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Beta Decay and Charge

  1. Oct 4, 2004 #1
    Okay, so one of my hall mates across the way asked this question about his Chemistry, and then I realised that I wasn't too sure why it was myself, so I figured I'd pass this on to you.

    Take the decay equation

    [tex]
    {}^{24}_{11}Na \rightarrow {}^{24}_{12}Mg + {}^0_{-1}\beta
    [/tex]

    His question pertained to the overall charge of the nuclide in question. He mentioned that since a neutron is converting to a proton, there should be a net charge of +1 now on the Magnesium (since the [itex]{}^{0}_{-1}\beta[/itex] comes from the conversion of a netron into the proton, that doesn't get factored in). Is it just that the charge is irrelevant to the decay that it's not bothered to be included?
     
  2. jcsd
  3. Oct 4, 2004 #2

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    A minor point is that the right side should also have an antineutrino. Physically speaking, the newly formed magnesium atom is likely to be a +1 ion for a while, the amount of time it takes to neutralize depending on its environment. But I think when they write [tex]{}^{24}_{12}Mg [/tex], the context is understood to be the nucleus itself, not the atom as a whole.
     
  4. Oct 4, 2004 #3
    Okay, that's what I figured, I was just wanting to make sure that there wasn't something I was missing. Thanks.
     
  5. Oct 5, 2004 #4
    To anticipate a possible extension, the electron emitted by the nucleus usually has a high enough energy to escape the atom, although it's also possible for the neutrino to carry the lion's share of the energy released in the decay. But it's not really relevant; what happens in the electron shells has no impact on the decay process.
     
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