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A How to calculate the energy of Gamma rays in fusion

  1. Nov 10, 2017 #1
    Lets say you have your basic fusion reaction

    D + HHe3 + γ + 5.49 MeV

    Now exactly how much of the energy is carried away by the gamma ray and how much by the Alfa particle ?
     
    Last edited: Nov 10, 2017
  2. jcsd
  3. Nov 10, 2017 #2

    mfb

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    You can use conservation of energy and momentum to calculate it. As the photon won't have much momentum, the He-3 nucleus won't move much, which means its energy will be very small. To a good approximation the photon gets the whole energy.

    Alpha particles are only He-4.
     
  4. Nov 10, 2017 #3
    This is what I feared, but perhaps there are some fundamental laws which limit or mitigate the energy released by Gamma rays in fusion reaction

    Do note this answer contradicts what is mention here.
     
  5. Nov 11, 2017 #4
    There is indeed such a fundamental law.
    Strong interaction is strong.
    Where it is possible to release energy by restructuring nuclei and carrying energy away by kinetic energy rather than by gamma ray emission, that is usually much faster.
     
  6. Nov 11, 2017 #5

    mfb

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    You can't violate energy or momentum conservation.
    (Highlight by me) That is correct, and you found one of the exceptions. Most fusion reactions of light elements produce two (sometimes three) nuclei, and then typically the energy is carried away as kinetic energy. Fusion reactions that only produce one nucleus are rare, and in this case the energy is carried away by photons.
     
  7. Nov 11, 2017 #6
    Actually, fusion reactions of light elements in which more than one nucleus is produced and energy is carried away as kinetic energy of nuclei are a minority.
    There are also fusion reactions in which energy is carried away as kinetic energy of positron and neutrino, or of neutrino alone.
    On average, production of an alpha particle takes 5 fusion reactions. Of which the minority of 1 out of 5 forms more that 1 nuclei - and in that case 3 not 2:
    He-3+He-3→α+p+p
    The other fusion reactions are:
    2 emit gammas:
    d+p→He-3+γ
    and 2 usually emit positron:
    p+p→d+e+e
     
  8. Nov 11, 2017 #7

    mfb

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    A sample of 5 reactions of your choice is not an argument.
    Check this list - all with two reaction products! What does this show? Nothing, because it is a biased list as well.

    Fusion reactions with more than two nuclei in the final state can happen via the strong interaction, while reactions with only one need the weaker electromagnetic interaction. That makes the first more likely, especially if both options are possible.
     
  9. Nov 11, 2017 #8
    But is not claiming to be:
    Actually, there are obvious reasons why most fusion reactions must be ones that produce one nucleus.
    "Fusion" by definition is formation of fewer, bigger nuclei out of more numerous smaller ones.
    Mostly this happens by two nuclei forming one nucleus.
    An alternative might be three nuclei forming two. But can you propose an actual common three-nucleus process?
    Triple alpha is not, actually. Because its true mechanism is a multistep one:
    α+α→Be-8+γ (Be-8 has actual half-life of 10-16 s - 106 times longer than the duration of nuclear collisions)
    Be-8+α→C-12*+γ (Hoyle state)
    C-12*→C-12*+γ
    C-12*→C-12+γ
    Due to selection rules, Hoyle state needs two steps to decay to C-12 ground state.
    There IS a somewhat common, but decidedly minority, true three-particle fusion process. But just two are nuclei:
    p+p+e→d+νe
    Many fusion reaction cannot proceed via weaker electromagnetic interaction either, and must happen via even weaker weak interaction. Like the simplest way to get 1 α particle out of 4 nuclei, in 3 steps all of which start with 2 particles:
    p+p→d+e+e
    d+p→He-3+γ
    He-3+p→α+e+e
    As you see, reactions of "fusion reactions of light elements". None of the 3 is a strong process. Just 1 of the 3 is an electromagnetic interaction, and most of the 3 are weak interaction fusion reactions.
     
  10. Nov 13, 2017 #9
    How much of the kinetic energy is caried by the positron and neutino?
     
  11. Nov 13, 2017 #10
    Like in beta decay. Most of the energy goes to positron and neutrino, and is freely distributed between them. A small fraction goes to the recoil of the nucleus.
     
  12. Nov 14, 2017 #11
    This remind me that whenever nature produces something cool there is a catch. So the only useful thing about this reaction is energenic positron and deuterium core which could be usefull for other fusion reactions.

    Question, lets say the energenic positron anihilates when it comes into contact with an electron, what happens with it kenetic energy stored in positron from p-p fusion?, Will the energy leased by Gamma be 1MeV + 0.5 E p-p
     
    Last edited: Nov 14, 2017
  13. Nov 14, 2017 #12
    Yes, roughly that. The energy carried away by the neutrino escapes.
    The net energy balance depends on whether the conditions are suitable for further reaction of He-3.
     
  14. Nov 15, 2017 #13

    mfb

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    Most of the time the positron will lose most of its kinetic energy before it annihilates.

    @snorkack: You are wrong with the two/three body discussion, but as it is off-topic anyway I won’t continue that discussion part.
     
  15. Nov 17, 2017 #14
    Wait a second. Exactly what do you mean by "freely distributed between them", the electron-neutrino is much less massive than the positron (which is at least 500000 more massive). Wouldn't this mean it would carry almost all of the 0.42 MeV energy?
     
    Last edited: Nov 17, 2017
  16. Nov 18, 2017 #15
    No. It is the nucleus which is the most massive and therefore has to take a tiny fraction of energy. The neutrino and the positron have very different rest masses, but since the momentum conservation is already met by the nucleus, this does not much constrain how the energy is divided between positron and neutrino.
     
  17. Nov 18, 2017 #16

    mfb

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    The distribution is not symmetric due to the electron mass, but the possible kinetic energy for both (individually) goes from zero to nearly the total released energy minus the electron rest energy.
     
  18. Nov 19, 2017 #17
    Am I right in guessing that while the distribution of electron and antineutrino energies is not symmetric, the distribution of electron and antineutrino momenta is?
     
  19. Nov 19, 2017 #18

    mfb

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    That is not symmetric either, but it is a better approximation.
    Here are plots.
     
  20. Nov 20, 2017 #19
    Fusion of two protons releases an average of 420 keV plus the kinetic energy of protons.

    How much of this on average goes to kinetic energy of positron, how much escapes with neutrino?
     
  21. Nov 21, 2017 #20
    actually I beleive is more than 0.42 MeV which is just the surplus power released, but it first aborbs energy

    The first step of the proton–proton chain reaction is a two-stage process; first, two protons fuse to form a diproton:

    H + H + 1.25 MeV → 2He

    followed by the immediate beta-plus decay of the diproton to deuterium:

    2He → D + e+ + νe + 1.67 MeV

    with the overall formula

    H + HD + e+ + ve + 0.42 MeV

    So irs not 0.42 MeV but 1.67 MeV which is divided over the positron and neutrino electron. If I applied the lesson leaned untill now, I would think most of the energy would be caried away with the nutrino
     
  22. Nov 22, 2017 #21

    mfb

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    Unless you shoot the two protons together in accelerators, you don’t have this initial 1.25 MeV around. The diproton has to be off-shell, and the total energy released is just 420 keV as required by conservation of energy (add a fee keV from the thermal energy of the protons). As this is smaller than the electron mass most of the energy goes into the neutrino (on average).
    Here is a plot, the peak is somewhere around 300 keV.
     
  23. Nov 22, 2017 #22
    I see there is some probability distribution from which the peak is around 300 keV and it can go up to almost 420 keV. Does this also mean on average its energy is about 300 keV as well?
     
  24. Nov 22, 2017 #23

    mfb

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    We would have to integrate the distribution to get the average, but it should be something between 250 to 300 keV.
     
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