<p>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 <em>many</em> times before they will fuse to deuterium and this means a higher temperature is required.</p>
<p>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 <span class="math-container">$3\times$</span> 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.</p>

CC BY-SA 4.0
2021-08-30T17:27:14.283Z
2021-08-31T06:58:05.287Z
2021-08-31T06:58:05.287Z
ProfRob
<p>Protium (<span class="math-container">$^1H$</span>) 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?</p>

CC BY-SA 4.0
2021-08-30T14:28:27.930Z
Protium has a nuclear binding energy exactly equal to zero.
User123
CC BY-SA 4.0
2021-08-30T14:41:36.900Z
You can add ~22 more nines to that percentage.
PM 2Ring
<star><stellar-evolution><nucleosynthesis>
Why is it that lithium and deuterium are burned before hydrogen?
WarpPrime
# Why is it that lithium and deuterium are burned before hydrogen?

(Asked by: username_0 on 2021-08-30)

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?

**Tags:** star, stellar-evolution, nucleosynthesis

> **Comments:**
> **username_1**: Protium has a nuclear binding energy exactly equal to zero.  
> **username_2**: You can add ~22 more nines to that percentage.  

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## 1 answers

### Answer by username_3 on 2021-08-30. Score: 8 | Accepted answer

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.


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