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.