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Confused about why beta plus decay happens

  1. Dec 29, 2012 #1
    For β- minus decay

    Q = m_x - m_y

    Whereas for β+

    Q = m_x - m_y - 2m_e

    Where the masses are atomic masses, x and y are the parent/daughter.

    The SEMF can be used to find a parabolic curve showing the mass excesses for an A isobar for particular values of Z, and thus show at which Z the atom has the least mass. However a textbook of mine says;

    "The mass-energy difference between two adjacent isobars is the energy available for a radioactive transition from the heavier to the lighter one". However, is this strictly true? The Q value for Beta minus has a -2m_e term, so it's not a simple case of the differences in atomic masses being the energy available.

    Which suggests to me that for a mass excess/Z diagram for an A isobar, even if (on the proton rich side [to the right of the minimum of mass excess]) the atom with Z protons is JUST above the atom with Z-1 protons (to which it would decay to), this doesn't necessarily mean the decay might happen, because the -1MeV arising from the 2m_e term might mean the Q value is less than zero for beta minus decay. So although it looks like the decay process should happen on a mass excess/Z diagram, it shouldn't. Is that right?
    Last edited: Dec 29, 2012
  2. jcsd
  3. Dec 29, 2012 #2


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    Why do you get the electron mass there? Do you mean electron capture instead of positron emission? This is favored by 2 electron masses.

    Where a part of this energy might be provided by the captured electron (or is required to produce an electron or positron).
  4. Dec 29, 2012 #3
    I do mean beta plus decay, although if I had given my Q value for electron capture it would be

    m_x - m_y - BE(captured electron).

    Where m_x and m_y are again the atomic masses of the parent and daughter. It might be worth noting that when I say atomic masses I'm referring to the atomic masses to the general atoms, not the atoms that are produced in this reaction process. If you want a full derivation;

    Q value is the difference in nuclear masses before/after reaction process

    [itex] Q = m_x^N - m_y^N - m_e - m_{neutrino} [/itex]

    Neglect neutrino mass.

    [itex] = (m_x - Zm_e + BE_{x(atomic)}) - (m_y - (Z-1)m_e + BE_{y(atomic)}) - m_e [/itex]

    Difference in atomic BE is minimal so ignore those.

    [itex] = m_x -Zm_e - m_y + (Z-1)m_e - m_e [/itex]

    [itex] = m_x - m_y - 2m_e [/itex]
  5. Dec 29, 2012 #4


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    You cannot treat electrons like that. In a beta+-decay, you might lose one of the outer electrons afterwards, but that electron does not magically vanish, its energy is still there after it leaves the atom. Take it into account, and you get

    ##Q=m_x-m_y-m_e## both for beta+ and beta- decays
    ##Q=m_x-m_y+m_e## for electron capture
  6. Dec 29, 2012 #5
    Outer electron? Are you referring to one of the atomic electrons of the daughter?

    My equations above are directly quoted from my lecture notes, but I don't see what is wrong with them.

    You are definitely certain that you know that when I refer to atomic masses, I am referring to the general atomic masses of the daughter and parent? As in, the atomic masses you would find in a periodic table, not the atomic masses of the atoms specific to this situation.

    So for instance, when Rubidium beta+ decays to Krypton. The Rubidium loses a proton and converts this to a neutron, a beta+ particle is emitted outside the atom. Now the Krypton has its normal atomic number Z-1, but it has one more electron than it should generally have in its atomic orbitals. Generally the Q value is the difference in masses before/after reaction, but this always reduces to the difference in nuclear masses. The Rubidium atom and Krypton atom in this case have the same amount of electrons, so the difference in their mass is their nuclear masses. Also remembering to subtract the beta+ particle mass.

    Q = [itex] m_n(Rb) - m_n(Kr) - m_e [/itex]

    Now there's just general terms for the nuclear masses of the Rb and Kr atoms. I can work backwards and introduce the general atomic masses of these atoms in to the equation, as I did before.

    There's no deleting or adding any electron masses in my equations at all. I think you are referring to the atomic masses of the atoms in the reaction, which I am not.

    So the difference being, for example in the reaction above;

    Your [itex] m_{atomic}(Kr) [/itex] is referring to Krypton with 36 protons and 37 electrons, the ionised Krypton atom in the reaction.

    My [itex] m_{atomic}(Kr) [/itex] is referring to Krypton with 36 protons and 36 electrons. The general Krypton atom.

    Hence the difference in one electron between our formulas.
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