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Calculating energy released from beta decay

  1. Sep 11, 2009 #1
    I have been learning particle physics lately but it's been mostly from a theoretical perspective and not a mathematical one so I have yet to come across any such math but my curiosity is peaked.

    From what I understand it, this is the process:

    [itex]
    n \rightarrow p + W^{-}
    [/itex]

    Followed by:

    [itex]
    W^{-} \rightarrow e^{-} + \bar{v}_{e}
    [/itex]

    Since [itex]m_{e} ~= .511 MeV/c^2[/itex] and [itex]m_{v_{e}} << m_{e}[/itex], there is about [itex]79.9995 GeV/c^{2}[/itex] missing. I am unclear where this energy goes. Does it go into the momentum of the electron (since it is ultrarelitivistic during the decay)?

    I've tried to solve the equation [itex]E = m^{2}c^{4} + p^{2}c^{2}[/itex] but I get a weird mass for the electron ([itex]5.68 x 10^{-12} kg[/itex]) and I am all around confused by the equation.

    If I have the wrong equation, which other one do I use for this type of math? If I am using the write equation, what is the best way of dealing with the units?
     
  2. jcsd
  3. Sep 11, 2009 #2

    Astronuc

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    The kinetic energy of the proton (or nucleus), electron and anti-neutrino comes from the difference in mass between the products [itex]p + e^{-} + \bar{v}_{e}[/itex] and the neutron. The partition of the energy is governed by conservation of momentum. If the neutron is in a nucleus, the nucleus acquires very little kinetic energy because it's some massive. Even the case of the bare neutron, the vast majority of kinetic energy is manifest in the electron and anti-neutrino, and the partition of energy is largely determined by the angle between the electron and anti-neutrino trajectories.

    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta2.html#c1
     
    Last edited: Sep 11, 2009
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