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Energy generated by proton-proton cycle

  1. May 1, 2013 #1
    Untitled.png

    My attempt

    Q Reaction 1:
    Q1 = 2(1.007825) - (2.014102 + 0.000549) = 9.99E-4 u

    Q Reaction 2:
    Q2 = 1.007825 + 2.014102 - 4.002603 = -0.980676 u

    Q Reaction 3:
    Q3 = 3.016030 - (4.002603 + 2(1.007825)) = 0.013807 u
    ---------
    2(Q1) + 2(Q2) + Q3 = -1.945547 u

    (-1.945547 u)(931.5 MeV) = -1812.28 MeV

    But it is not supposed to be negative? Thanks for any help.
     
  2. jcsd
  3. May 1, 2013 #2
    The second equation given in the problem is wrong. See if you can figure out why and how to correct it.
     
  4. May 1, 2013 #3
    One of the isotopes in equation 2 is wrong.

    Also note that ion's fuse not neutral atoms! You should really be using the rest mass of the ion, not the rest mass of the atom. You can approximate this by subtracting the mass of the neutral atom by the mass of an electron.

    To see why this important we should really write equation 1 as
    [itex]\stackrel{1}{1}{H^{+}}+e^- +\stackrel{1}{1}H^{+}+e^-=\stackrel{2}{1}H^{+}+2e^-+e^+ +\nu[/itex]

    From this we can calculate Q1
    [itex]Q1=2mass(\stackrel{1}{1}{H^{+}}) +2mass(e^-) -mass(\stackrel{2}{1}H^{+}) - 2 mass(e^-) - mass(e^+)[/itex]

    If we approximate the mass of the ion as the sum of the masses of the electron and the neutral atom we then get
    [itex]Q1=2mass(\stackrel{1}{1}{H} ) -mass(\stackrel{2}{1}H) - mass(e^-) - mass(e^+)[/itex]
     
  5. May 1, 2013 #4

    mfb

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    Staff: Mentor

    In the core of our sun, nearly all hydrogen and helium atoms are ionized.
     
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