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Proton fusion, Beta + decay

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider;
    [tex]p+p \rightarrow d + e^+ + \nu_e[/tex]

    Assume the binding energy of deuteron is 2.2MeV, calculate the maximum energy that the positron can have.
    2. Relevant equations
    [itex]m_p=938.28MeV/c^2[/itex]
    [itex]m_n=939.566MeV/c^2[/itex]
    [itex]m_d=1875.6MeV/c^2[/itex]
    3. The attempt at a solution
    Assuming the neutrino was at rest after the colision for maximum positron energy.

    So basically two process's go on, the [itex]\beta^+[/itex] decay of one of the protons, then the fusion of the neutron and remaining proton.

    [tex] (1) \quad p \rightarrow n + e^+ +\nu_e [/tex]
    [tex](2) \quad p+n \rightarrow d [/tex]

    The Q value of the decay:
    [itex] Q=(m_p)c^2-(m_n)c^2=-1.286MeV [/itex]
    this is the minimum energy the proton would need to decay to a neutron?

    so the energy i got was;
    [tex] T_{e^+}=2.2-1.286=0.914MeV [/tex]
    where the 2.2 is the energy released from (2)?

    But im not sure i think this is wrong, but dont quite understand.
     
  2. jcsd
  3. May 11, 2015 #2

    mfb

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    Fusion plus decay is one single process. If you want to split it into parts (which does not help here), fusion would have to happen first, otherwise the decay could not happen at all.
     
  4. May 11, 2015 #3
    Ah thank you, I just don't quite understand where to begin, would I need to find the energy from the decay of the bound protons, to the bound neutron and proton?
     
  5. May 11, 2015 #4

    mfb

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    All you need is the binding energy of 2.2 MeV and particle masses for kinematics. Neglect the initial kinetic energy of the protons, and find a way to distribute the 2.2 MeV over the reaction products for the maximal positron energy.
     
  6. May 11, 2015 #5
    Oh I see, so treat it as a relativistic kinematic problem;

    Energy conservation;
    [tex] Q=\frac{1}{2}m_d v_d^2+\gamma_{e^+} m_{e^+} v_{e^+} [/tex]
    Because the deuteron mass would be larger than the portion it would get from 2.2MeV.

    Conservation of momentum;
    [tex]m_d v_d = m_{e^+}v_{e^+}[/tex]
     
    Last edited: May 11, 2015
  7. May 11, 2015 #6

    mfb

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    Check the formula for the positron energy.
    Apart from that, the approach is right.
     
  8. May 12, 2015 #7
    Oh yes it's [itex]c^2[/itex] not [itex]v^2[/itex] thank you!
     
  9. May 12, 2015 #8
    Okay so I got;

    [tex] E_{e^+}=Q-\frac{m_{e^+}^2v_{e^+}^2}{2m_d} [/tex]

    where i used momentum conservation to substitute for [itex]v_d[/itex], I'm not sure how to get the velocity of the positron or how to get rid of it in the equation.

    I could substitute for the momentum using [itex]E^2=(pc)^2+(mc^2)^2[/itex] but that gives a quadratic, where i could take the larger solution for the total energy of the positron?
     
    Last edited: May 12, 2015
  10. May 12, 2015 #9

    mfb

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    Right (I guess one solution will be negative anyway),
     
  11. May 12, 2015 #10
    Thank you, got a reasonable solution.

    Im just curious do you think it would've been okay to say that the second term for the equation for [itex] E_{e^+} [/itex] would be much less than Q and therefore;

    [tex] E_{e^+} \approx Q [/tex]

    Because the actual solution i got was [itex] E_{e^+} = 2.199MeV[/itex]
     
    Last edited: May 12, 2015
  12. May 12, 2015 #11

    mfb

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    It is certainly a reasonable approximation if you don't care about keV precision, but then the question is too easy.
     
  13. May 12, 2015 #12
    Ah yes that makes sense always better to be precise when appropriate. Thank you for your help it was much appreciated :)
     
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