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Fermi/Coulomb correction factor for beta decay

  1. May 21, 2015 #1
    Hi all,

    Fermi introduced a multiplicative correction factor in order to correct for the effect of the Coulomb field on the resulting energy distribution of ejected beta particles (i.e., either electrons or positrons) following a beta decay reaction. The result is that positrons are accelerated as they exit the nucleus, while electrons are slowed down. My question is, why does this correction factor have to be a multiplicative term, as opposed to simply subtracting the energy lost or gained for the electron or positron respectively? Is there a way that I can derive the momentum-dependent energy loss/gain, and arrive at the final corrected energy distribution?

    Kind regards
    Last edited: May 21, 2015
  2. jcsd
  3. May 26, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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