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Radioactive capture of proton and neutron

  1. Nov 11, 2012 #1

    I'm considering the following process

    n+p→d+[itex]\gamma[/itex] where d is the deuteron and [itex]\gamma[/itex] a photon.

    I want to find out the energy of the photon. I know it will be much less than the rest mass of the deuteron (1875.666 MeV/c2). Can I simply use conservation of energy here?

    i.e E[itex]\gamma[/itex] = Ei - Ef?

    Thanks :)
  2. jcsd
  3. Nov 11, 2012 #2

    Simon Bridge

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    Energy is certainly conserved - yep.
    Momentum is also conserved and photons carry momentum.
  4. Nov 11, 2012 #3
    So is it really as simple as E[itex]\gamma[/itex] = (En+Ep) - Ed?
  5. Nov 11, 2012 #4

    Simon Bridge

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    Pretty much. On the scale of nuceons physics gets real simple ... and then it gets weird.
    Anyway, the deuteron mass-deficit is termed it's "nuclear binding energy". This energy is released in fusion and you have to supply this energy for fission.
  6. Nov 11, 2012 #5
    thanks for clearing that up :)
  7. Nov 12, 2012 #6
    just one more thing, if considering the particles to be relativistic, is it still sufficient to apply simple energy conservation?

    Or should the 4-vector formalism be used?
  8. Nov 12, 2012 #7

    Simon Bridge

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    It is usual to go right for the conservation laws - the math is normally done in center-of-mass coordinates. You use ##E_{tot}=\gamma mc^2## etc. It will be a linear transformation to the lab frame to get what your equipment is supposed to see.

    In your example, if the proton and neutron are effectively at rest, then the photon and deuteron will have to go in opposite directions (to conserve momentum) ... so you do have a kinetic energy term to consider.

    In general, the energy of the interaction can do all kinds of things ... for instance, there need not be only one photon. High energy photons can pair-produce ... all sorts of things. That's why when you look at the CERN stuff, the detectors are always shown with a shower of tracks inside them.
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