Why does deuterium fusion create helium, and when does the sequence stop?

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Recently i watched a video in Khan academy about the birth of a star and i got kind of confused, a deuterium fuses with an H to create He right? then what happens if a He fuses with an H and form an isotope of
Li ? When this sequence stops?

ps: how can i apply Latex? i can't find the icon to put those [/tex] symbols
 

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In general, certain fusion reactions occur only when the conditions, like a high enough pressure or temperature, are present in the core of a star. In a star like the sun, for example, the predominant fusion reaction which occur are called the proton-proton chain, which has three main branches:

http://en.wikipedia.org/wiki/Proton–proton_chain_reaction

The reactions listed in the first branch, where protons fuse to form deuterium and then eventually helium, is the process which primarily occurs in the sun's core now. After helium forms, it essentially becomes inert, as far as fusion reactions are concerned. Helium gradually accumulates in the core of the sun, which currently is about 73.5% hydrogen and about 25% helium, with the rest being trace amounts of various elements up to and including iron. If the temperature gets hot enough, eventually branch two and three fusion reactions occur. In all of these reactions, elements like lithium, boron, and beryllium will form and then break down again constantly.

The helium will stay in the core until the sun begins to run out of hydrogen to fuel the fusion reactions (don't worry, this won't happen for billions of years), at which time the core will begin to contract, raising the temperature and pressure above the levels seen currently. When the pressure and temperature go above a certain threshold, then the helium will start fusing in a different set of fusion reactions.
 
The sequence can go on as long as energy allows it to...
the final stage is thus determined by the temperatures in the core of the star and the particle abundances (the star's mass)
After the formation of He, the star can undergo the Hellium burning period, where 3 He nuclei can be combined to give a Carbon (triple alpha process), and so on...
When the energy of the star is not enough to keep up the reactions, they stop. Ideally this can happen after Iron (Fe), because for every heavier nuclei the energy produced is less than the energy needed for the reaction. So the star in that case can't survive the collapse of its core, and will undergo a supernova. Of course not all stars can reach the Iron but "die" soon after the Helium burning is over [like the machine stops working], but when they stop depends on their cores' energy/temperature.
 
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oksuz_ said:
If a 3He fuses with 1H then 4Li can be created, but 4Li is an extremely unstable isotope of Li. Therefore it emits proton just after it is formed.
Actually, it is unbound. Most of time, the proton simply bounces off, just like it bounces off another proton.
 
Kostas Tzim said:
Li ? When this sequence stops?
No, stellar metal production stops at Fe57. Beyond that requires nova conditions.
 
Thanks all of you the information was really helpful...i understood a lot of this..im a high school student and i got confused a little bit :) One more question are there any stars that they can "work" only with Hydrogen and Deuterium?, if the energy conditions are not ideal

Happy christmas
 
If the star burned only Hydrogen, or even Deuterium, then it would be a really faint star with very small mass... (look for substellar objects)
 
thanks Chris :) i'll look for it...
 
ChrisVer said:
If the star burned only Hydrogen, or even Deuterium, ... [ ... ]
To what does a star 'burn' H and H2? Surely you are not suggesting that quenches the reaction.
 
Doug Huffman said:
To what does a star 'burn' H and H2? Surely you are not suggesting that quenches the reaction.

What do you mean? There are certain phases called eg Deuterium burning. The Deuterium is "burned" (fused) with another proton to give Helium-3
For a better put phrase, you can check here:
http://en.wikipedia.org/wiki/Substellar_object
 
ChrisVer said:
If the star burned only Hydrogen, or even Deuterium, then it would be a really faint star with very small mass... (look for substellar objects)
Our sun (as most stars) burns hydrogen only (deuterium is an intermediate product in the reaction) and converts it to helium. In the very distant future, it will burn helium to carbon (starting with a helium flash), but that is billions of years away.
 
ChrisVer said:
What do you mean? There are certain phases called eg Deuterium burning. The Deuterium is "burned" (fused) with another proton to give Helium-3 For a better put phrase, you can check here:
http://en.wikipedia.org/wiki/Substellar_object
Oh, no, the Wikipedia article is fine, and so is the one titled Deuterium burning.
 
mfb said:
Our sun (as most stars) burns hydrogen only (deuterium is an intermediate product in the reaction) and converts it to helium. In the very distant future, it will burn helium to carbon (starting with a helium flash), but that is billions of years away.

I thought the question was concerning stars that cannot reach the Helium in their core, but reach up to H or Deuterium burning phase,,,our sun so, is not an example for these type of "stars"... In general it can take more time than the Universe's age for those stars to finish their supplies...
 
SteamKing said:
In general, certain fusion reactions occur only when the conditions, like a high enough pressure or temperature, are present in the core of a star. In a star like the sun, for example, the predominant fusion reaction which occur are called the proton-proton chain, which has three main branches:

http://en.wikipedia.org/wiki/Proton–proton_chain_reaction

The reactions listed in the first branch, where protons fuse to form deuterium and then eventually helium, is the process which primarily occurs in the sun's core now. After helium forms, it essentially becomes inert, as far as fusion reactions are concerned. Helium gradually accumulates in the core of the sun, which currently is about 73.5% hydrogen and about 25% helium, with the rest being trace amounts of various elements up to and including iron. If the temperature gets hot enough, eventually branch two and three fusion reactions occur. In all of these reactions, elements like lithium, boron, and beryllium will form and then break down again constantly.

No, helium is not inert in Sun.
The first reactions of pp chain are
1)p+p->d+e+
estimated lifetime of p on Sun 10 milliards of years.
Alternative branch pep
2)p+p+e->d
about 0,23 % of the first
Either way to form d is followed by deuterium fusion
3)d+p->3-He
d lifetime 4 s
And then that 3-He is, in Sun, not inert. Main fate, pp-I (85 % in Sun):
4) 3-He+3-He->4-He+p+p

The end result, 4-He, is not inert in Sun, either. The 15 % fate of 3-He is pp-II:
5)4-He+3-He->7-Be
6)7-Be+e->7-Li
7)7-Li+p->4-He+4-He
As you see, 4-He reacted - but you get the initial 4-He back. So the net result is that you have turned a 3-He and p into the second 4-He.

But it is said that pp-II is favoured by high temperature.
What becomes of He-3 in stars that are colder than Sun?
 
It is true that helium is involved in the various parts of the proton-proton reaction chain, but there must come a time where some of the helium begins to accumulate in the stellar core and is no longer directly involved in the proton-proton reactions. Eventually, the proton-proton reactions cease, because there is no more hydrogen left to fuse in the core, and helium burning then begins in the triple-alpha process.

The reactions in branches I and II of the p-p chain are predominant in the sun now, due to its estimated core temperature of 15.7 million K.

In the proton-proton chain, I should have said there are four branches rather than three. The reactions in branch III only become significant when the core temperature exceeds 23 million K; the reactions of Branch IV are predicted to occur but have not yet been observed.
 
SteamKing said:
It is true that helium is involved in the various parts of the proton-proton reaction chain, but there must come a time where some of the helium begins to accumulate in the stellar core
It accumulates from the beginning, but as you see the net result is to make one 4-He back into two 4-He.
SteamKing said:
and is no longer directly involved in the proton-proton reactions. Eventually, the proton-proton reactions cease, because there is no more hydrogen left to fuse in the core, and helium burning then begins in the triple-alpha process.
Triple-alpha takes very much higher temperature and density than pp chain. Hydrogen is nearly exhausted long before that.
SteamKing said:
In the proton-proton chain, I should have said there are four branches rather than three. The reactions in branch III only become significant when the core temperature exceeds 23 million K; the reactions of Branch IV are predicted to occur but have not yet been observed.
pp reaction has never been observed either.
Branch IV is
3-He+p->4-He+e+
or a branch
3-He+p+e->4-He
 
i wonder, why is Branch IV so suppressed if there are a lot of protons around? :(
 
None of the fusion processes which occur in the sun's core have been directly observed, but by measuring the neutrinos coming from the sun's core, one may make reasonable deductions about what occurs there, given our state of knowledge about fusion reactions. Much of what we think we know about the sun comes from modeling its behavior, and of course, it goes without saying that models can be adjusted if their predictions do not agree with observations.
 
ChrisVer said:
i wonder, why is Branch IV so suppressed if there are a lot of protons around? :(
Because it is a weak process!
Compare:
p+p->d+e+
It is massively improbable. Diproton is not a bound state. Protons that collide with another proton just bounce almost all time.
Now, if you look at a later step
p+3-He->4-He+e+
It is even more improbable. 4-Li is not a bound state either, so a proton colliding with 3-He bounces almost all time, and for all the same reasons.
But an additional complication: 3-He has twice the charge of a proton, so a proton of a given energy bounces from twice the distance.

As you see, Branch I dominates because the reaction between two 3-He nuclei is a strong process, not weak.
 
So much information thanks...And when the sun starts producing carbon then what's happening?
 
but the other branches (II,III) contain weak inrweactions
 
ChrisVer said:
but the other branches (II,III) contain weak inrweactions
Yes - electron capture or beta decay of bound nuclei. If He-3 bounces off a proton without a weak interaction, it produces no result, and it takes long time until the next collision that finally has a result. Whereas a Be-7 produced in pp-II is bound and never returns to He-3 - it stays around till it either captures an electron in pp-II or a proton in pp-III.
 
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Kostas Tzim said:
And when the sun starts producing carbon then what's happening?
Nearly no hydrogen is left then (and at the high temperatures and densities, it fuses quickly). Three He-4 nuclei can combine to C-12 if they come together within a very short timescale (the intermediate state Be-8 is very short-living). This is called triple-alpha-process.
 
and then the star is transforming into something bigger?...thanks for information
 
Kostas Tzim said:
and then the star is transforming into something bigger?...thanks for information

The star's core is always collapsing as the supplies are burned, its surface is expanding. If the temperatures achieved are not enough to keep up the burning, the star stops working and turns into a white dwarf...That moment the collapse is evaded by the core's degeneracy [quantum effects over fermions] instead of nuclear burning.
 
understood thanks chris :)
 

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