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B Account for Coulomb repulsion in nuclear fission energy?

  1. May 3, 2017 #1
    Hi,

    I have learned that, in a nuclear fission or an alpha decay, the available energy released is the difference of masses between the initial nuclide (+ eventually some other initial particles) and the output particles, times c^2 (as E = mc^2).

    I'm fine with that. But I wonder : what about the coulomb repulsion ? For instance, an alpha particle, once "ejected", will gain energy due to the coulomb repulsion with the other nuclide produced, I suppose ? Same thing with fission. I have found the following explanation related to this question, from the Feynman Lectures :

    "The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all of the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just “tapped” lightly (as can be done by sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called “nuclear” energy, but it is really “electrical” energy released when electrical forces have overcome the attractive nuclear forces."

    Strangely, he doesn't talk about the difference of masses.

    Anyway, my question is : should we take into account that coulomb energy in the energy output from the nuclear fission (or alpha decay), in addition to the difference of masses, if we want to be really precise ? Why is the coulomb repulsion ignored in the computations ? I have the feeling I am missing something here.
     
  2. jcsd
  3. May 3, 2017 #2

    mfb

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    The Coulomb repulsion is taken into account already. The decay energies apply to the particles at large distances.
     
  4. May 3, 2017 #3
    Thank you for answering. I dont get it, it is taken in account in the masses ? How ?
     
  5. May 3, 2017 #4

    mfb

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    Just consider the total energy before and afterwards.

    Before: You have a nucleus with mass M, the total energy is Mc2.
    After: You have two nuclei far apart with masses m1 and m2, the total energy is (m1+m2)c2 + kinetic energy.
    Both energies are the same. The kinetic energy of the two nuclei together is then given by (M-m1-m2)c2. Measuring the masses gives you the kinetic energy.
     
  6. May 3, 2017 #5

    ChrisVer

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    In general I don't like Feynman's description there, since it's very simplistic and it's not a rule... For example an unstable particle which doesn't have a lot of protons is Be-8. Be-8 is very unstable against alpha decay. The addition of a neutron can turn it stable (eg the stable isotope of Be-9).
    Now we can say that alpha decay is not a fission process [discussed in other recent posts https://www.physicsforums.com/threads/a-decay-and-nuclear-reaction.913465/], but for Be it fulfills the general requierments :biggrin:
     
  7. May 4, 2017 #6
    Mmh I still don't see it...To me Mc² = (m1+m2)c² + KE simply says that the missing mass have been converted into KE. Where is the KE gained from coulomb repulsion in this ?
     
  8. May 4, 2017 #7

    mfb

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    It doesn't make sense to try to split the released energy into components like that.

    There is a distance between the nuclei where the nuclei would not have any kinetic energy. And there is even a distance where they cannot exist at all classically. They have to tunnel through this potential barrier.
     
  9. May 12, 2017 #8
    Are you saying the ejected particle tunnels too far from the potential to take it into account ?
     
  10. May 12, 2017 #9

    mfb

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    I don't understand your question.

    Tunneling happens in the region that is classically forbidden because the potential is higher than the particle energy.
     
  11. May 13, 2017 #10
    Ok, I think I just misunderstood your previous answer.

    I'm trying to go back to simple energy conservation. So we have :
    1. Initially, taking the initial nuclide of mass M at rest : E1 = Mc2
    2. Right after fission occured, but before the ejected nuclide flew away, we should have : E2 = U + (m1+m2)c2, U being the potential energy due to coulomb repulsion between the protons of the initial nuclide and the protons of the ejected nuclide
    3. When the ejected is far away, U has been converted into KE, and we have : E3 = KE + (m1+m2)c2
    4. And of course by energy conservation we have E1 = E2 = E3
    First of all, is this correct ? I simply added the intermediate step with potential energy to the initial and final steps you gave earlier.

    If it is indeed correct, here is what bugged me. In step 2, we have that potential energy U. I thought we had this potential energy, AND, a NON RELATED decrease of the masses of the two nuclides produced after fission. So I was thinking "there are two released energy, one is potential, the other is deacreased mass". The thing I was probably misunderstanding is that this missing mass has been converted into the potential U. So that the KE in step 3 comes from the potential U, which itself "comes from" the decreased mass.

    Is this correct ? This is probably supposed to be very simple, but it's hard for me to picture that mass has disapeared and "turned" into the coulomb potential. How do physicists think of it ? Do they have an understanding of how this "conversion" process occurs ? Or am I wrong in my explanation ?
     
  12. May 13, 2017 #11

    mfb

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    U > KE. Step 2 with the nuclei too close cannot occur as proper state. You would need a negative kinetic energy for that step, which doesn't exist.
    Step 2 with the nuclei a tiny bit separated already works.

    Assigning masses to every part of a system is not always useful, especially as they don't add up nicely or you need negative masses or other weird things. Assigning energies is more reasonable.
     
  13. May 13, 2017 #12
    Oh right, because the particle tunnels through the barrier, I think ? It can't appear inside the barrier. But where the particle appears, after the barrier, is there still some U (compensating the lost mass ?), which is converted into the KE the particle has when it is far away ?

    I though mass and energy were the same thing :D (appart from a c2 factor).
     
  14. May 13, 2017 #13

    mfb

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    Right.
    There are many cases where energy is a useful concept, but assigning a mass via E/c2 to it is not so useful. This includes everything that moves.
     
  15. May 13, 2017 #14
    Awesome, thank you very much for your help @mfb.
     
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