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Positron-electron annihilation in matter

  1. Sep 4, 2007 #1
    I have seen some seemingly differing statements in various sources about the dynamics of annihilation when the positron has some kinetic energy and is travelling through a medium.

    Two sources I've read state or imply that a positron in motion through matter must first lose it's kinetic energy through ionization or bremsstrahlung before it can annihilate with an electron, and even then, only through an intermediate positronium, and will therefore only annihilate with the signature 511 kev photons. Is all this true?

    Another source (the Wiki on annihilation) seems to say that annihilations can take place involving higher energy photons or even heavier particles if the kinetic energies are high enough.

    So, what really happens with an energetic positron in matter? I'm especially interested in the case of positrons emitted in radioactive decay and the physics behind PET scanning.

  2. jcsd
  3. Sep 4, 2007 #2


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    Be careful with Wikipedia since it does not have a rigorous control system on the quality of information.

    Positrons used in PET are of low energy (keV and low MeV), so they slow down in matter, primarily by ionization. Once they slow down to very low energy, they will combine with an electron (positronium) and then annihilate with the electron, which results in the formation of two 0.511 MeV gamma rays.

    15O and 18F, have respective maximum positron energies of about 1,730 keV and 630 keV. Ref: http://hps.org/publicinformation/ate/q6655.html [Broken]

    See also - http://www.np.ph.bham.ac.uk/pic/physics.htm [Broken]

    Radiation Basics — Beta Particles (negative and positive) and Electrons
    http://hps.org/publicinformation/ate/q6240.html [Broken]

    Other common positron emitters are:
    22Na, 65Zn, 68Ga, and 114In.
    http://www.ehs.ucsf.edu/Manuals/RSTM/RSTM%20Chap1.htm#2.%20POSITIVE%20BETA%20PARTICLES [Broken]

    Higher energy positrons must be energized in an accelerator. They will not produce heavy particles unless their kinetic energy is on the order of high MeV or GeV.
    Last edited by a moderator: May 3, 2017
  4. Sep 4, 2007 #3
    As usual, you nailed my question. So, it appears that annihilation can occur either after thermal equilibrium or through positronium on the way down. I need to research positronium, that's an interesting phenomenon in itself. In either case, it's always 2 x 511kev in lower energy situations, which certainly explains the reliable and useful signature. Very curious behavior, I wonder what the explanation is for having to come to rest before annihilation.

    I know what you mean about Wiki. I always try to "consider the source," but with Wiki, you don't usually know the source.

    Thanks for the reply and references.
  5. Sep 19, 2010 #4
    Hi...sorry, this may be a closed topic now, but just wanted to know if the photons formed in this anhilation process carry any charge with them ? cause what will happen to the negative charge of electron and the positive charge of the positron ? they are neutralized for the system as a whole, but charges wont be destroyed rt ?
  6. Sep 20, 2010 #5
    Nope, photons do not possess electric charge. All that matters is that the total charge of the system be the same (zero) before and after annihilation.
  7. Sep 20, 2010 #6
    Heitler "The Quantum Theory of Radiation" Third edition (1954) plots the integrated probability of positron annihilation in flight vs. positron kinetic energy on page 385. For a 1-MeV positron, the integrated probability of annihilation in flight (before stopping) is ~4%.

    Bob S
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