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Triple alpha reaction

  1. Sep 18, 2010 #1
    Hi guys,
    I posted this in the homework subforum but since its more intensive on the astrophysics side of things I thought I'd give it a shot here. Hope this isn't a problem mods.

    Helium is burnt via the triple alpha reaction: 3He4 -> C12 at a rate of λ(He4)3, where He4 denotes the number abundance of He4. Once there is some C12 present then we can get C12(α,γ)O16 which occurs at the rate λ12αHe4C12. Then, once O16 has been produced, we also get O16(α,γ)Ne20 which occurs at the rate λ16αHe4O16.

    Write down DE's for He4,C12, O16, Ne20 abundances as a function of time.

    Basically here's what I did.
    d(He4)/dt = production - destruction = - λ(He4)3
    d(C12)/dt = production - destruction = λ(He4)3 - λ12αHe4C12
    d(O16)/dt = production - destruction = λ12αHe4C12 - λ16αHe4O16
    d(Ne20)/dt = production - destruction = λ16αHe4O16

    This is most likely wrong, but I'd thought I'd just give it a shot.
    Any thoughts?
    Thanks in advance
  2. jcsd
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