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If you take any large nucleus and add protons to it, the electrostatic repulsion between them will make the nucleus more unstable, because the electrostatic force between them is more repulsive at a greater distance than the strong force is attractive

So how come if you add more neutrons, which don't have a charge and so there is no electrostatic force, the nucleus still becomes more unstable?

Also why aren't there groups of neutrons bound together, so purely neutron nuclei (because they are neutral, I'm guessing they couldn't form atoms, because of the electrons needed for an atom)

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    $\begingroup$ Stability depends on many other things than just electrostatic repulsion $\endgroup$ – Lapmid Feb 14 '17 at 13:53
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In a nucleus whose N/Z ratio is too large, the Pauli exclusion principle forces many of the neutrons to be in states with high energies. This makes the system less stable. For a fixed N, adding protons also makes such highly neutron-rich systems more stable, because the interaction between the protons and the neutrons is attractive, and the protons can go into low-energy states.

There is no Coulomb barrier for neutrons, so if a neutron has a high enough energy to escape, it just does -- no tunneling is required. Even if the system is bound, the system undergoes beta decay toward the line of stability.

There are nevertheless some systems with very high N/Z that are stable against neutron emission. E.g., 8He is bound and has a half-life of 119 ms.

The two pure-neutron systems that theorists predict might have the best chances of holding together are N=2 (the dineutron) and N=4. Experimental searches for dineutrons over a period of decades have failed to find any, so we're pretty sure they're unbound. These people claimed to have detected the N=4 system, with a lifetime of at least ~100 ns, which means it would have to be bound, although not stable with respect to beta decay. Whether they're right is a whole different question. I wouldn't bet a six-pack on it.

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  • $\begingroup$ I thought the Pauli exclusion principle just applied to electrons around the nuclei. So such principle also applies to neutrons within the nuclei? $\endgroup$ – João Pimentel Ferreira Aug 27 '16 at 13:00
  • $\begingroup$ I don't think the sentence "There is no Coulomb barrier for neutrons, so if a neutron has a high enough energy to escape, it just does -- no tunneling is required." is correct. Sure having no Coulomb repulsion doesn't make it easier for the neutron to escape. $\endgroup$ – Virgo Feb 14 '17 at 18:19
  • $\begingroup$ You say adding protons to a neutron rich atom will make it more stable, but isn’t the strong force of a proton weaker than a neutrons? $\endgroup$ – user73837 Mar 21 at 14:41
  • $\begingroup$ @Virgo: Sure having no Coulomb repulsion doesn't make it easier for the neutron to escape. It's counterintuitive but true. $\endgroup$ – Ben Crowell May 1 at 23:04
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    $\begingroup$ @JoãoPimentelFerreira: The exclusion principle applies to all fermions, which includes neutrons and protons. $\endgroup$ – Ben Crowell May 1 at 23:04
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A very n-rich nucleus is unstable to beta decay. The neutron is more massive than the proton, there are therefore lower energy proton states available for neutrons to decay into (emitting a beta decay electron at the same time). Filling these states with protons (i.e. reducing the N/Z ratio) blocks this beta decay channel because the Pauli exclusion principle prevents two protons occupying the same quantum state.

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protected by Emilio Pisanty Apr 15 '17 at 17:37

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