"The answer lies in their one part in a thousand extra mass. Adding neutrons costs a little in mass, and in Einstein $E=mc^2$ implies that mass equates to energy. So it costs energy to add neutrons(due to their mass) and it costs energy to add protons(due to their electrical repulsion). This qualitatively explains why the elements have N larger than P, but not by too much.

This extra energy locked into the neutron's mass also leads to instability. As nature seeks the state of lowest energy, like water running downhill to sea level, so a neutron left to itself will eventually experience beta decay: $n\to p+e+\nu$, which converts it to the marginally lighter proton, the excess energy being transformed into the electron and neutrino. So while a neutron in a nucleus can be stabilised, if you gather too many together, they will undergo beta decay, increasing the number of protons at the expense of neutrons. Conversely trying to put too many protons together and their net electrostatic repulsion destabilises them; in this case it is possible to lower the net energy by "inverse beta decay" where one of the protons converts to a neutron: $$ \rm p\text{ (in nucleus)} \to n \text{ (in nucleus)} + \text{positron} + \nu. $$ The net effect is when collections of N and P get too big a mismatch, beta decay or inverse beta decay moves the whole back toward the "valley of stability" where the number of neutrons N tends to exceed the number of protons P."

(The above is from a book called "Cosmic onion. Quarks and the nature of universe")

From my understanding (not too sure as I am new on this), it took much more energy to gather protons together as compared to gathering neutrons together. What holds them together is the strong force and since the strong force holding them together is the same, there will be "a lot of strong force" left over when holding just neutrons together. As nature seeks to minimise the energy state of the nucleons, some of the neutrons will decay to proton such that there will not be "any strong force left over", hence ensuring that the energy state of the necleons are minimise. If what I say above is true, the neutron and proton "produce" equal amount of force carrier particles as is it because they themselves are made of quarks held together by the same colour charge and thus each of them produce the same amount of residual strong force?


The strong force has an approximate symmetry called isospin, which is a rotation in an abstract space spanned by the neutron and proton. This symmetry is broken by the Coulomb force, which only acts on protons, and by the slight mass difference between up quarks and down quarks, which makes the neutron slightly heavier than the proton.

Consider possible bound states of neutrons and protons. These are fermions, so the wave function has to be anti-symmetric. Neutrons and protons are heavy, so we can analyze the situation using non-relativistic quantum mechanics. If a bound state exists, the ground state is in an s-wave (symmetric) because the centrifugal barrier is purely repulsive. This means that the spin-isospin wave function has to be anti-symmetric. This leaves two possibilities: $$ (I=0,S=1) \;\;or\;\; (I=1,S=0) $$ where $I$ and $S$ are the total isopsin and spin of the state. Note that $I,S=0$ is anti-symmetric in isospin/spin, and $I,S=1$ is symmetric. The $I=0$ state corresponds to $$ |I=0\rangle = \frac{1}{\sqrt{2}}\left(|np\rangle -|pn\rangle\right) $$ and $I=1$ has three isospin components $$ |I=1,I_3=+1,0,-1\rangle = \left\{|pp\rangle ,\frac{1}{\sqrt{2}}\left(|np\rangle +|pn\rangle\right),|nn\rangle\right\} $$ The potential in the $I=0,1$ is determined by complicated interactions between quarks and gluons.

Empirically (or from numerical calculations in lattice QCD) we know that both the $I=0$ potential and the $I=1$ potential are attractive, but the $I=0$ interaction is slightly more attractive. As a consequence, there is an $I=0$ bound state, called the deuteron (binding energy 2.2 MeV), but the di-neutron (and its isospin partners) fall just short of being bound. Due to the Coulomb force, there is a little extra repulsion in $pp$, but the $nn$ channel is almost bound.

This can be quantified using the scattering length. The $nn$ scattering length is about 20 fm, and the corresponding energy scale is $$ B=\frac{1}{2ma^2}\simeq 50 \, keV $$ so the di-neutron comes within 50 keV of being bound. This also means that if one could change the quark masses slightly, there would presumably be a bound di-neutron (this can be checked in lattice QCD).

It is possible in principle to have a three neutron bound state even if there is no two neutron bound state (states like that are called Borromean), but this is disfavored by the Pauli principle (not all three neutrons can be in an s-wave). The same applies to 4n bound states (there is a recent claim of a 4n resonance http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.052501).

In practice $^3H=pnn$ and $^3He=ppn$ exist, but not $3n$.

Note that neutron stars, clusters of $10^{57}$ neutrons, do exist. Of course, these objects are gravitationally bound.

  • $\begingroup$ Neutron stars are also not just neutrons. $\endgroup$ – Rob Jeffries Mar 10 '16 at 21:38
  • $\begingroup$ Yes (but to first approximation they are). Also note that $10^{57}$ neutrons would form a bound state. This state is not the ground state, and about $10^{55}$ neutrons would decay into $pe\nu$. $\endgroup$ – Thomas Mar 10 '16 at 22:28
  • $\begingroup$ Agreed, but that is key. The lowest energy state cannot just include neutrons. $\endgroup$ – Rob Jeffries Mar 10 '16 at 23:33

There are two forces acting in a nucleus - the coulomb repulsion between protons and the strong force attraction between nucleons (neutrons and protons) which as you have pointed out is the same for nn, pp and np.

The neutrons and protons are in two sets of energy levels within the nucleus. This is called the shell model of the nucleus. Because of the coulomb repulsion the proton energy levels are slightly higher than that of the neutrons.

So you start filling the energy levels.

1 proton $\Rightarrow ^1_1\text{H}$ stable
1 p + 1 n $\Rightarrow ^2_1\text{H}$ stable
1 p + 2 n $\Rightarrow ^3_1\text{H}$ stable
1 p + 3 n $\Rightarrow ^4_1\text{H}$ unstable as it is energetically favourable to convert a neutron into a proton with the emission of a $\beta^-$
1 p = 4 n $\Rightarrow ^5_1\text{H}$ very unstable and ejects neutron
2 p + 1 n $\Rightarrow ^3_2\text{He}$ stable
2 p + 2 n $\Rightarrow ^4_2\text{He}$ a very stable nucleus because the lowest energy levels are full
2 p + 3 n $\Rightarrow ^5_2\text{He}$ very unstable and ejects neutron

Adding too many neutrons in not favoured as they would have to go into much higher energy levels than if a proton was added.

For the smaller nuclei the pattern is that the number of protons is about the same as the number of neutrons as the energy levels are filled up.

However as the nuclei get bigger the proton energy levels becomes higher and higher relative to the corresponding neutron energy levels due to the coulomb repulsion.
So it becomes more favourable to add more neutrons.

Another way of looking at this is to note that the range of the strong nuclear force is relatively short and when there is a large nucleus not all the nucleons will feel the attractive force of all the other nucleons, whereas the long range coulomb repulsion force is felt by all the protons. Adding extra neutrons increases the attractive force whilst at the same time “diluting” the coulomb repulsive force.

In the end the coulomb repulsive forces win and there are no stable isotopes above atomic number 81 and then the naturally occurring isotopes run our at atomic number 92.

  • 1
    $\begingroup$ You haven't directly addressed the question, which was about states like $2n$, $3n$ and so on. And while beta decay is the answer to why they don't appear over the long term they also don't appear in the detritus of heavy nuclei interactions. $\endgroup$ – dmckee Mar 10 '16 at 16:38

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