When two deuterium nuclei are shot at each other they have some

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    Deuterium Nuclei
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SUMMARY

When two deuterium nuclei collide, they have a probability of fusing into helium nuclei, transitioning from a potential well of 1 MeV to a final well of 7 MeV. During this process, the emission of a 24 MeV photon occurs, but the intermediate states at 2 MeV or 3 MeV do not emit lower energy photons. The emission of a single high-energy photon is the most probable outcome due to the nature of quantum interactions, while the formation of helium-4 is rare compared to tritium plus proton or helium-3 plus neutron outcomes, which occur via the strong interaction.

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edpell
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When two deuterium nuclei are shot at each other they have some probability of fusing into a He nuclei. At the start we have a proton and a neutron in a potential well 1 MeV deep and another proton and neutron in another well 1 MeV deep. At the end we have two protons and two neutrons all in a well 7 MeV deep. At which point a 24 MeV phton can be emitted.

My question is what happens in between? Is there a time when the four particles are in a well 2MeV deep, 3MeV, etc. And why don't they radiate lower energy photons as the well forms? Why do they wait until the end to radiate one high energy photon?
 
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Classical descriptions don't work well for such a process.

The emission of a single photon is the lowest-order process and therefore the most likely one, but it can happen that more photons are emitted.
The formation of He-4 is a rare outcome anyway, tritium plus proton or He-3 plus neutron are more common - this process can happen via the strong interaction.
 

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