Given hydrogen is a lighter and more abundant element, why do most baby stars start off burning deuterium?

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    $\begingroup$ Deuterium is relatively easy to burn, compared to protium (plain hydrogen). And the first steps in the proton-proton chain actually produce deuterium, which is quickly converted to He-3. $\endgroup$
    – PM 2Ring
    Commented May 27, 2019 at 15:32

1 Answer 1


The threshold temperature at which D-burning becomes significant is significantly lower than that to produce significant energy from the pp H-burning chain.

Deuterium is found as a trace element in the gas from which stars form. It is depleted from the entire star well before the hydrogen burning pp chain (or CNO cycle in more massive stars) begins. Essentially, D-burning is the faster second step in the pp-chain, the slowest initial step being the formation of new deuterium in the first place.

Thus, whilst D-burning takes place at temperatures of $10^6$ K, the pp chain does not start significant energy production until temperatures exceed $10^7$ K. The underlying reason that D-burning happens more easily, is that getting two singly positively charged nuclei together is only part of the problem and is similar for isolated protons or deuterons - however a diproton is unstable and falls apart before fusion can take place unless one of the protons can turn into a neutron via a weak-force interaction to form a stable deuteron. If the deuterium nuclei are already formed, then their fusion with another proton occurs relatively quickly without having to go through the slow first step.


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