Why does deuterium burn at lower temperature than H

In summary, nuclear fusion reactions occur at higher temperatures for heavier elements and have a probability proportional to exp(-mrZ1Z2), where mr is the reduced mass of colliding particles and Zn are the atomic numbers. However, deuterium burning occurs at a lower temperature of 106K compared to regular hydrogen burning at 107K. This is because creating deuterium involves a weak force reaction, while fusion of deuterium only requires the nuclear force. The creation of 2H is a limiting step in the p-p chain of 1H fusion, and having 2H already present allows for fusion to occur at lower temperatures.
  • #1
beee
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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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  • #2
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.
 
  • #3
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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