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Beta decay mass condition

  1. May 3, 2009 #1
    In beta decay, positron emission, how come the condition for decay is:

    M_p > M_d + 2m_e

    Thats: atomic mass of parent > "daughter + twice the mass of an electron.

    I'm sure there is some simple way of showing it, but I cant seem to find it!!

    Also, is the most stable isobar on an atomic mass parabola the most stable one? It's a question my lecturer posed to us, and I have been thinking about it a while.

    Thanks!
     
  2. jcsd
  3. May 3, 2009 #2

    clem

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    It is 2m_e because it is the atomic and not the nuclear mass.
    One m_e is because of the electron emitted by the nucleus.
    The second m_e is the extra electron that must be added to the outer shell of electrons in the atom.
     
  4. May 3, 2009 #3

    malawi_glenn

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    that is since the daughter nucleus will have one electron less than the parent, and what is listed is ATOMIC masses. So if you calculate it with NUCLEAR masses it should of course be:

    M_p > M_d + m_e

    if now M is nuclear masses.

    The other question is a homework - course work question, and should not be adressed in this forum but in the homework forum, with an attempt to solution.
     
  5. May 3, 2009 #4

    clem

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    "Also, is the most stable isobar on an atomic mass parabola the most stable one?"
    As you write this question it is a tautology: Is the most stable the most stable?
     
  6. May 3, 2009 #5
    For the parent p to decay to the daughter d by positron decay, the parent would have to create both a positron and electron (lepton number is conserved) before having sufficient energy to decay by positron emission. Example: could a proton decay to a neutron plus positron? What would the minimum proton mass be?
     
  7. May 4, 2009 #6

    malawi_glenn

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    nope, the electron number is conserved by emitting a neutrino

    beta+ decay on subnuclear level is:

    p -> n + e(+) + neutrino

    You are proposing:

    p-> n + e(+) + e(-)

    which violates electric charge conservation.
     
  8. May 4, 2009 #7
    Mea Culpa. There are three types of beta decay exibited by nuclei: Here are 3 examples, all from Cu64
    A) Positron emission (by proton), with an antineutrino
    B) Electron emission (by neutron) , with a neutrino
    C) K-shell electron capture (by proton) with only an antineutrino emission

    (From Wiki):
    64Cu has a half-life of 12.701 ± 0.002 hours and decays by 17.86 (± 0.14)% by positron emission, 39.0 (± 0.3)% by beta decay, 43.075 (± 0.500)% by electron capture
     
  9. May 5, 2009 #8
    where does the electron that must be added to the shell then come from??? if its from outside the atomic system in question, then surely it has no place in the mass condition?
     
  10. May 5, 2009 #9

    malawi_glenn

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    WHERE it comes from is not important, the issue is that you are using atomic masses.. as I explain to you.
     
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