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Inside the stars

  1. Dec 25, 2014 #1
    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 cant find the icon to put those [/tex] symbols
     
  2. jcsd
  3. Dec 25, 2014 #2
  4. Dec 25, 2014 #3

    SteamKing

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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.
     
  5. Dec 25, 2014 #4

    ChrisVer

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    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.
     
    Last edited: Dec 25, 2014
  6. Dec 25, 2014 #5
    Actually, it is unbound. Most of time, the proton simply bounces off, just like it bounces off another proton.
     
  7. Dec 25, 2014 #6

    Doug Huffman

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    No, stellar metal production stops at Fe57. Beyond that requires nova conditions.
     
  8. Dec 25, 2014 #7
    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
     
  9. Dec 25, 2014 #8

    ChrisVer

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    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)
     
  10. Dec 25, 2014 #9
    thanks Chris :) i'll look for it...
     
  11. Dec 25, 2014 #10

    Doug Huffman

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    To what does a star 'burn' H and H2? Surely you are not suggesting that quenches the reaction.
     
  12. Dec 25, 2014 #11

    ChrisVer

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    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
     
  13. Dec 25, 2014 #12

    mfb

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    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.
     
  14. Dec 25, 2014 #13

    Doug Huffman

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    Oh, no, the Wikipedia article is fine, and so is the one titled Deuterium burning.
     
  15. Dec 25, 2014 #14

    ChrisVer

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    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...
     
  16. Dec 25, 2014 #15
    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?
     
  17. Dec 25, 2014 #16

    SteamKing

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    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.
     
  18. Dec 25, 2014 #17
    It accumulates from the beginning, but as you see the net result is to make one 4-He back into two 4-He.
    Triple-alpha takes very much higher temperature and density than pp chain. Hydrogen is nearly exhausted long before that.
    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
     
  19. Dec 25, 2014 #18

    ChrisVer

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    i wonder, why is Branch IV so suppressed if there are a lot of protons around? :(
     
  20. Dec 25, 2014 #19

    SteamKing

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    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.
     
  21. Dec 26, 2014 #20
    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.
     
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