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Calcyulating number of protons and neutrons in early universe

  1. Feb 25, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    At very high temperatures (as in the early universe), the proton and the neutron can be thought of as two different states of the same particle, called the “nucleon”. (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron’s mass is higher than the proton’s by 2.3 × 10−30 kg, its energy is higher by this amount times c2. Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 1011 K. What fraction of the nucleons at that time were protons, and what fraction were neutrons?

    2. Relevant equations

    [tex] \frac{P(s_2)}{P(s_1)} = \frac{e^-E(s_1)/k_BT}{e^-E(s_2)/k_BT} [/tex]

    3. The attempt at a solution

    I have done most of this question.

    I have said (s_1) is the neutron, and (s_2) = the proton

    S_1 = 2.3 * 10^{-30}C^2 = 2.07*10^{-13}

    S_2 = 0

    I have inserted these values, as well as T = 10^11, into the above equation and have got:

    [tex] \frac{P(s_2)}{P(s_1)} = \frac{1}{0.98} = 1.015 [/tex]

    I have now arranged it to say:

    P(protons) = 1.015 P(Neutrons)

    from this, how can I calculate the fraction of the nucleons that were protons, and what fraction were neutrons?

    Many Thanks,

    TFM
     
  2. jcsd
  3. Feb 25, 2009 #2

    Delphi51

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    Homework Helper

    Well, the part you did is beyond my knowledge, but I think I can help you with the last bit.
    You found that the number of protons is 1.015 times the number of neutrons, right?
    If x is the number of neutrons, then you have 1.015x protons. The fraction of the total that is neutrons is x/(x + 1.015x).
     
  4. Feb 25, 2009 #3

    TFM

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    Would they want the answer in terms of x, or do I need to find out what x is?
     
  5. Feb 25, 2009 #4
    I'm not sure you have gone about this the right way. You have calculated the ration of probabilties correctly, but they want the fraction of neutrons and protons. So you need something like,

    P(s1)/[P(s1)+P(s2)] and P(s2)/[P(s1)+P(s2)]
     
  6. Feb 25, 2009 #5

    TFM

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    so that will be:

    [tex] \frac{e^{E(s_2}/k_BT}{E(s_2}/k_BT + E(s_1}/k_BT} [/tex]

    and

    [tex] \frac{e^{E(s_1}/k_BT}{E(s_2}/k_BT + E(s_1}/k_BT} [/tex]

    Right?
     
  7. Feb 25, 2009 #6
    Yeah, but those energies over kt should have exponentials. Also I have noticed there is an arithmetic error in your original calculation. For one of the energies you get 0.98, this is incorrect.
     
  8. Feb 25, 2009 #7

    TFM

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    Okay, so:

    [tex] P(N) = \frac{e^{E(s_1)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} [/tex]

    and:

    [tex] P(P) = \frac{e^{E(s_2)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} [/tex]


    Now:

    for Proton:

    [tex] e^{E(s_2)/k_BT} [/tex]

    E(s2) = 0

    so

    [tex] e^{0)/k_BT} = 1 [/tex]

    Now for the neutron:

    [tex] e^{E(s_1)/k_BT} [/tex]

    E = 2.3 * 10^30 * C^2 = 2.3 * 10^30 * (3 * 10^8)^2 = 2.07*10^-13

    kB = 1.38*10^-23
    T = 10^11

    thus:

    [tex] k_BT = 1.38 * 10^{-11} [/tex]

    so:

    [tex] e^{E(s_1)/k_BT} = 0.015 [/tex]

    So:

    [tex] P(N) = \frac{e^{E(s_1)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} = \frac{0.015}{0.015 + 1} = \frac{0.015}{1.015} = \frac{15}{1015} [/tex]

    and

    [tex] P(P) = \frac{e^{E(s_2)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} = \frac{1}{0.015 + 1} = \frac{1}{1.015} = \frac{1000}{1015} [/tex]

    Does this look right? There seems to be a lot more protons to neutrons?

    Then again, The Early Universe would have been mostly hydrogen, which is just protons.
     
    Last edited: Feb 25, 2009
  9. Feb 25, 2009 #8
    Fine
    This should be 1.38*10^-12. If you make the correction here the rest should come out fine.
     
  10. Feb 25, 2009 #9

    TFM

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    Okay, so:

    [tex] e^{2.07 * 10^{-13}/1.38 * 10^{-12} = 0.150} [/tex]

    So:

    [tex] P(N) = \frac{e^{E(s_1)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} = \frac{0.15}{0.15 + 1} = \frac{0.15}{1.15} = \frac{15}{115} [/tex]

    and:

    [tex] P(P) = \frac{e^{E(s_2)/k_BT}}{e^{E(s_1)/k_BT} + e^{E(s_2)/k_BT}} = \frac{1}{0.15 + 1} = \frac{1}{1.15} = \frac{100}{115} [/tex]

    Better?
     
  11. Feb 25, 2009 #10
    Surely the exponential power shuold be negative here? And the above is wrong even for the values you have given, it should be equal to 1.1618, not 0.15? At the risk of being rude I would say bin your calculator, it is dragging you down!
     
  12. Feb 25, 2009 #11

    TFM

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    Technically, it is excel, not a calculator...:tongue:

    Although, it sometimes requires the operator to remember to actually do the exponential, not just the division.

    So:

    [tex] e^{-E/k_BT} = 0.86 [/tex]

    so:

    P(P) = 1/(1+0.86) = 1/1.86 = 100/186

    P(N) = 86/186

    Is this better now?
     
  13. Feb 25, 2009 #12
    Perfect! Excel sucks! Matlab is much better, you don't have to remember silly things like = (although it does have other annoyances).

    Best
     
  14. Feb 25, 2009 #13

    TFM

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    Excellent, Thanks

    TFM
     
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