Protium ($^1H$) has a nuclear binding energy near 0. Meanwhile, lithium and deuterium have higher binding energies. Why is it that they are burned first (or at lower temperatures)? Is it perhaps related to the fact that hydrogen fusion is very difficult to achieve since over 99.99% of fusion attempts result in the diproton to undergo fission instead of beta decay?
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$\begingroup$ Protium has a nuclear binding energy exactly equal to zero. $\endgroup$– User123Aug 30, 2021 at 14:28
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1$\begingroup$ You can add ~22 more nines to that percentage. $\endgroup$– PM 2RingAug 30, 2021 at 14:41
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1 Answer
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Deuterium burns first because its combination with a proton or another deuteron is a pure strong nuclear force interaction, with the minimum of Coulomb repulsion. It occurs at lower temperatures than protium burning because although getting two protons together requires tunneling through a similar Coulomb barrier, a weak interaction is also required to turn the unstable diproton into stable deuterium after fusion. In other words, you need to bring two protons together many times before they will fuse to deuterium and this means a higher temperature is required.
Lithium burning is intermediate (i.e. it occurs at temperatures between that of deuterium and protium burning). Although it is a strong nuclear force interaction that is required to form a couple of stable helium nuclei, it does initially require tunneling through a $3\times$ higher Coulomb barrier to bring a Li nucleus and a proton together and that is why it needs higher temperatures (roughly by a factor of 3) than deuterium burning.
