# Eli5: Why do too many neutrons turn a nucleus unstable?

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Hi,

Not necessarily on the level of a 5 year old, but would be great if anybody could explain this on the level of a mid-schooler.

I learned about the instability of nucleus and found out that the difference in the number of Protons and Neutrons, both being more or less than the other turns a nucleus unstable.

The Proton here makes sense. As having less Neutrons leads to less force to bind the Protons together in the Nucleus, resulting in the Proton’s repulsion force overpowering it and turning the nucleus “Unstable”

But by the same logic, having more Neutrons would then result in a stronger force to bind the Protons, thus making it harder for the repulsive force of the Protons to succeed here.

But even those nuclei somehow manage to turn unstable. How?

I searched forums and they brought up advanced stuff to explain this thing that were clear as mud to me.

Can anybody on this subreddit explain it to me in the simplest manner? Or is it something that must require further knowledge to be understood?

If so, can you tell me what do I need to learn about first?

Thanks a lot

In: 6

Imagine holding a foam cup in your hand. As long as you squeeze lightly, it is no problem. But what happens if you squeeze very hard? You end up squeezing the cup hard enough to rip it, possibly ripping it to pieces.

Adding more neutrons is kind of like that. At a certain point, you’re squeezing so hard that you rip the nucleus apart into smaller, more stable nuclei.

Your logic works on the assumption that the force driving the nucleus apart is just the electric charge repulsion between protons. That is *one of* the relevant forces, but it is not the only one. Warning: this answer gets pretty technical, and it has to to give any meaningful explanation for what you’re talking about.

Most notably, the dineutron – no protons, two neutrons – is unstable. In fact, it’s not even unstable in the way you think. Neutrons are already unstable in isolation and decay to protons, so you might imagine the dineutron would at most have one of its ~~protons~~ (edit: neutrons, sorry, typo) decay, resulting in stable deuterium (1 proton, 1 neutron). And if the strong force were a bit stronger, that is indeed what would happen…

…but in fact, the dineutron doesn’t even exist as a bound state (that is, there’s no actual net attractive force between the neutrons).

—-

To get into why, we have to investigate the other forces at work in a nucleus.

Protons and neutrons in nuclei fall into “shells” of a sort, just like electrons do in an atom. The exact dynamics of those shells are more complicated because:

* There are two types of particle involved, not just one.
* There are three forces (electromagnetic, weak, *and* residual strong) involved, not just one, and those forces have different dynamics.

…but nevertheless, they still have a notion of “energy levels”, of how “occupied” those levels are, etc.

The Pauli exclusion principle tells us that fermions – a class of particles that includes protons, neutrons, and electrons – can’t occupy the same state as one another. That is, at least one detail of their states must differ. For two electrons to sit in the same orbital, they have to have some different value. Since an orbital by definition fixes three of the four numbers that describe an electron, they must differ in the fourth number (their *spin*). The same goes for protons and neutrons. For two neutrons to both have the lowest energy level, they have to have different spins.

It turns out that having two particles with opposite spins creates a state with slightly higher energy than just the sum of each of their individual energies. (If you’ve had reasonably advanced chemistry, you might recall that Hund’s rules tell us that electrons spread out across different orbitals before they start pairing up in the same ones – this is [part of] the reason for that.)

In this case, the desire of the neutrons to take different spins creates a repulsive force when two neutrons get too close. And it turns out that repulsive force is larger than the strong-force attraction between the two neutrons, resulting in a net repulsion between the two. So neutrons won’t stick together with one another, even temporarily.

—-

But you weren’t asking about the dineutron. Let’s go back to your question, which deals with nuclei that *are*, for the most part, bound. That is, they can stick around for a while, they just eventually transform into a different nucleus.

The critical question here is whether the nucleus has a lower-energy state available to it. And for that, we need to return to the idea of “shells”.

Consider, for example, a helium-4 nucleus. It has two protons and two neutrons, both (in a sense) occupying the lowest possible energy levels. The energy of the nucleus is given by:

* Proton-proton repulsion via electromagnetism (raises energy = repulsive)
* Pairs-of-either-with-the-same-spin binding via the strong force (lowers energy)
* Pairs-of-either-with-opposite-spin repulsion via the strong force (raises energy)
* The base mass of each proton and neutron (raises energy)
* The energy of the state each proton and neutron is currently in via the exclusion principle (raises energy)

Now, add a neutron, forming helium-5.

The problem with this extra neutron is that the lowest-energy state for neutrons is occupied in that nucleus. It has to go to the “second shell”. And that makes this extra neutron add a lot of energy when it’s added to the nucleus. Our energy levels change as follows:

* Proton-proton repulsion stays the same.
* # Pairs with same spin goes up -> lower energy
* # Pairs with opposite spin goes up -> higher energy
* The base mass of the protons and neutrons -> goes up, but is the same relative to the neutron hanging out by itself.
* The energy of the state each proton and neutron is currently in -> goes **way** up because the new neutron is in a high energy state.

The combination of these effects makes a helium-5 nucleus have more energy than a helium-4 nucleus plus a neutron. That causes the helium-5 nucleus to be able to emit a neutron, decaying back to helium-4. We don’t necessarily know how quickly this will happen (it turns out that in this case it happens very fast, but that wasn’t a guarantee), but it *can* happen, and the nucleus is unstable as a result.

You can think of this as a sort of nuclear analog of the metals on the left side of the periodic table. The last electron in those elements is in a very high energy state, and would rather be just about anywhere else. (In this case, though, the electron still reduces the energy of the atom, just not by very much, so it has to have something to react with.) It has this behavior because we’re one particle above a “magic number”, where a shell (electron shells in the case of chemistry, nuclear shells in the case of nuclear physics) is completely full.

—–

Now, consider nitrogen-16 (7 protons, 9 neutrons). Why is this unstable when nitrogen-14 (7 of each) is stable?

Well, it turns out that the first two “shells” in the nucleus have room for 2, then 6, particles, so 8 is a “magic number” where one is closed. The 9th neutron here is in a higher energy state than the first 8, and there’s a particularly *low*-energy slot open for an 8th proton. Since neutrons already have slightly higher mass than protons, they already want to turn into protons given the chance, and this particular neutron can easily decay and slot itself neatly into that 8th proton slot.

And in fact, it does so. Nitrogen-16 decays very rapidly to oxygen-16 (8 protons, 8 neutrons) with a half-life of a few seconds. To go back to our energy, in so doing, it:

* Slightly increases proton-proton repulsion -> higher energy
* More paired spins -> slightly higher energy
* Base mass -> significantly lower energy, since we’ve converted a neutron to a proton
* The energy of the state each proton and neutron is currently in -> much lower, since the neutron has dropped from the 3rd “shell” to the 2nd “shell”.