<p>The slowest reaction rate in the pp chain determines how quickly hydrogen can &quot;burn&quot; in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay.</p>
<p>The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of the PPII series of reactions that convert helium 3 into helium 4. It does not involve weak interactions and therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants. It is therefore initiated at lower temperatures - as you say, Li &quot;burning&quot; occurs in low-mass pre main sequence stars well before hydrogen &quot;burning&quot; commences (but after deuterium burning).</p>
<p>At typical densities in the interiors of (proto)stars, the ignition temperatures for deuterium, lithium and protium are <em>around</em> <span class="math-container">$10^6$</span> K, <span class="math-container">$3\times 10^{6}$</span> K and <span class="math-container">$10^7$</span> K respectively.</p>

CC BY-SA 3.0
2016-06-05T12:53:26.027Z
Hmm...so if a star's core is hot enough for lithium burning, but not for hydrogen burning, could it still be considered a protostar?
Sir Cumference
CC BY-SA 3.0
2016-06-05T13:26:33.707Z
@SirCumference A star doesn't know it is a star until it starts hydrogen burning. There is little physics that distinguishes between a 0.1 solar mass "star", that will eventually burn H after contracting for about 200 million years, and a 0.02 solar mass brown dwarf, when both are 10 million years old. I suppose pedantically you could distinguish a protostar from a brown dwarf based on its mass.
ProfRob
CC BY-SA 3.0
2016-06-13T05:10:13.523Z
So why does helium require such high temperatures to fuse, even when it has a lower Coulomb repulsion than lithium? It doesn't require $\beta^+$ decay.
Sir Cumference
CC BY-SA 3.0
2016-06-13T07:40:47.813Z
@SirCumference (a) Coulomb repulsion between two He nuclei is larger than between Li and a proton. (b) You have to get THREE in the same place at once - it's the "triple alpha reaction".
ProfRob
CC BY-SA 3.0
2016-06-13T11:38:42.523Z
Why is that? It's two protons vs three protons.
Sir Cumference
CC BY-SA 3.0
2016-06-13T12:00:40.183Z
@SirCumference The repulsive potential is $q_1 q_2/r$.  What is two protons vs three protons? You are comparing $1\times 3$ with $2\times2$.
ProfRob
CC BY-SA 3.0
2017-08-31T08:08:50.677Z
@RobJeffries: When you say that weak interactions are involved, do you speak about the process $Be + e^{-} \rightarrow Li + \nu$, which I guess would be a bottleneck?
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-08-31T09:16:59.087Z
@AtmosphericPrisonEscape the only weak interaction I mention is the diproton decaying to a deuteron. First step in the pp chain.
ProfRob
CC BY-SA 4.0
2016-06-05T12:26:16.223Z
2023-11-29T15:59:46.067Z
2023-11-29T15:59:46.067Z
ProfRob
<p>This is a basic question, but it's been bugging me. In the Wikipedia article for <a href="https://en.wikipedia.org/wiki/Lithium_burning" rel="noreferrer">lithium burning</a>, it states that:</p>

<blockquote>
  <p>Stars, which by definition must achieve the high temperature (2.5 × 10^6 K) necessary for fusing hydrogen, rapidly deplete their lithium. This occurs by a collision of lithium-7 and a proton producing two helium-4 nuclei. The temperature necessary for this reaction is just below the temperature necessary for hydrogen fusion.</p>
</blockquote>

<p>I would imagine that with lithium having more protons, it would have a stronger Coulomb repulsion and require higher temperatures to fuse with hydrogen. Now, this article is pretty sketchy since it cites no sources, and I would normally dismiss this. But according to <a href="http://iopscience.iop.org/article/10.1086/304151/fulltext/35775.text.html" rel="noreferrer">here</a>, lithium burning occurs in protostars, before hydrogen fusion even takes place. How could lithium fuse with hydrogen at temperatures that low?</p>

<star><temperature><stellar-astrophysics><hydrogen><nucleosynthesis>
Why does lithium fuse at lower temperatures than hydrogen?
Sir Cumference
# Why does lithium fuse at lower temperatures than hydrogen?

(Asked by: username_0 on 2016-06-05)

This is a basic question, but it's been bugging me. In the Wikipedia article for [lithium burning](https://en.wikipedia.org/wiki/Lithium_burning), it states that:

> Stars, which by definition must achieve the high temperature (2.5 × 10^6 K) necessary for fusing hydrogen, rapidly deplete their lithium. This occurs by a collision of lithium-7 and a proton producing two helium-4 nuclei. The temperature necessary for this reaction is just below the temperature necessary for hydrogen fusion.

I would imagine that with lithium having more protons, it would have a stronger Coulomb repulsion and require higher temperatures to fuse with hydrogen. Now, this article is pretty sketchy since it cites no sources, and I would normally dismiss this. But according to [here](http://iopscience.iop.org/article/10.1086/304151/fulltext/35775.text.html), lithium burning occurs in protostars, before hydrogen fusion even takes place. How could lithium fuse with hydrogen at temperatures that low?

**Tags:** star, temperature, stellar-astrophysics, hydrogen, nucleosynthesis


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## 1 answers

### Answer by username_1 on 2016-06-05. Score: 12 | Accepted answer

The slowest reaction rate in the pp chain determines how quickly hydrogen can "burn" in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay.

The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of the PPII series of reactions that convert helium 3 into helium 4. It does not involve weak interactions and therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants. It is therefore initiated at lower temperatures - as you say, Li "burning" occurs in low-mass pre main sequence stars well before hydrogen "burning" commences (but after deuterium burning).

At typical densities in the interiors of (proto)stars, the ignition temperatures for deuterium, lithium and protium are *around* $10^6$ K, $3\times 10^{6}$ K and $10^7$ K respectively.

> **Comments:**
> **username_0**: Hmm...so if a star's core is hot enough for lithium burning, but not for hydrogen burning, could it still be considered a protostar?  
> **username_1**: @username_0 A star doesn't know it is a star until it starts hydrogen burning. There is little physics that distinguishes between a 0.1 solar mass "star", that will eventually burn H after contracting for about 200 million years, and a 0.02 solar mass brown dwarf, when both are 10 million years old. I suppose pedantically you could distinguish a protostar from a brown dwarf based on its mass.  
> **username_0**: So why does helium require such high temperatures to fuse, even when it has a lower Coulomb repulsion than lithium? It doesn't require $\beta^+$ decay.  
> **username_1**: @username_0 (a) Coulomb repulsion between two He nuclei is larger than between Li and a proton. (b) You have to get THREE in the same place at once - it's the "triple alpha reaction".  
> **username_0**: Why is that? It's two protons vs three protons.  
> **username_1**: @username_0 The repulsive potential is $q_1 q_2/r$.  What is two protons vs three protons? You are comparing $1\times 3$ with $2\times2$.  
> **username_2**: @RobJeffries: When you say that weak interactions are involved, do you speak about the process $Be + e^{-} \rightarrow Li + \nu$, which I guess would be a bottleneck?  
> **username_1**: @username_2 the only weak interaction I mention is the diproton decaying to a deuteron. First step in the pp chain.  

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