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Homework Help: Kinetic energy of daughter nucleus and alpha particle from alpha decay

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data
    In [itex]\alpha[/itex] decay a nucleus X at rest decays to a daughter nucleus Y and an [itex]\alpha[/itex] particle. Conservation of momentum and kinetic energy gives:
    [itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

    [itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

    Where the Q value is the available energy found through [itex]Q=(M(X)-M(Y)-M(\alpha))c^{2}[/itex]

    Show the kinetic energy of the two decay products are given by

    [itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

    [itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]


    2. Relevant equations
    [itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

    [itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

    [itex]Q=(M(X)-M(Y)-M(\alpha))c^{2}[/itex]

    [itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

    [itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]

    3. The attempt at a solution
    I have tried rearranging the energy equation to get [itex]E^{\alpha}_{k}=Q-\frac{1}{2}M_{Y}v_{Y}^{2}[/itex]
    since
    [itex]E^{\alpha}_{k}=\frac{1}{2}M_{\alpha}v_{\alpha}^{2}[/itex]

    Then rearranging the conservation of momentum equation and substituting in for various variables but I can't get anything that looks like the required expressions. I know this is a fairly simple algebra exercise but I just can't figure out what to do so any advice or suggestions would be appreciated.
     
  2. jcsd
  3. Sep 2, 2011 #2
    I managed to figure this out. I rearranged the expression for Q to get [itex]E^{\alpha}_{K}[/itex]like in my initial attempt but then I took Q out as a common factor which was the key. It was then just a matter of doing about 3 pages of algebra on what was inside the brackets.
     
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