Another discussion on this site (Why is there a scarcity of lithium?) as well as the wikipedia page on Lithium burning (https://en.wikipedia.org/wiki/Lithium_burning), have helped me understand why stellar nucleosynthesis doesn't generally produce a net amount of Lithium out of He-H reactions: because the Li thus produced quickly absorbs more protons (H nuclei), turning into unstable forms of Be, and then back into 2 He nuclei. But since stars also are about 1/4 He, why doesn't some of the Li absorb, not a H but a He nucleus, turning into a stable form of Boron (B)?
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2 Answers
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The usual lithium burning reaction with protons is extremely temperature sensitive - something like $T^{20}$ in typical stellar interiors (Bildsten et al. 1997).
The Coulomb barrier for fusion with He, rather than protons, is twice as high and could only be initiated at higher temperatures.
Therefore, at the temperatures required for Li+He fusion, the Li has already been destroyed.
Lithium is produced inside stars (in very small quantities) as part of the pp chain. Specifically, it is part of the pp-II chain, which occurs above temperatures of $15\times 10^6$ K. At these temperatures the rate of Li destruction by protons is incredibly rapid, and even if a very small amount could fuse with He to produce Boron, then at these temperatures, the boron itself would also be destroyed by proton capture if $T>5\times 10^{6}$K. Thus there is no route to boron creation in stellar interiors and the majority of cosmic boron is thought to arise by spallation reactions between CNO nuclei and cosmic ray protons (or vice-versa).
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$\begingroup$ Interesting. How is Boron produced then? Spallation? $\endgroup$– user154997Commented Aug 5, 2017 at 14:17
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$\begingroup$ @LucJ.Bourhis Yes, spallation. Interaction of cosmic rays with C,N,O nuclei. $\endgroup$– ProfRobCommented Aug 5, 2017 at 14:47
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$\begingroup$ @ScottForschler Temperature to the power of 20. Standard maths notation! Boron destruction takes place at higher temperatures than Li burning. No, I am not saying it is unlikely because of the reaction rate; stellar cores easily get hot enough; I am saying that one of the reactants is not present at those temperatures. $\endgroup$– ProfRobCommented Aug 5, 2017 at 18:45
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$\begingroup$ @ScottForschler There is no Li in any region of a star where the temperature exceeds $3\times 10^{6}$K. It is produced as part of the pp chain of nuclear burning in main sequence stars, but since this happens at higher temperatures there is negligible Li available to react with He and even I guess the miniscule trace that could produce Boron, well the Boron would also be destroyed. I'll put an edit in. $\endgroup$– ProfRobCommented Aug 6, 2017 at 18:45
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$\begingroup$ Thank you Rob, the edit makes the whole picture much clearer; I'll delete my intermediate comments/questions as they are now answered. $\endgroup$ Commented Aug 6, 2017 at 21:52
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There is no fundamental reason the reaction can't happen. The reaction $Li~+~He~\rightarrow~B$ is however not energetically that favorable. In a supernova core fusion of elements into the actinide series can happen, but they are endothermic. Fusion processes that generate nuclei with an odd number of nucleons tend as a rule not to be as energetically favorable. The binding curve of energy permits one to see this somewhat for low mass elements.
answered Aug 5, 2017 at 14:13
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1$\begingroup$ It is not a question of energetics. It is that one of the reactants (Li) is not present at the temperatures required to initiate fusion. $^{10}B$ does not have an odd number of nuclei and could be produced by alpha capture onto $^6Li$, if any such Li existed. $\endgroup$– ProfRobCommented Aug 5, 2017 at 18:57