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Another Nuclear

  1. Mar 7, 2009 #1
    Another Nuclear Physics problem: Minimum photon energy for deuteron dissociation

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

    What is the minimum photon energy necessary to dissociate a deuteron. Take the binding energy to be 2.224589 MeV

    2. Relevant equations

    [tex]\gamma[/tex] + 2H [tex]\rightarrow[/tex] 1H + n

    [tex]\vec{P}[/tex] = [tex]\vec{P_n}[/tex] + [tex]\vec{P_p}[/tex]

    where P = momentum of incident gamma ray
    P_p = momentum of recoiling proton
    P_n = momentum of recoiling neutron

    E + m(2H)*c2 = E_p + E_n

    where E = energy of incident gamma ray (what I need to find) = P/c, c = speed of light
    E_p = energy of recoiling proton = [m(1H)2*c4 + P_p2*c2]1/2
    E_n = energy of recoiling neutron = [m(n)2*c4 + P_n2*c2]1/2

    B = [m(1H) + m(n) - m(2H)]*c2 = 2.224589 MeV

    3. The attempt at a solution

    I tried solving all the above equations simultaneously, but then of course ran into problems writing the magnitudes of the momentum for the neutron and proton in terms of the momentum of the incoming photon. How do I know the neutron and proton recoil at the same angle and how could I go about calculating that angle?

    Thanks for any help.
    Last edited: Mar 7, 2009
  2. jcsd
  3. Mar 7, 2009 #2


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    The minimum energy is found when the proton and neutron are created at rest in the centre of mass frame. Use the concept of invariant mass.
  4. Mar 7, 2009 #3
    Thank you very much Malawi Glenn! You are turning out to be a real life saver.

    Ok, I just want to check that I understood the physical situation properly.

    I think what happens here is that the centre of mass of the system is the same in the lab frame as in the centre of mass frame? So, because the neutron and proton pick up no kinetic energy for a minimum energy photon, this is true both before and after the reaction? So, the rest mass energy before the reaction equals the rest mass energy after the reaction. Does this make sense? I'm not too sure, but here is the answer I got anyway: If B = 2.224589 MeV, then the minimum photon energy to dissociate the deuteron is 2.225908 MeV.

    Thanks again for the time and effort you put in to help newbies to Nuclear Physics like me.

  5. Mar 7, 2009 #4


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    Try to put it more stringent, using the formalism of invariant mass.

    Invariant mass of a system is:

    [tex]s = (\sum _i E_i )^2 - (\sum _i \vec{p}_i)^2[/tex]

    And this quantity is INVARIANT, same in ALL frames!

    So first evaluate the invariant mass before reaction, in "lab.frame" where the deuteron is at rest. Call the photon energy = E.

    Then set this equal to the final state, in CENTRE OF MOMENTUM frame (i.e. where the sum of momenta = 0. This is 'almost' centre of mass frame, since proton and neutron do not have same mass, if they would, centre of momentum will also be centre of mass frame).

    Now the mass of the deuteron is what? well it is m_neutron + m_proton - Binding energy, as you had.

    Try to do this in the invariant mass formulation, which is the standard way to do things such as this, called 'threshold energy problems', the minimum energy for a process to occur.
  6. Mar 7, 2009 #5
    Thank you, now I understand it much better. I didn't know that the expression you gave me is true no matter what reference frame I use. I read something on the internet about it being invariant under the Lorentz transformation in Special Relativity.

    I was exposed to Special Relativity very generally in first year and we were supposed to do some more in second year in Classical Mechanics, but then they changed the syllabus, my lecturers probably thought that what we had done in first year was enough (we used Halliday, Resnick and Walker's Fundamentals of Physics), and I've forgotten everything. This has been a good wake up call for me to go and read up some more about it, because I using it in QM as well.

    Anyway, back to the problem in question. I got the same answer I did before though, 2.225908 MeV. So, maybe I'm doing something wrong?

    For the expression before the reaction, I used the rest energy of the deuteron plus the energy of the photon. For the momentum, I used just the momentum of the photon which I then rewrote in terms of energy via E = Pc.

    For the expression after the reaction, I used the rest energy of the neutron plus the proton's. The momentum was zero because there is no kinetic energy.

    Finally, I equated the two expressions and solved for the energy of the photon.

    Does it sound right to you?

    Thanks again.

  7. Mar 8, 2009 #6


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    IF you show me how you done, I can tell if you did it correctly. The answer IS 2.225908 MeV, but since you said "the neutron and proton pick up no kinetic energy for a minimum energy photon" I have to give you this tutorial. Due to momentum conservation, the proton and the neutron HAVE to get momentum in the lab-frame.

    Lorentz transformations TAKES you to different references frames.

    Imagine a proton with lab kinetic energy 100GeV hitting a fixed carbon nuclei, which does not move in lab. Now there is one unique lorentz boost which takes that reference frame (the lab) to a frame where the total momentum of those two nuclei (the proton and the carbon) is zero. This frame is called the Centre of momentum frame.

    Now same here, in lab system, you have a fixed deuteron and one photon with energy coming in. You want to evaluate what energy the photon needs to have in order to make the reaction p + n.

    s_init = (E_photon + E_d)^2 - (p_photon + p_d)^2 = d at rest = (E_photon + m_d)^2 - p_photon ^2 = 2* E_photon * m_d + m_d^2

    s_final = centre of momentum frame, and where we create the particles at rest, will give the min. E to disintigrate D - there is one such lorentz transformation which takes you to this frame. = (m_p + m_p)^2

    If you did it like this, you did it the right way
  8. Mar 8, 2009 #7
    Yes, I was definitely wrong when I said that the proton and neutron pick up no momentum in the lab frame, thank you very much for pointing that out to me and also insisting that I understand it right. It's in the centre of momentum frame where the total momentum of the proton and neutron is zero, and that's what I used in the end to find the answer.

    So, I can use the Invariance of mass principle to relate the before and after of a nuclear reaction without having to worry about which reference frame I'm in on either side of the equation, that's a very powerful and helpful method! I definitely won't forget it!

    Thank you again for all your help.

  9. Mar 8, 2009 #8


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    yup, invariant mass is really good to use.

    As an exercise, consider the reation

    proton + proton = 2 protons + 2 pi^0

    With the proton as fixed target. Calculate the threshold kinetic energy of the proton beam.

    If you want ;-)

    The pi^0 is a meson, a relative to the proton, which has rest mass 135MeV/c^2. Take the proton mass to be 940MeV/c^2 in this problem.
  10. Mar 9, 2009 #9
    Hi there,

    Thanks for the additional example!

    Let me just check with you if the expression I wrote down for the invariant mass of the left side of the reaction (which I took to be in the lab frame) is correct:

    {Ep + (Ep2 + P2c2)1/2}2 - {([tex]\vec{P}[/tex]c)}2

    where Ep = rest mass energy of a proton
    [tex]\vec{P}[/tex] = momentum of incident proton

    The expression I wrote down for the right side of the reaction (taken in the centre of momentum frame because this is a "threshold" problem) is:

    {Ep + Ep + E[tex]\pi[/tex] + E[tex]\pi[/tex]}2 - {[tex]\vec{P}[/tex]c + [tex]\vec{P}[/tex]c + [tex]\vec{P\pi}[/tex]c + [tex]\vec{P\pi}[/tex]c}2
    = {Ep + Ep + E[tex]\pi[/tex] + E[tex]\pi[/tex]}2 - 0

    because the total momentum in the centre of momentum frame is zero.

    where E[tex]\pi[/tex] = rest mass energy of the mesons
    [tex]\vec{P\pi}[/tex] = momentum of a meson (which is not the same for both of the mesons in general, same for the two proton momenta)

    so, then I equate these two expressions and solve for Pc, the kinetic energy of the incident proton.
  11. Mar 9, 2009 #10


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    Try to use standard notation, m for rest-mass and T for kinetic energy, p for momentum.

    Recall that

    p^2 = T^2 + 2*T*m.

    So your statement "solve for Pc, the kinetic energy of the incident proton" is not quite true ;-)

    Otherwise it is quite fine
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