Why does Nickel-62 have the highest BEPN?

[Jimmy87]

Hi,

I have been looking for an answer to this question and came across this PF article (https://www.physicsforums.com/threads/why-is-iron-56-the-most-stable-nuclei.635183/) which explains most of the type of answer i was looking for.

So as I understand ( Quantum Pion's thread from above) if you have a small nucleus and add more nucleons you increase the binding energy per nucleon as each new nucleon creates more strong force in the nucleus. For larger nuclei since they have a much larger diameter the strong force dies off inside the nucleus causing the electrostatic proton-proton repulsion to dominate thus reducing the BEPN. It was pointed out in the thread that this answer would mean that adding more neutrons for heavier nuclei would solve the problem and goes on to talk about nuclear energy levels.

Could someone please explain how adding excess neutrons and thus forcing them into higher energy shells reduces the BEPN of the nucleus. I get that it makes the nucleus more unstable as the neutrons will decay into protons (i.e. radioactive nucleus) but I don't see how this process lowers the energy required to separate the nucleus?

Thanks

 

[mfb]

The nucleus has more energy. It needs less additional energy to transform to something else - or it might even have enough energy to decay on its own.

 

[Jimmy87]

The nucleus has more energy. It needs less additional energy to transform to something else - or it might even have enough energy to decay on its own.

I see why it decays with excess neutrons but why does it require less energy to separate the nucleons in a nucleus i.e. why does it lower the BEPN? Say you have Nickel which requires the most energy to separate the nucleons in a nucleus, why does it require less energy to separate the nucleus if you added extra neutrons? This should increase the strong nuclear force and reduce electrostatic repulsion so BEPN should increase? Or is it just that extra neutrons always decay into protons if there are too many and this increases the electrostatic repulsion?

 

[mfb]

Consider a somewhat similar situation for electrons in atoms, e.g. lithium. You add one electron, it goes to the deepest available energy state. You add another electron, its binding energy will be smaller - the average goes down. You add a third electron, it now has to go to the second shell and is quite weakly bound - the average gets even lower.

 

[snorkack]

Right.
If you add only neutrons to a nucleus then the extra neutrons occupy higher orbitals and are more weakly bound than an extra proton would be, which is why it tends to undergo beta decay. If you add too many neutrons, they are not bound at all and reach neutron dripline like H-4 or He-5.
If you add just protons, you have exact same problem as with neutrons, plus additional problem with Coulomb repulsion.

If you add both protons and neutrons then for small nuclei, up to Ni-62, balanced addition will result in stronger strong interaction.

But as you go further, you cannot add neutrons only (Fermi repulsion raises their energy and forces them to beta decay) and you cannot add neutrons and protons in a balanced amount (because the remaining Coulomb repulsion of protons, forced by Fermi repulsion of neutrons, is stronger than strong force attraction).