Why does deuterium burn at lower temperature than H

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    Deuterium Temperature
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SUMMARY

Deuterium fusion occurs at approximately 106K, while hydrogen fusion requires around 107K due to the differences in the forces involved. The fusion of two hydrogen nuclei (1H) to form deuterium (2H) involves a weak force reaction, which includes the transformation of a proton into a neutron, emitting a positron and an electron neutrino. In contrast, deuterium fusion relies solely on the nuclear force, allowing it to occur at lower temperatures. This makes the creation of deuterium a rate-limiting step in the proton-proton (p-p) chain reaction of hydrogen fusion.

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In general, nuclear fusion happens at a higher temperature for heavier elements. In my astrophysics class we derived an equation for probability of nuclear reactions, and without quoting it exactly, it boils down to have this probability proportional to exp(-mrZ1Z2), where mr is the reduced mass of colliding particles, Zn are the atomic numbers.

So how come then, that deuterium burning occurs at roughly 106K while regular hydrogen burns at 107K?
 
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Fusing two 1H to form deuterium also involves a weak force reaction, as one of the protons has to transform itself into a neutron (giving off a positron and an electron neutrino). Fusion of deuterium only involves the nuclear force.
 
Makes perfect sense, thank you. As I see now, creating 2H is one of the steps in p-p chain of 1H fusion, and rate limiting one at that. If you already have 2H, the fusion can take place at much lower temperatures.
 
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