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Formation of elements under conditions of thermodynamic equilbrium

  1. Apr 21, 2008 #1
    [Sorry about the long post]

    I'm doing an essay where I am exploring whether or not elements were, for the most part, formed under conditions of thermodynamic equilibrium. We know that some element clearly aren't (such as Potassium 40 and uranium which decay radioactivly). I've scoured the web but information on this is suprisingly hard to come by! I'm hoping someone can help!

    Basically, I am assuming elements were formed by a simple nuclear build up model thus:

    [A-1,Z]+neutron--> [A,Z]

    where [A,Z] is an element (or isotope) with mass number A and atomic (proton) number Z. This is plausible because there is no colomb barrier for neutrons to pass through, unlike protons.

    If elements WERE formed under conditions of thermodynamic equilibrium, we can use simple Maxwell Boltzmann statistics to work out equilbrium temperatures. (The probability of a particle being in a state Ea is simply exp(-Ea/KT) so we can use the relative abundances of isotope pairs to work out a temperature at which both [A-1,Z] AND [A,Z] are at equilibrium:

    Equilbrium Temperature=[tex]\frac{-Dc^{2}}{kln([A-1,Z]/[A,Z]}[/tex]

    D= Mass Difference = M[A,Z] - {M[A-1,z]+M[neutron]}
    k is boltzmann's constant
    c is speed of light
    and where square brakets [..] now represent the concentration, or relative abudance of the isotope in question.

    NOW, on calculating equilibrium temperatures for isotope pairs ([A,Z] and [A-1,Z] of all the elements in the periodic table (please see attached file), I get negative temperatures in some instances! Is there a physical explanation for this?

    Furthermore, the equilibrium temperatures seem very high (~10^10 to 10^11 kelvins). As far as I am aware, at such temperatures there is plasma - a "cosmic soup", can there be thermodynamic equilbrium under such conditions?

    (if you've got this far, thank you!)

    Attached Files:

  2. jcsd
  3. Apr 22, 2008 #2

    any thoughts would be greatly appreciated!
  4. Apr 22, 2008 #3
    The explanation is that the formation of elements isn't done under conditions of thermodynamic equilibrium. This reaction [A-1,Z]+neutron--> [A,Z]
    doesn't run backwards.
  5. Apr 22, 2008 #4
    Thanks for your reply kamerling. Why in principle can it not run backwards?
  6. Apr 22, 2008 #5

    This is what I am thinking:

    If elements were formed under conditions of thermodynamic equilibrium, the equilibrium temperatures (*as I have defined it*) for the different elements should show some consistency, right?

    HOWEVER some elements were NOT formed under conditions of thermodynamic equilbrium.
    For example:

    -nuclei created in supernovae explosions, where nuclei are made much faster than the rate of reaction (and hence no equilbrium temperature is established).

    -nuclei which decay by radioactivity (eg. potassium 40)

    -some elements that were formed by CNO processes as well as H, C, O, Si burning in stars (WERE these "equilibrium processes"? In anycase, they don't involve neutron capture)

    when I try excluding nuclei pairs corresponding to these elements, I still get equilibrium
    temperatures which still vary somewhat.

    To remind you, I need to determine whether or not most elements were formed under conditions of thermodynamic equilbrium. Can anyone suggest how best to proceed?
  7. Apr 22, 2008 #6
    I advise you to abandon the attempt to connect cosmic abundances to an equilibrium temperature. You CAN connect cosmic abundances to the stability of the various isotopes or to the neutron densities you might find in a star.
  8. Apr 23, 2008 #7
    I wish I could (it doesn't make sense to me), but this the approach that my tutor suggested.

    Can anyone help with the questions I have raised on this thread?
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