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Relativistic Energy/Momentum?

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data

    Screen Shot 2017-02-09 at 8.15.44 PM.png
    2. Relevant equations


    3. The attempt at a solution
    I'm not sure how to go about working on this. What I have tried is this line of reasoning:

    1x4He needs 2x3He. 2x3He needs 2x(2H + p)

    Energy released is therefore 12.98 + 2(5.51) _+ 2(0.41) = 24.82 MeV.

    However, it does seem that there are no p's in the result of the first reaction to stick into the second. Could anyone shed some light on the proper way to go about solving this? Thanks.
     
  2. jcsd
  3. Feb 9, 2017 #2

    kuruman

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    The proton in the second reaction is gotten from the same pool that provided the two protons in the first reaction. Look at the final result. You have six nucleons coming out so you need six protons going in.
     
  4. Feb 10, 2017 #3
    I'm not sure what you mean as I don't know the names of these particles to begin with. Could you analyse the 1st and 2nd reactions as an example?
     
  5. Feb 10, 2017 #4

    kuruman

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    Before you attempt to use an equation, first you need to understand what the symbols in it mean and second you need to understand what the equation is saying to you and translate that in English. I will tell you the names of the symbols, particles in this case, but you will have to look them up and see what their properties are and what they are made of.
    p = proton, 2H = deuteron, e+ = positron, ##\nu## = neutrino, 3He = helium 3 nucleus, 4He = helium 4 nucleus a.k.a. alpha particle, γ = photon.
    After you look up these particles, study the reactions and verify that the following are true
    1. The number of protons plus neutrons before the reaction is the same as the number of protons plus neutrons after the reaction
    2. The total charge before the reaction is the same as the total charge after the reaction

    Reaction 1 says that 2 protons fuse and out comes a deuteron, a positron a neutrino and some energy.
    Reaction 2 says that the deuteron produced in reaction 1 fuses with a third proton to produce a helium 3 nucleus a photon and some energy

    Now imagine two sets of reaction 1 and 2 happening in parallel producing two helium 3 nuclei plus two of all the other stuff. These two fuse together to form an alpha particle, 2 protons, a photon and some energy.

    To get to this last stage, consider, overall, what is going in and what is coming out.
     
  6. Feb 11, 2017 #5
    Points 1 and 2. The bottom equation seems to have 4e ----> 4e, which means i need to multiply both the equations by 2, am I right?

    Also, could you explain what you mean by "happening in parallel" and how it affects the MeVs released?

    Thank you for you patience
     
  7. Feb 11, 2017 #6

    kuruman

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    Study the picture below. It is worth 1000 words, give or take a few. I have only indicated the protons going in the dashed box where the reaction takes place. It should be easy for you to figure out what I mean by "in parallel" and how much energy and what else comes out of the dashed box.
    Fusion.png
     
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