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Homework Help: 4-momentum conservation (Deuteron + Photon)

  1. Jan 1, 2014 #1
    1. The problem statement, all variables and given/known data
    See attachment.

    2. Relevant equations

    3. The attempt at a solution


    [tex]\gamma + D \rightarrow p + n[/tex]

    [tex](E_1,\underline{k}_1) + (E_2,\underline{0}) = 2(E_3,\underline{k}_3)[/tex]

    where I have assumed E3 ≈ E4 and k3k4 as mn≈ mp and vp = vn

    then splitting E and K components up and solving for photon energy E1 I end up with,

    [tex]E_1 = \frac{4m_p^2}{2E_2}-\frac{E_2}{2}[/tex]

    this is where I start to get confused, so the binding energy of the Deuteron is given in the question, would I need to calculate the rest energy of the Deuteron E2 as,

    [tex]E_2 = [m_p + m_n + m_e]c^2 - B(^2_1D_1)[/tex] (c=1)

    if I do this and then sub values in I get E1 = 0.4236MeV, ok so I realize the question is asking to show that the minimum energy of the photon needed is when vp=vn, the way I understand this is that the minimum energy required to break the deuteron apart would be the deuterons binding energy. Therefore my value of E2 should be equal or at least close to 2.2MeV but it's not so I have obviously interpreted something badly, can anyone help?

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    Last edited: Jan 1, 2014
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  3. Jan 1, 2014 #2


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    Your equations look ok to me except I don't see the reason for including the mass of the electron in your deuteron energy.

    Did you mean E1 here? Anyway, I don't see how you got this value.
  4. Jan 1, 2014 #3
    Yes I did mean E1, so I solved the 4-momentum conservation for Energy of the photon, then used the Deuterons binding energy to calculate its rest energy, subbed in the expression for the photon energy and ended up getting that value. Could you elaborate which part you don't see? I have obviously done something wrong so I don't see how I got to that value apparently either, is my methodology correct? If so then I must have made an error somewhere along my calculations.

    EDIT: My bad the deuteron is just the nucleus of the deuterium right? I'll recalculate without the electron energy. . . :)
  5. Jan 1, 2014 #4


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    I don't see anything wrong in your equations. Just in the numerical result.
  6. Jan 1, 2014 #5
    ok so neglecting any electron energy I end up with deuteron rest energy as 1875.625411MeV, plug this into the E1 expression and get E1 as 0.9348MeV, thats still not really a good result right? Any ideas? :s

    I'm using mass of proton as 938.28MeV/c^2 and mass of neutron as 939.57MeV/c^2 btw
  7. Jan 1, 2014 #6
    Surely it's not possible for the photon to split the Deuteron with less energy than its binding energy?
  8. Jan 1, 2014 #7


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    That's right.

    I think the problem is using the approximation mp + mn ≈ 2mp in E3 but not making that approximation in E2.

    The result for E1 involves small differences and so you need to be careful with any approximations.
  9. Jan 1, 2014 #8
    Haha, I managed to suss that but forgot to reply, anyway thanks for pointing this stuff out to me it all works :)
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