say 70 MeV or something. Why it does not produce 2 electron and 2 positrons (2 pair or 4 body decay) instead of only 1 pair?

There will be a probability for this to happen, equally complicated as the one in the link , appresicably smaller because there will be more electromagnetic vertices.

For high enough energy the strong interaction takes over , and hadron pairs can be produced, see this for example.

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