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Dissociation of deuteron by gamma rays

  1. May 25, 2007 #1
    1. The problem statement, all variables and given/known data
    I have a doubt in the following problem. I get an answer which is slightly different from the book answer.
    1) A 6 MeV gamma ray is absorbed and dissociates a deuteron into a proton and a neutron. If the neutron makes an angle of 90 degrees with the direction of gamma ray, determine the kinetic energies of proton and neutron.



    2. Relevant equations

    E=B.E + K(p) + K(n)


    3. The attempt at a solution

    I solved it in the following way:
    Let E be the energy of the gamma ray. Let K(p) and K(n) be the kinetic energies of proton and neutron respectively. Let m(p) and m(n) be the rest masses of proton and neutron respectively.
    We assume that the gamma ray moves along positive X-direction, neutron moves along positive Y-direction and proton moves at an angle theta to the positive X axis.
    Applying law of energy conservation,
    E=B.E + K(p) + K(n)
    Where B.E=Binding energy
    K(p) + K(n) = 3.78 MeV --------(1)
    Applying law of momentum conservation along X direction,
    E/c = [2m(p)K(p)]^(1/2)cos(theta) ------------- (2)
    Applying law of momentum conservation along Y direction,
    0= [2m(n)K(n)]^(1/2) - [2m(p)K(p)]^(1/2)sin(theta)
    => [2m(n)K(n)]^(1/2) = [2m(p)K(p)]^(1/2)sin(theta) ---------- (3)

    Squaring and adding equations (1) and (2) we get,
    [E/c]^2 + 2m(n)K(n) = 2m(p)K(p)
    On simplification I get the following equation,
    K(p) – K(n) = 0.019 MeV ---------(4)

    Solving equations (1) and (4),
    K(p) = 1.89 MeV
    K(n) = 1.87 MeV
    But the answer is,
    K(p) = 1.91 MeV
    K(n) = 1.86 MeV
    Please help.
     
  2. jcsd
  3. May 25, 2007 #2

    Dick

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    And I get slightly different answers from both of those. But you should note that the masses of the particles are very large compared compared to the kinetic and binding energies. A small change in the mass numbers can produce a pretty large change in the kinetic energies. I doubt you are doing anything really wrong. If you want to send me the exact mass numbers you are using I can try it with those.
     
  4. May 25, 2007 #3
    Neglecting the answers, is my method right?
     
  5. May 25, 2007 #4

    Dick

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    Your method is a nonrelativistic calculation, it that what you were supposed to do? Its justifiable in this case, but it could lead to the sort of small errors you are seeing. The momentum formula sqrt(2*m*KE) is only valid for small velocities. I used the full 4-vector method.
     
  6. May 26, 2007 #5
    Thanx buddy.
     
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