# Why is an atom with excess neutrons unstable? [duplicate]

In this chart, I can see that stable nuclides (other than hydrogen) have a neutron count greater than or equal to their proton count, and that the neutron:proton ratio for stable nuclides increases with atomic number.

I can also see that in many cases, radioactive nuclides which have excess neutrons undergo beta minus decay, converting one of the excess neutrons to a proton, and moving toward that stable ratio.

I can see why having excess protons might make an atom unstable: they have positive charge and push each other apart. But it seems to me that excess neutrons would make it more stable, adding to the strong force between the nucleons.

Why is an atom with excess neutrons unstable?

• Let's put it this way: why are free neutrons unstable? Jun 19 '20 at 11:39
• OR, why are not there any stable neutron only kernels ? Even 3H should be by your idea more stable than 3He, but the opposite is true. Jun 19 '20 at 11:57
• Cross-site detailed Q&A: physics.stackexchange.com/questions/78107/… Jun 19 '20 at 14:11
• chemistry.stackexchange.com/questions/21848/… Jun 19 '20 at 20:56

## 2 Answers

Pulling out Theoretical Nuclear Physics by Blatt and Weisskopf (published in 1952, so a bit dated in areas but still a good introduction), one finds (in Chapter VI.1 The Systematics of Stable Nuclei) that the stability of isobaric nuclei (all the same $$A$$) is determined by:

(1) The symmetry effect: the charge independence and the exchange character of nuclear forces, together with the Pauli exclusion principle, strongly depresses the energy of nuclei with equal or nearly equal numbers of protons and neutrons.

(2) The charge effect: the effect of the Coulomb repulsion of the protons favors nuclei with fewer protons than neutrons. The neutron-proton mass difference also enters the charge effect, but only as a small correction. The charge effect increases in importance with increasing nuclear charge.

(3) In a few special cases a third effect is important: the spin dependence of nuclear forces favors parallel spin over anti-parallel sin of a pair of (extra) nucleons.

So, yes, you picked up on the Coulomb bit (number 2 above), but it is the symmetry effects of the strong force that are more important, particularly for light nuclei. Until one gets to some nuclear physics, the characteristics of the weak and strong forces just aren't familiar to us humans.

Neutrons are fermions; they obey the Pauli exclusion principle. With only a few protons, most baryons in a nucleus would be neutrons, and there would be just a few low-energy states for those neutrons. Hence, most would be in higher-energy states.