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Radiotracer Decay - What's Left?

  1. Feb 27, 2008 #1
    Greetings all. I'm a just a chemist banging my head about some what seems to me as fundamentally a nuclear physics question, so please bear with me.

    If you have a molecule bearing a radioactive atom, a PET radiotracer, say F-18 FDG or some such beast, and it decays (to O-18 correct?) what exactly happens to the entire molecule? Another way of putting it is: As the positron-neutrino pair is released, is the rest of the tracer molecule blown apart by kinetic energy from the decay event?
    Or am I thinking too much in a classical sense and you are left with a molecule happily bearing O-18 rather than F-18?
    And lastly, does the answer depend on what type of decay is occurring?
  2. jcsd
  3. Feb 27, 2008 #2


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    It depends on the electron configuration of the radioactive elements daugher.

    You can look on wikipedia for example, on all decay modes. Then there are things like "Internal conversion", "electron capture"; which changes the electron configuration of the daugher atom.
  4. Mar 12, 2008 #3
    Thanks for the reply! So it looks like this was the wrong place to post (sorry about that). I have asked people here in radiochemistry and medical physics and I still am left without an answer, though. Oh well.
  5. Mar 12, 2008 #4
    I'm not a radiochemist, but I believe that most of the time the daughter product dissociates from the tagged molecule/pharmaceutical (which remains intact)
  6. Mar 14, 2008 #5
    Hello Chyral,

    I would say that the recoiling of the nucleus is negligible (specially for F) wrt the energy of positron-neutrino pair. In some way, that's why you can do imaging.
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