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Deuterons used up before reaching the A=5 and 8 bottlenecks?

  1. Nov 16, 2012 #1

    bcrowell

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    I've seen A=5 and 8 described as bottlenecks in nucleosynthesis, because there are no stable nuclei with these mass numbers. One of my students suggested that you could, e.g., fuse 7Li+2H to produce 9Be, thereby bypassing the A=8 bottleneck. I've read that A=8 is actually surmounted by doing 4He+4He->8Be, then 8Be+4He (during the extremely short time that the 8Be holds together). Am I correct in thinking that 7Li+2H doesn't happen because by the time a star has built up an appreciable amount of 7Li, it has already used up all its 2H? What about the A=5 bottleneck?
     
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  3. Nov 16, 2012 #2

    Drakkith

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    I'm no expert, but I believe that neither deuterium nor lithium are significant reaction products at the same time as the other, and neither build up to significant amounts because they react with protons or helium soon after production.

    As for the A=5 bottleneck, well it too seems to be surpassed by the triple alpha process as you described in your post.
     
  4. Nov 16, 2012 #3

    Vanadium 50

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    Yes, all the D is burned early in the star's life.
     
  5. Nov 16, 2012 #4

    Chronos

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    Deuterium is believed to fuel stars in the brown dwarf sized mass range, so it appears highly probable most of it is consumed by the time a protostar approaches the mass required to initiate hydrogen fusion.
     
  6. Nov 17, 2012 #5

    jimgraber

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    I vaguely remember that those two bottlenecks are more important during big bang nucleosynthesis than during later stellar burning.
    Best,
    Jim Graber
     
  7. Nov 18, 2012 #6

    mathman

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    The A = 8 bottleneck is important for getting past Li. The 3 He4 reaction (2 + 1 or all 3) takes place in the interior of stars.
     
  8. Nov 19, 2012 #7

    bcrowell

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    Thanks, all, that's very helpful.
     
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