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Proton-proton I chain and mass defect reactants and products

  1. Sep 4, 2015 #1
    Hello there,

    I'm starting S382 astrophysics with the OU. The course book says "The proton-proton chain converts four hydrogen nuclei (protons) into a ^4_2He nucleus, two positrons that quickly collide with electrons and are annihilated, and two neutrinos. Hence, branch I of the p-p chain may be summarised as:

    2e- + 4p goes to ^4_2He + 2v_e + 2gamma_pd + 4gamma_e "

    I don't quite understand where all of the reactants and products are coming from. The 4p is obvious but the 2e^- I can only think are the two electrons that annihilate with the two positrons. If this is the case then should they really be included as part of the p-p I chain? What is also throwing me is the four gamma rays associated with electrons. Is this gamma ray a product of the electron-positron annihilation? If so why are there four? I would have thought there should only be two!

    Any help would be greatly appreciated. Sorry about the layout. I would have used latex but I'm on my tablet and it won't work.

    Thanks.

    Brian
     
  2. jcsd
  3. Sep 4, 2015 #2

    Astronuc

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    A positron-electron annihilation produces two gammas of at least 0.511 MeV. Think of conservation of energy and momentum.

    Note the intermediate steps - http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html#c1
     
  4. Sep 4, 2015 #3
    Thank you for replying so quickly. That's great.
     
  5. Sep 4, 2015 #4

    mfb

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    Right.
    A more "charge-neutral" way to write it down would include 4 electrons on the left side (corresponding to the 4 protons - so the sum is equivalent to four neutral hydrogen atoms) and two on the right side (so one could form a neutral helium atom out of them). As those two additional electrons don't participate in the reaction they are ignored.
     
  6. Sep 4, 2015 #5
    At least two gammas. A positron is equally likely to encounter a same spin electron, in which case at least three gammas are produced.
     
  7. Sep 4, 2015 #6

    ChrisVer

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    The 3 photon decays is not as likely as the 2 photon decay... Even if the angular momenta were not considered, the 3 gets supressed by a factor of ~100.
     
  8. Sep 4, 2015 #7

    mfb

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    That difference just means orthopositronium lives much longer than parapositronium - a factor of about 1000. There is no state where there would be a (relevant) choice between the two decay mechanisms. The fraction that decays to two photons is given by the initial formation, and both types are formed with a similar probability.
     
  9. Sep 4, 2015 #8

    ChrisVer

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    ?
    Wouldn't the 3 photons have an [itex]\alpha^3[/itex] vs the two photons with [itex]\alpha^2[/itex]?
     
  10. Sep 4, 2015 #9

    mfb

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    Consider 1000 positrons produced and slowed to thermal energies.
    About 500 will form parapositronium. Those decay to their ground state, and decay to 2 photons after (on average) 125 picoseconds.
    About 500 will form orthopositronium. Those decay to their ground state, and decays to 3 photons after (on average) 140 nanoseconds.

    The power of alpha gives a rough estimate for the relative lifetimes, the details of the decay (phase space, ...) are responsible for details. It does not influence how often which type of positronium is formed.
     
  11. Sep 4, 2015 #10

    Astronuc

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    This is interesting because in PET and PAS, the emphasis is on the two (collinear) gammas, rather than two or three.
    http://www.princeton.edu/~romalis/PHYS312/Positron.pdf

    Perhaps the collinear gammas are easier with respect to discrimination.
     
  12. Sep 5, 2015 #11

    mfb

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    The scanners usually just cover a small fraction of the full solid angle (too expensive, and also for practical reasons - the patient has to be somewhere). Finding two photons is easy - you know the energy, and you know they fly in opposite directions so if you detect one you also find the other one with a symmetric detector.
    Finding three is much harder - the probability that one does not fly into your detector is very large, and you don't know the energy.
    To make it worse, you would have to detect the flight directions or the detection time ("Time-of-Flight PET") to reconstruct the production point. More modern scanners can do this, but older ones cannot.
     
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