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We know that neutrons exert short ranged nuclear forces over other nucleons in a nucleus, and these forces are only attractive in nature. Also this force is universal and doesn't differentiate between charged or uncharged bodies. So why doesn't a nucleus-like body exist, purely made up of neutrons? That would be highly stable because there is no coulombic repulsion between neutrons as they don't have any charge, and they will still be strongly bounded by nuclear forces.

So why doesn't this exist? What am I missing?

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2 Answers 2

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The reason for this is that unlike the electrostatic force the nuclear force depends on how the spins of the two particles are aligned. The force is stronger when the spins are in the same direction than when they are in opposite directions. To see why this causes a problem imagine trying to assemble some number of neutrons into a nucleus. We expect there will be energy levels like the energy levels for electrons in an atom, though they'll be more complicated (this is the nuclear shell model) and each energy level will contain two particles.

We try to put the first two neutrons into the first energy level, but the problem is the strong force wants the spins to be in the same direction and the Pauli exclusion principle doesn't allow this. We would have to flip one of the spins to make the two spins opposite, but this reduces the nuclear force between the particles and it raises the energy of the level. Then we try to add the third and fourth neutron into the second lowest energy level, and we run into the same problem. We can do it, but the energies will be much higher than they would be if the spins were parallel, so the nucleus would be much less strongly bound.

Now this doesn't mean the collection of neutrons wouldn't be bound, but there's a problem. Neutrons freely convert to protons, and you can put a proton and neutron together into a single orbital with their spins in the same direction because they have different isospins. So if you put two neutrons into the lowest orbital with their spins opposite one of the neutrons will turn into a proton. Now the two particles can have their spins parallel, which lowers the energy of the orbital and therefore makes the nucleus more strongly bound. The same will happen with the next lowest orbital, then the next and so on. Your nucleus made up of neutrons would spontaneously convert into a 50/50 mixture on neutrons and protons because it lowers the energy.

This argument implies all nuclei should be a 50/50 mixture of protons and neutrons, and this is approximately right but only approximately. This is because binding in nuclei is more complicated than the rather simple model I've described above. But while the model is wrong in detail it does capture the general principle involved.

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    $\begingroup$ This is the clearest exposition of this topic I have read. Thanks for posting it- NN $\endgroup$ May 17 at 15:43
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    $\begingroup$ @CGS A neutron star is about 1% protons. See e.g. $\endgroup$
    – rob
    May 18 at 0:23
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    $\begingroup$ @lalala Yes, exactly! If a neutron decayed to a proton in the nucleus it would increase the total energy since the orbital would now contain two protons and they'd have to align their spins antiparallel. That makes it energetically unfavourable for the neutron to decay. $\endgroup$ May 18 at 10:33
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    $\begingroup$ @RBarryYoung the decay releases about 1 MeV of energy and even the most strongly bound electrons have energies of only around 10 keV. So the emitted electron hightails it for parts unknown and is never seen again. $\endgroup$ May 18 at 15:09
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    $\begingroup$ @JohnDoty still way less than a Mev though ... $\endgroup$ May 18 at 15:23
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  • The strong nuclear force conserves the number $$A~=~N+Z$$ of nucleons at low energy modeled by Yukawa pion exchanges. In this answer we will focus on stability under neutron/proton transformations, i.e. processes that keep $A$ fixed, cf. OP's title question.

  • In Weizsäcker's semi-empirical binding-energy formula, the 1st term (the volume term) and the 2nd term (the surface term) depend only on $A$ and are therefore constant.

  • The 3rd term (the Coulomb repulsion) term obviously favors a surplus of neutrons.

  • OP's title question is related to the 4th term (the asymmetry/Pauli energy term) which yields a tendency to have the difference $$\Delta~=~N-Z~\in~[-A,A]$$ be zero.

    To a first approximation, the nucleus can be modeled as 2 independent proton and neutron Fermi gasses confined within a 3D volume, with total kinetic energy$^1$ $$\begin{align} E_{\rm kin}~=~~&C\left(N^{5/3}+Z^{5/3}\right)\cr ~=~~&C\left(\left(\frac{A+\Delta}{2}\right)^{5/3}+\left(\frac{A-\Delta}{2}\right)^{5/3}\right)\cr ~\stackrel{\text{convex}}{\geq}&C\left(\left(\frac{A}{2}\right)^{5/3}+\left(\frac{A}{2}\right)^{5/3}\right). \end{align}$$ This clearly has a minimum kinetic energy (corresponding to maximum binding energy) at the diagonal $$\Delta~=~0$$ because $N\mapsto N^{5/3}$ is a convex function. This is the standard answer to OP's title question. (If the function had instead been concave, then the 4th term would have favored nuclei consisting of only neutrons or of only protons.)

  • More features of the valley of stability are provided by the 5th term (the pairing term), and the nuclear shell model.


$^1$ The number of neutrons $N_{|\vec{n}|}\sim |\vec{n}|^3$ in 3D grows with the 3rd power, while the kinetic energy $E_{|\vec{n}|}\sim |\vec{n}|^2\sim N_{|\vec{n}|}^{2/3}$, so that the total kinetic energy of neutrons $E_{\rm kin}=\int_0^N E_{|\vec{n}|} \mathrm{d}N_{|\vec{n}|}\sim N^{5/3}$ is a convex function of the number $N$ of neutrons. There is an identical argument for protons.

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