Energy of a deuteron

  1. 1. The problem statement, all variables and given/known data
    A proton (1.6726×10-27 kg) and a neutron (1.6749×10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "heavy hydrogen". In this process, a gamma ray (high-energy photon) is emitted, and its energy is measured to be 2.39 MeV (2.39×106 eV). Keeping all five significant figures, what is the mass of the deuteron? (Assume you can neglect the small kinetic energy of the recoiling deuteron.)


    2. Relevant equations
    Rest Energy = m c^2
    Ebefore=Eafter



    3. The attempt at a solution

    I found the energy before the combining by

    c^2 (1.6726×10-27+1.6749×10-27) = 3.0086*10^-10 J

    this should equal the energy of the deuteron (mc^2) + the energy of the photon

    energy of the photon in joules: 3.8292*10^-13 J

    therefore:

    3.0086*10^-10 = mdc^2 + 3.8292*10^-13

    md=3.3433*10^-27 kg

    However, this isn't right
     
  2. jcsd
  3. Do you know the correct answer? You should be right, unless there is a rounding error, but I know that the mass of deuteron is just about what you've stated. Make sure you're giving the answer in the correct units (usually we talk about particle masses in terms of rest mass and give the units in MeV).
     
  4. you're right, it's wrong- binding energy is negative so the above equation should read:

    3.0086*10^-10 = mdc^2 - 3.8292*10^-13,

    so

    md=3.34318*10^-27 kg

    only a fraction out.

    What gets me is that, where Ed is the deuteron binding in joules, Up and Un are the proton and neutron magnetions respectivey and Uo is the magnetic permeability of free space:

    (UpUnUo/Ed)^1/3 = 0.746 fm

    in other words a coupling distance between proton and neutron forming a deuteron of about half a nuclear radius from simple magnetic dipole binding, without having to invoke an extra-nucleonic strong force at all. Protons will also bind nicely once the magnetic attraction overcomes electrostatic repulsion at about 1.8 fm, at an energy of about 532Kev. Of course there are no quantum numbers involved here so the digits are subject to change, but not the scale, which is one of nuclear interactions.
     
  5. vela

    vela 12,780
    Staff Emeritus
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    Actually, you have it backwards.
     
  6. Last edited: Mar 20, 2010
  7. I know what you happened. You got all the steps right, but you probably need to use the speed of light as 3*10^9 in the program. I did this same problem for my class and i figured it out by your post since it didn't except my answer wither until i did that.
     
  8. although i now realise the formula is wrong by a factor of 4pi, and the separation distance is also wrong- I have been looking at Yukawa potentials, but formalism constantly gets in the way of a formula which is actually useul when it comes to replicating numbers on a calculator. It's nice to know people are out there to help and don't just ignore wrongheaded ideas in favour of stating the equally incorrect (for example that the deuteron is heavier than its constituents), or repeating indeciperable jargon. :| Einstein would be ashamed!
     
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