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Protons and neutrons in a nucleus are both in a well of $\sim$ 50 MeV (obviously that depends on the specific nucleus), but the shape of the quantum well is different because there is the Coulomb force. If protons repel each other, why the Coulomb interaction between them acts as a barrier during a fission, like a $\alpha$-decay? Shouldn't it facilitate the fission?enter image description here

My professor and one of the book she follows, says: "The emitted nucleons are primarily neutrons since they are not hindered by the Coulomb threshold".

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The barrier is not due to the Coulomb force, but due to the interplay between the short-range nuclear forces and longer range Coulomb force:

The variation in specific binding energy with atomic number is due to the interplay of the two fundamental forces acting on the component nucleons (protons and neutrons) that make up the nucleus. Nuclei are bound by an attractive nuclear force between nucleons, which overcomes the electrostatic repulsion between protons. However, the nuclear force acts only over relatively short ranges (a few nucleon diameters), since it follows an exponentially decaying Yukawa potential which makes it insignificant at longer distances. The electrostatic repulsion is of longer range, since it decays by an inverse-square rule, so that nuclei larger than about 12 nucleons in diameter reach a point that the total electrostatic repulsion overcomes the nuclear force and causes them to be spontaneously unstable. For the same reason, larger nuclei (more than about eight nucleons in diameter) are less tightly bound per unit mass than are smaller nuclei; breaking a large nucleus into two or more intermediate-sized nuclei releases energy.

See Origin of the active energy and the curve of binding energy

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  • $\begingroup$ I still don't understand the role of the barrier. I get it that the barrier is not due solely to the Coulomb force, but it appears only for protons. Neutrons are in a well due to the nuclear forces, instead protons are in a well like neutrons AND they have to face a barrier, which is due to their charge. This way it seems that it's harder to free a proton than a neutron, which is false. I expect the Coulomb force to act as a facilitator rather than a barrier, as it should promote fission. I guess it's something about the depth of the well.... $\endgroup$
    – Ako
    Commented Sep 21, 2023 at 17:17
  • $\begingroup$ @Ako If we have only one proton and one or two neutrons, then there is no Coulomb interaction - that is there is only potential well, in which the particles are confined. Coulomb force does act as a facilitator in the sense that a system has higher bound state energies than it would have without the Coulomb interaction - e.g., you could compare tritium and Helium-3, which have the same number of nucleons, but different number of protons. $\endgroup$
    – Roger V.
    Commented Sep 21, 2023 at 18:19
  • $\begingroup$ I guess I didn't ask correctly. My book ("Particles and Nuclei" by Povh et al.) says, about separation energy and nuclear stabilty, that "The emitted nucleons are primarily neutrons since they are not hindered by the Coulomb threshold". $\endgroup$
    – Ako
    Commented Oct 5, 2023 at 9:07

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