Why are nuclides with an even number of protons and neutrons more stable?

Question

I am currently learning about Radioactivity and Nuclear Physics. Why are nuclides with an even number of protons and neutrons more stable?

I have read that nuclides with an even number of protons or neutrons or both, are more stable than those with an odd number of nucleons.

Is this something that is based on theoretical reasoning? Or just something taken from observation?

I understand that nuclides with more neutrons would be more stable due to more nuclear forces to act against the electrostatic repulsion between protons (to an extent), but I don't know why odd/even counts matter. Is there more stability due to possible greater symmetry if the nuclide has an even number of nucleons?

Related: What makes the number of neutrons the number of proton similar?

Some info on even-even stability: https://physics.stackexchange.com/a/340678/123208 — PM 2Ring, Mar 13 '21 at 14:18
BTW, while having more neutrons generally does help to offset the electrostatic repulsion of the protons (with a higher neutron:proton ratio required for heavier elements), an excess of either isn't good for stability. Note that tritium with 1 proton & 2 neutrons decays to stable helium-3, with 2 protons & 1 neutron. There's some info on the neutron:proton balance at https://physics.stackexchange.com/a/323358/123208 & https://physics.stackexchange.com/a/354855/123208 — PM 2Ring, Mar 13 '21 at 14:28
Thanks @PM2Ring . That makes sense. Yes, I agree that an excess of protons/neutrons is detrimental to stability. — m-Xylene, Mar 13 '21 at 15:58
@Nihar Karve Interesting. I wasn't aware that nucleons followed similar (to an extent) rules to electrons in orbitals. — m-Xylene, Mar 13 '21 at 15:58
Nucleon orbitals & electron orbitals are quite similar, but there are several important differences. — PM 2Ring, Mar 13 '21 at 16:09
Does this answer your question? What do we know about the interactions between the protons and neutrons in a nucleus? — Peter Duniho, Mar 13 '21 at 23:58
Does this answer your question? What makes the number of neutrons the number of proton similar? — John Rennie, Mar 14 '21 at 17:24
@JohnRennie Yes, thank you; Is it correct to think that even numbers of protons and neutrons providing more symmetry for inter conversion is somewhat analogous to the greater symmetry (and thus stability) of half filled orbitals due to exchange energies of electrons? — m-Xylene, Mar 15 '21 at 01:01
@PeterDuniho I read that answer, I don't think that was what I was looking for, but it was certainly very interesting. — m-Xylene, Mar 15 '21 at 01:02

score 5 · Accepted Answer · answered Mar 13 '21 at 13:57

Most stable nuclides have both even $Z$ and even $N$ called "even-even" nuclides.

We can understand this in terms of Pauli's exclusion principle. Neutrons and protons are distinguishable fermions; hence they separately obey the exclusion principle. Only two neutrons (or protons ) may coexist in each spatial orbital (quantum-state), one with spin up and the other with spin down. Each nuclear energy level is thus able to hold two particles, the spins of which are paired to $0$. This configuration of opposite spins is particularly stable because placing the same number of particles in any other arrangement will produce a (less stable) state of higher energy. Therein lies the preference for even $N$ and $Z$.

The above is taken care of with an extra term in The von Weizsacker semi-empirical mass formula $$B(X^A_Z)=a_VA-a_AA^{2/3}-0.72Z(Z-1)A^{-1/3}-a_S\frac{(N-Z)^2}{A}+\delta$$ The last term is due to the pairing energy and reflects the fact that the nucleus is more stable for even-even nuclides.

$$\text{Pairing} \ \delta =\begin{cases} +\Delta & \text{even-even nuclei}\\ 0 & \text{for odd-A ( even-odd, odd-even nuclei}) \\ -\Delta & \text{odd-odd nuclei} \end{cases} $$

That makes sense, thanks. Although I haven't yet learned about the orbitals in a nucleus, I can relate them to the orbitals of the electron shells, which I think would follow similar rules. But I don't think rules like greater symmetry results in greater exchange energy apply here, right? — m-Xylene, Mar 13 '21 at 15:54
I don't think so as the potentials in both cases are different. — Young Kindaichi, Mar 13 '21 at 17:21
Ahh, okay, thank you. Nuclear 'orbitals' (I'm not sure of the right word) sounds like a really interesting topic to learn about. — m-Xylene, Mar 13 '21 at 17:23

score 2 · Answer 2 · answered Mar 13 '21 at 14:35

The answer by Young Kindaichi is completely incorrect, which can be seen from the fact that it doesn't use any facts about the strong nuclear force. It therefore reads as an argument that would apply equally well to atomic physics, and yet the odd-even differences in binding energy do not exist for atoms.

The correct answer to this question requires knowing not just the exclusion principle but also two facts about the nuclear interaction: (1) it is a short-range attractive force, and (2) it has a big spin-orbit coupling.

The spin-orbit coupling results in the fact that orbitals are not states of good intrinsic spin (spin-1/2). They are states of good total angular momentum. These states occur in degenerate multiplets of $2j+1$. Within one of these multiplets, there will be states that have equal values of $|j_z|$, for example a state with $j_z=+3/2$ and one with $-3/2$. A pair like this has maximum spatial overlap.

We now use the fact that the nuclear force is short-range and attractive. Because of this property, a pair of states with high spatial overlap experiences a strong binding. When you have an odd nucleon, it can't be paired in this way.

Thanks. Sorry, but I can't see where @Young Kindaichi is wrong though. Why wouldn't the even binding energy be more? Also, sorry about this but I'm not familiar with the term 'j'. My guess is that it's a number for nuclear orbitals, but I've only learned atomic physics as far as the electron numbers (m,n,l etc.). — m-Xylene, Mar 13 '21 at 16:01

Why are nuclides with an even number of protons and neutrons more stable?

2 Answers2