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Homework Help: Beta plus decay

  1. Dec 14, 2007 #1
    1. The problem statement, all variables and given/known data
    The nuclear reaction [tex]^63_29[/tex]Cu(p,n) [tex]^63_30[/tex]Zn is immeadiately followed by [tex]/Beta^+[/tex]. Write the equation for this [tex]/Beta^+[/tex] decay.

    If the maximum energy of the positrons is 2.36 MeV, find the Q value of the original (p,n) reaction.

    2. Relevant equations
    [tex]Q=[M(A,Z) - M(A,Z-1) -2m_e]c^2[/tex]

    3. The attempt at a solution
    Decay chain:
    First: Copper + proton ---> Zinc + neutron
    Second: Zinc ---> Copper + positron + neutrino

    So Zinc can be substituted into the first equation to give that:
    proton ---> neutron + positron + neutrino

    Now I get confused. I pulled the above equation from my notes, and I know that its the energy of the system, but other than that I don't understand.

    I've spoken to my lecturer about it, but he didn't make much sense with it so I'm hoping one of you could help me better with it.

    As I understand it [tex]Q_{tot} = Q_1 + Q_2[/tex] where Q_2 is the second decay, given by the previous equation, but can be assumed to be the max KE of the positrons (as the neutrion is massless etc). But then I get stuck, as I can't see how I'd find the total energy of the system, much less the energy of the original (p,n) reaction.

    Could someone point me in the right direction?


  2. jcsd
  3. Dec 14, 2007 #2
    Ok, first let's understand the energetics of what is going on. In your original reaction you are shooting protons at copper nuclei and this is causing a proton capture reaction. You could calculate the Q value for this reaction using masses only if the proton didn't carry kinetic energy. Since the proton must be carrying a fair amount of kinetic energy to overcome the Coulomb barrier (positive charges repel), simply using:


    will not work.

    Secondly, we know that the energy of the positron is 2.36 MeV. Since the big copper nucleus is unlikely to carry away much kinetic energy (because of its big mass), and the neutrino is unlikely to carry away much kinetic energy (because of its small mass) we can assume (as you already stated) that ALL of the total reaction energy is found in the positron. The final energy is the Q value for the whole process. Thus, the Q value for the total reaction is 2.36 MeV.

    Next, you have a formula for the regular Q-value of the beta plus decay. Compute it. It should be smaller than 2.36 MeV by some amount. Where could this energy have come from?
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