Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quick question about positive beta decay and mass defect?

  1. Feb 3, 2010 #1
    Hi,
    I was wondering about something in my notes that I don't quite understand. In positive beta decay in nuclear physics, we have a proton becoming a neutron and kicking out a positron and an electron neutrino (which is assumed massless here).

    In the expression for the energy released, Q = [M(A,Z) - M(A,Z-1) - 2me]c2, we have the kinetic energy of the products is equal to the mass-energy of the original nucleus minus the mass energy of the new nucleus minus TWO electron masses.
    This factor of two is what confuses me.

    All my notes really say is that this is due to an excess of electrons. Now I assume what this means is that originally the atom is electrically neutral, with equal numbers of protons and electrons. After this decay, we are one proton short, so there is on excessive electron in an atomic electron shell.

    I still do not see where this extra electron mass-energy comes from though, because the excess electron in question is not being created, it was always there. All that has been created is one positron, and we are assuming the neutrino to be massless.

    Can anyone explain this?

    Thanks.
     
  2. jcsd
  3. Feb 3, 2010 #2

    jtbell

    User Avatar

    Staff: Mentor

    If we were to use nuclear masses in the calculation, there would be no factor of 2:

    [tex]Q = M_{nuc}(A,Z) - M_{nuc}(A,Z-1) - m_e[/tex]

    However, the masses that we actually find in the standard tables are atomic masses of neutral atoms, which include the mass of Z electrons: [itex]M(A,Z) = M_{nuc}(A,Z) + Zm_e[/itex]. Substituting for the nuclear masses in the first equation:

    [tex]Q = [M(A,Z) - Zm_e] - [M(A,Z-1) - (Z-1)m_e] - m_e[/tex]

    Remove the brackets and parentheses, collect terms, and you end up with your equation with the "extra" factor of 2.
     
  4. Feb 3, 2010 #3
    ah cheers for that jtbell.

    can I ask what the justification is for using the atomic mass? its just that the title of my module is "nuclear physics" after all.
     
  5. Feb 3, 2010 #4

    jtbell

    User Avatar

    Staff: Mentor

    Probably because what we measure in mass spectrometers etc. is the atomic mass, or something close to it, e.g. a singly-charged ion. It's kind of hard to strip all the electrons off, say, an iron or a uranium atom.
     
  6. Aug 9, 2010 #5
    I am as confused as the thread starter =)

    I get that the M(A,Z) - M(A,Z-1) - 2Me is from the atomic mass..

    I just don't know how to relate these graphically..
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook