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Compute the energy of the emitted alpha for Na20 decaying to an excited state

  1. Mar 20, 2017 #1
    I think I got this right, I just want a second opinion to know if my concepts are correct

    1. The problem statement, all variables and given/known data

    20Na decays to an excited state of 20Ne through the emission of positrons of maximum kinetic energy 5.55 MeV. The excited state decays by ##\alpha## emission to the ground state of 16O. Compute the energy of the emitted ##\alpha##

    2. Relevant equations
    ##T_{\alpha}=\frac{Q}{1+\frac{m_{\alpha}}{m_{X'}}}## where ##T_{\alpha}## is the kinetic energy of the alpha particle, ##m_{\alpha}## is the mass of that alpha particle, and ##m_{X'}## is the mass of the daughter nucleus.

    ##Q=\Delta mc^2##

    3. The attempt at a solution
    For 20Na →20Ne,
    ##Q_0## = ##\Delta mc^2## = 13.8885 MeV. Of that 13.8885 MeV, 5.55 MeV goes to positrons so 13.8885-5.55 = 8.3385 MeV is left, which is the excitation energy.
    For 20Ne*16O + 4He, where * denoted an excited state,
    ##Q_1## = 8.3385 + ##\Delta mc^2## = 8.3385 + (-4.72987252) = 3.6086295 MeV = ##Q_1##
    ##T_{\alpha} =\frac{Q_1}{1+\frac{m_{\alpha}}{m_{X'}}} = 2.88634 MeV = T_{\alpha}##

    Now this would be the kinetic energy. Would I have to take the square root of the kinetic energy squared plus the rest mass energy squared? Or is the 2.88 my answer? I think it is the 2.88 because it should not be relativistic so rest mass energy should not matter here in this case. What are your thoughts?
     
  2. jcsd
  3. Mar 20, 2017 #2

    mfb

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    I'm quite sure the question asks for the kinetic energy of the alpha particle, not the total energy.
     
  4. Mar 20, 2017 #3
    Nah, it just says "energy"
     
  5. Mar 20, 2017 #4

    mfb

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    Yes, and I'm quite sure that refers to the kinetic energy.
     
  6. Mar 20, 2017 #5
    Oh okay. So in that case do you think I did it correctly with 2.88 MeV as my answer?
     
  7. Mar 20, 2017 #6

    mfb

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    I didn't check the numbers, the approach is right.
     
  8. Mar 20, 2017 #7
    Okay great, thank you!
     
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