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Kinetic Energy of a daughter Particle in Alpha decay.

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    How much kinetic energy does the daughter have when Samarium 148 undergoes alpha decay from rest?


    2. Relevant equations

    Q=(Mparent + Mdaughter + Malpha)c^2=Kd+Kalpha

    Kalpha=MdQ/(Md+Malpha)

    3. The attempt at a solution
    I started with the fact that samarium 148 would become Neodymium 144 and a Helium Nucleus after alpha decay. Then, I used the above equation for disintegration energy to get 3.00800145MeV. Then I used the fact that 3.00800145-Kalpha=Kd but I am having trouble finding the kinetic energy of the alpha particle. I thought it was Kalpha=MdQ/(Md+Malpha)but when I submit my answer on WebAssign using this method it says the answer is wrong so there must be a problem somewhere.
     
  2. jcsd
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