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- Thread starter BillKet
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You can read about the semi-empirical mass formula that predicts $$\frac{N}{Z}\approx1+0.02(N+Z)^{2/3}$$ where ##N,Z## are the number of neutrons and protons respectively.

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Oh I see. So basically the extra mass of the neutron must add more energy than the mass of the proton plus the coulomb repulsion (roughly), in order for beta decay to happen, right?

You can read about the semi-empirical mass formula that predicts $$\frac{N}{Z}\approx1+0.02(N+Z)^{2/3}$$ where ##N,Z## are the number of neutrons and protons respectively.

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Is one argument, another argument, slightly more accurate, is using the Pauli exclusion principle, that says that two identical fermions (the proton and the neutron are fermions) cannot be in the same quantum state.Oh I see. So basically the extra mass of the neutron must add more energy than the mass of the proton plus the coulomb repulsion (roughly), in order for beta decay to happen, right?

Therefore since the energies are quantized, if you add a lot of neutrons they will go to higher energy states, while the protons will go to states with lower energies (since if there are few protons, they will not be occupied), so adding a neutron will be less efficient than adding a proton.

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You have Fermi repulsion. Since the strong force potential hole is small, with only very shallow tails, you can only have a finite number of states in it - unlike electrostatic monopole-monopole attraction whose long tail of attraction allows an infinite number of states. With fermions, you can fill all of them. Example He-5. Neither an extra proton nor an extra neutron can be bound to an alpha.

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