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Energy in positron emission

  1. Nov 7, 2005 #1
    In positron emission, a proton is converted into a neutron, a positron, and a neutrino:

    p --> n + e + v

    this conversion occures only in a nucleus, because this process consumes energy (the neutron and positron have a higher mass than the proton).

    The energy needed is twice the rest mass energy of the electron: 1.022 MeV.

    Why exactly 2*mass(electron).
    I would excpect it to be something like
    (mass of neutron + mass of electron)-mass of proton.

    Thanks in advance
     
  2. jcsd
  3. Nov 7, 2005 #2

    ZapperZ

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    I am not so sure that you haven't mixed apples with oranges.

    If a positron is emitted via nuclear decay, such as those used as the positron source in PET scans, for example, there is no need for an "energy needed" to initiate such an event. The nuclear is unstable, and it'll decay when it wants to.

    However, if the positron is generated from pair production, then the energy of the photon needed would require a minimum twice the electron's rest mass energy, for obvious reason.

    Zz.
     
  4. Nov 7, 2005 #3
    We studied this in a pretty good bit of detail this earlier this semester, so I think I can shed a little light on your question. For a positron emission to happen, the parent atom must mass at least 1.022 MeV (= 2 X me) greater than the daughter atom. This seems like a strange result, and indeed it is.

    The 2 X me comes in because we use atomic masses rather than nuclear masses most of the time. This is partly because the nuclear mass is difficult to calculate exactly, but mainly because it's accurate enough most of the time. If you look at regular beta (electron) decay, you'll see that the difference cancels out. For example,

    60Co -> 60Ni + electron + antineutrino.

    The mass of cobalt atom includes 27 electrons and the mass of the nickel atom includes 28, *but* since an electron doesn't magically appear in the outer shell when the atom decays, we've got to account for the discrepancy. Fortunately, since there's another electron on that side of the equation, it cancels out the discrepancy and the Q-value is just the difference between the mass of the two atoms.

    In positron decay, the situation is reversed. If 26Al decays into 26Mg, the mass of the Aluminum atom accounts for 13 electrons, but the mass of the Magnesium only accounts for 12. There's an extra electron from the Aluminum atom we have to account for. So when calculating the Q-value for the decay, you have to add the mass of the positron *and* the mass of the extra electron on the daughter side of the equation. So the Q-Value for this equation would be:

    Q = {mAl - mMg - 2 X me}c2

    So, when using atomic masses, the parent must mass at least 1.022 MeV greater than the daughter, or a positron decay is impossible. In that case, the atom can still decay by electron capture, which is a competing process.
     
  5. Nov 7, 2005 #4
    thank you grogs, things are clearer to me now :smile:

    Do i understand it correctly : When i would use nuclear masses for my computation, a difference of 511 MeV would be sufficient for positron emission?

    good bye...
     
  6. Nov 7, 2005 #5
    Exactly. As long as what's on the left side (the parent) had more mass than the products (daughter nucleus + positron), it would be sufficient. You would also see the same restriction on the beta- decay, which makes sense.
     
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