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Astrophysics - triple alpha reaction

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Helium is burnt via the triple alpha reaction: 3He4 -> C12 at a rate of [tex]\lambda[/tex](He4)3, where He4 denotes the number abundance of He4.

    Write the differential equation for He4 as a function of time.

    3. The attempt at a solution

    d(He4)/dt = production - destruction.

    I'm unfamiliar with how to find production and destruction rates...
    Any help appreciated
     
    Last edited: Sep 17, 2010
  2. jcsd
  3. Sep 16, 2010 #2
    Is there He4 production here? Btw, C12 is a stable isotope :smile:
    It's like a motion problem: you know v(x), you can find x(t) :smile:
     
  4. Sep 17, 2010 #3
    ok, since there is no He4 production I'm left with..

    d(He4)/dt = 0 - destruction.

    and destruction is occuring at a rate of [tex]\lambda[/tex](He4)3

    Therefore...

    d(He4)/dt = - [tex]\lambda[/tex](He4)3

    Thus, this would be my answer?
     
  5. Sep 17, 2010 #4
    I don't know what [tex]\lambda[/tex](He4)3 means but basically it's just like that.
     
  6. Sep 17, 2010 #5
    I don't what what it means either, but we are told that's the rate so what can you do!

    part b) and c) gets a little harder...

    Following part a) and 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 C12, O16, Ne20 abundances as a function of time.

    I think this is right, but I want to double check:
    d(C12)/dt = production - destruction = λ(He4)3 - λ12αHe4C12

    d(O16)/dt = production - destruction = λ12αHe4C12 - λ16αHe4O16

    d(Ne20)/dt = production - destruction = λ16αHe4O16

    The only thing that bothers me is that the abundances of some of the other elements is dependent on the amounts of other elements....
     
    Last edited: Sep 17, 2010
  7. Sep 17, 2010 #6
    If λ... is some constant, the DE can be solved for He4, then plug that into the first DE, you have DE for C12. Solving it and plugging it into the 2nd one...
    Anyway I don't get what the question expects.
     
  8. Sep 17, 2010 #7
    I understand what you mean. If lambda is a constant, I can use seperation of variables to solve for He4, which can then be subbed into the d(C12)/dt equation to solve C12 using same method so on and so forth until the end.

    I have a somewhat similar example, but I have trouble understanding it.
    It goes as such...

    p + p -> d + β+ + νe occurs at a rate: rpp=np2σpp/2
    p + d -> He3 + γ occurs at a rate: rpd=npndσpd
    He3 + He3 -> He4 + 2p occurs at a rate: r33 = n332σ33/2

    where rij = ninjσij/(1+ δij), therefore rpp is the reaction rate between 2 protons, rpd is the reaction rate between a proton and deuteron etc

    E.g.
    The DE for deuterium is then..
    d(D)/dt = formation rate - destruction rate = rpp - rpd = λppH2/2 - λpdHD

    I understand how they get formation rate - destruction rate = rpp - rpd, but how they do the next step to get the lambdas I have no idea.

    from what I gather, lambda = de broglies wavelength.

    Still very, very confused!! :surprised
     
    Last edited: Sep 17, 2010
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