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Quick question on the mass condition for beta decay

  1. Mar 27, 2015 #1
    For β- we have:

    ##M(A,Z)>M(A,Z+1) + m_{e} - m_{e}##

    An electron is removed from the atom and therefore we need to take that away from the M(A,Z+1) term

    But for β+ we have been given:

    ##M(A,Z)>M(A,Z+1) + m_{e} + m_{e}##

    What is this saying? A positron is emitted, therefore shouldn't we minus the mass of the positron from the atomic mass? which would give the same expression as for β-? I'm confused as to why it is plus, it's probably something really simple but I can't figure it for some reason.

    Thanks for any help/ideas

    edit also a quick side question, is β+ decay the same as electron capture? As they give the same daughter nucleus.
     
    Last edited: Mar 27, 2015
  2. jcsd
  3. Mar 27, 2015 #2
    Probably those who have given the equation are confused in explaining what the equation is for.
    Positron decay and electron capture are not the same - daughter nucleus is the same, but other ingredients are not. They therefore have different mass conditions.
     
  4. Mar 27, 2015 #3
    So is the mass condition that we have been given for β+ decay incorrect?


    Thanks! also i realised the spectrum is different, beta decay being broad and Auger giving discrete emission lines.

    thanks for the reply
     
  5. Mar 27, 2015 #4

    mfb

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    The beta+ equation is wrong in some way - if the left side is supposed to be the initial state (the ">" suggests this) then the charge at the right side is wrong. Fix that and the equation makes sense. Note that the number of electrons for the final atom (which is included in those mass values) is different for the two processes.
     
  6. Mar 28, 2015 #5
    Thanks. Would you mind writing out the correct equation? I'm unsure what you mean by "the charge at the right side" when they are masses. thanks again for your help.
     
  7. Mar 28, 2015 #6

    mfb

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    Z+1 in the brackets is the charge of the nucleus. If you emit a positive particle (a positron) the proton number has to decrease.
    $$M(A,Z)>M(A,Z-1) + m_{e} + m_{e}$$The two more electron masses compared to beta- decay come from the additional positron and the fact that you suddenly have an additional electron that does not "belong to" the atom - so you have the new neutral atom plus one electron plus one positron, and the original atom has to have enough energy to produce all of them.
     
  8. Mar 28, 2015 #7
    That's great thanks. Although I think i'm getting confused; Z is the number of protons, after decay it turns to a new element with Z-1 with an emitted positron, so then why are we adding the mass of an electron?
     
  9. Mar 28, 2015 #8

    mfb

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    Let's consider an example:
    $${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^+ + \nu_e$$
    Potassium on the left side normally has 19 electrons, so we still have 19 electrons on the right side - but Argon just has 18 protons, so one electron is surplus. We can write the reaction as
    $${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^- + e^+ + \nu_e$$
    to get neutral atoms, because the mass numbers refer to those neutral atoms.
     
  10. Mar 28, 2015 #9
    Of course, number of protons = number of electrons, so 1 electron is surplus. Sorry it's been 20 years since my GCSE chemistry, but still should know this lol

    Many thanks!!!
     
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