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I Beta decay and mass defect

  1. Mar 25, 2016 #1
    Hi, I was reading the wiki article here on beta decay and energy release: https://en.wikipedia.org/wiki/Beta_decay#Energy_release
    In obtaining a condition on the masses of the initial and final atoms we neglected the difference in electron binding energies of the two atoms. Does this mean we compare the electron binding energies of one atom to the binding energies of the other and say the difference between the two is negligible? The article says this is small for high Z atoms, which I don't understand really understand? Later, in the section on electron capture, the article goes on to say we can no longer neglect the electron binding energy for the captured electron, which again I don't fully understand.
    Sorry if I wasn't clear: I hope someone more knowledgable than me can help!
     
  2. jcsd
  3. Mar 26, 2016 #2

    Simon Bridge

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    Yes.
    What don't you understand about it? Have you tried a dictionary?
    Why not? What is it that you don't understand?

    Try expressing what you are after as a question.

    Guessing your confusion:
    Oversimplifying: the energy that the electron carries with it depends on where it comes from.
    In beta decay, the electron comes from the nucleus so it carries it's energy from there and not from the electronic environment of the atom, which, in any case, is typically 5-6 orders of magnitude smaller.
    In electron capture, the electron being captured comes from the electronic environment, so it carries energy from there, so the electron binding energy can be expected to be important.
    The bottom line in whether anything is negligible, though, is whether we can measure the difference.
     
    Last edited: Mar 26, 2016
  4. Feb 7, 2017 #3
    Is antineutrino mass negligible? What about the energy? Do antineutrinos travel at c? Need to know that in order to compute electron velocity, right?
     
  5. Feb 7, 2017 #4

    jtbell

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    In beta decay, it is an excellent approximation to assume that the outgoing antineutrino is massless and moves at speed c.

    I have never read about a case in which the mass of the antineutrino has a measurable effect in beta decay.
     
  6. Feb 7, 2017 #5
    What about the wavelength? I read that it depends on 'flavour' of neutrino / antineutrino?
     
  7. Feb 7, 2017 #6

    jtbell

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    Where? It would help if you could provide some context so we don't have to try to guess what you're thinking about. :smile:
     
  8. Feb 8, 2017 #7
  9. Feb 8, 2017 #8

    jtbell

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    I get "This question does not exist or is under review."

    The electron does not have a fixed velocity. The energy released in the decay is distributed randomly between the (anti)neutrino and the electron (also the recoiling nucleus, but its kinetic energy is very small so we normally ignore it).

    The electron's kinetic energy can range from nearly zero up to the total decay energy.
     
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