Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 dont have a charge and so there is no electrostatic force, the nucleus still becomes more unstable?

Also why arent there groups of neutrons bound together, so purely neutron nuclei (because they are neutral, im guessing they couldnt form atoms, because of the electrons needed for an atom)

share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.