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In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that?

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You probably know that the electrons in atoms occupy a series of energy levels, the $1s$, $2s$, $2p$, etc orbitals. Although the structure of nuclei is complicated, basically the same idea applies to nuclei as well as atoms.

This happens because you can't put more than one fermion into the same quantum state. The electrons in atoms are fermions, and so are the protons and neutrons in nuclei. However because the proton and neutron are different particles you can put a proton and neutron into the same energy state. Well, that's not really true but the point is that the energy states of the protons and neutrons overlap. Let me try to show this with a diagram:

Nuclear energy levels

Suppose we have eight nucleons and we arrange them, one per energy level, as shown in the diagram (a). This has some total energy $E_a$. Suppose now that we can pair up the nucleons as shown in (b). This obviously has a lower energy $E_b$. But remember that protons and neutrons can share energy levels because they are different particles. So if we have a nucleus with all protons or all neutrons it would look like (a), while an equal number of protons and neutrons would look like (b) and therefore have a lower energy.

But protons and neutrons can interconvert by beta or positron emission or electron capture. So if we started with eight protons as in (a) four of those protons can convert to neutrons to give the energy levels in (b) and this will overall reduce the energy. The nucleus with equal numbers of protons and neutrons will be the one with the lowest energy.

Now this is very oversimplified because protons and neutrons don't share exactly the same energy levels and the energy levels aren't equally spaced. But it gives you a feel for why we have approximately equal numbers of protons and neutrons in a nucleus.

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    $\begingroup$ "if we have a nucleus with all protons" you cannot have a nucleus with all protons because the electromagnetic repulsion does not allow the protons to get close enough for the strong force to dominate. Only if they were neutral but still separate entities as far as the pauli exclusion the argument would hold $\endgroup$ – anna v Apr 4 '17 at 6:13
  • $\begingroup$ Though $^8Be$ is an exception and has a very short halflife. $\endgroup$ – badjohn Feb 7 '18 at 18:32
  • $\begingroup$ @annav: you cannot have a nucleus with all protons because the electromagnetic repulsion does not allow the protons to get close enough for the strong force to dominate. This doesn't make sense. All heavy nuclei are formed by fusion, and they all require an input of energy to overcome the Coulomb barrier. John Rennie's answer is correct. $\endgroup$ – Ben Crowell Jan 13 at 19:15
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    $\begingroup$ @BenCrowell to have a nucleus it needs stability, an all proton nucleus might form in a nova explosion for a d(t) because of the pressures, but it will disintegrate very fast because of the coulomb repulsion. Neutrons are necessary to allow for the strong force to dominate at the close distances of fusion. $\endgroup$ – anna v Jan 13 at 19:18
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    $\begingroup$ Hydrogen has a nucleus that's all protons. $\endgroup$ – Ryan_L Sep 9 at 20:46
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A phenomenological model can be described in terms of the liquid drop model and semi-empirical mass formula.

This ascribes a nucleus a binding energy of $$B(A,Z) = a_v A - a_s A^{2/3} - a_c Z^2A^{-1/3} - a_a \frac{(A-2Z)^2}{A},$$ together with a smaller "pairing term", which I am going to ignore.

Here, $A$ is the total number of nucleons and $Z$ is the proton number. The four terms are the volume binding energy associated with the short-range strong nuclear force; a "surface term" which corrects the volume binding energy for the fact that a proportion of the nucleons are at the surface of the nucleus and not surrounded by other nucleons; a coulomb term that accounts for the (long-range) repulsion experienced by protons and an asymmetry term, which is the key to your entire question.

If consider a nucleus with a fixed $A$, and I was to ignore the asymmetry term, you can see that the binding energy is maximised (most stable nucleus) if $Z=0$. i.e. No protons and therefore no negative Coulomb energy. But this would mean the nucleus is made up entirely of neutrons and this isn't the case.

The explanation of this does not lie in the instability of the neutron, otherwise we would be able to build such nuclei that were stable on the decay time of the neutron. It lies with the asymmetry term.

If I now include the asymmetry term and differentiate with respect to $Z$ at fixed $A$. $$ \frac{\partial B}{\partial Z} = -2a_c Z A^{-1/3} +4a_a (A-2Z)$$ At a maximum I would set this to zero and hence find that neutron to proton ratio $$ \frac{(A-Z)}{Z} = 1 + \frac{a_c A^{2/3}}{2a_a}.$$

The empirical values of the constants are $a_c \simeq 0.7$ MeV and $a_a\simeq 23$ MeV, so $$ \frac{(A-Z)}{Z} \simeq 1 + 0.015 A^{2/3}.$$

This means that unless $A$ is much greater than 1, then the neutron to proton ratio is roughly unity and $A-2Z \simeq 0$. In physical terms it is because the asymmetry term would become large if $A \neq 2Z$ for small $A$, but when $A$ (and $Z$) are large, the Coulomb repulsion between protons shifts the balance towards neutrons a bit.

This is a sort-of-explanation because it evades explaining where the asymmetry term comes from! John Rennie's answer handles this. You can think of the protons and neutrons in a nucleus as consisting of two fermion gases (neutrons and protons) made of indistinguishable particles. Such particles cannot occupy the same quantum states (because of the Pauli exclusion principle), and they must therefore stack up, filling increasingly higher energy levels (2 per momentum/energy state, with opposite spins). The energy levels are roughly similar for the neutrons and protons because the strong nuclear force is blind to whether neutrons are protons are involved and the differences are smaller than the gap between adjacent energy levels.

The total energy of such an arrangement will be the sum of the energies of all the particles. This sum will be approximately minimised when there are the same number of neutrons and protons. If there are more neutrons, then it is possible to achieve a lower energy by beta decay to a proton and an electron and electron anti-neutrino (which escape); if there are more protons, then a lower energy state can be achieved by turning a proton into a neutron and emitting a positron.

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  • $\begingroup$ I'd give this answer a bounty if I could. $\endgroup$ – stafusa Feb 8 '18 at 13:46
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This is my first answer, be nice please.

A different approach (previous answers are already very good) that might give you a better intuition:

Back in the day, Weizsäcker proposed a parametrization of the binding energy of nuclei by analogy with a liquid drop such as

$BE(A,Z) = a_VA - a_SA^{2/3} - a_CZ^2A^{-1/3} - a_A(A-2Z)^2A^{-1}$ (not considering the pairing term here)

where the coefficients $a_V,...$ are fitted on nuclear masses tables (e.g. AME2012), Z is the number of protons and A is the total number of nucleons (protons + neutrons).

The different terms in this relation respectively correspond to the bulk energy, the surface energy, the coulomb energy, and the symmetry energy.

Considering $BE(A,Z)>0$, the most stable nuclei for a given baryonic number A corresponds to $(A,Z^{eq})$ with $Z^{eq}$ associated to the higher value of the binding energy. You could "reverse" the problem to find the value of A that maximize the binding energy for a given Z. For heavy nuclei, such as uranium, there is a significant difference between the number of protons and the number of neutrons (which is higher). Assuming that our previous relation is true, this is because the balance between the repulsive force between protons and the asymmetry tilts in favor of Coulomb.

This is the basics of nuclear phenomenology. There are a lot of microscopic approaches (e.g. DFT, ab initio) that consider the nucleon-nucleon interaction in a different way.

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  • $\begingroup$ Typo in your formula. But does this answer really explain anything? If I have no symmetry term, then the most stable nucleus has $Z=0$. $\endgroup$ – Rob Jeffries Apr 4 '17 at 13:21
  • $\begingroup$ Corrected, thanks. Obviously, what I explained is only pertinent if we consider the symmetry term which is here to take Pauli into effect. The previous answers are more pertinents but I thought it would be more intuitive with this phenomenological approach (maybe I was wrong!). $\endgroup$ – T. Auerrac Apr 4 '17 at 13:30

protected by Qmechanic Apr 4 '17 at 14:02

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