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

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?

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? Mar 13, 2021 at 15:54
• I don't think so as the potentials in both cases are different. Mar 13, 2021 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. Mar 13, 2021 at 17:23

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.). Mar 13, 2021 at 16:01