radioactivity - Stability variation in heavy nuclei

The heavier the atom, the more unstable it gets, right?

That is not true about Uranium and we know it. I wondered why. A brief explanation stated that since more neutrons are there in the nucleus there is more nuclear force to ²³⁸U than ²³⁵U. Seems legitimate, but then that should have been true about the other radioactive species. Again, we know that's not true.

So, either the explanation about the Uranium exception is not totally true or the principle of the heavier-atom instability (of course, exceptions occur at elements like U) is not a principle. Or maybe there is a relation between these two (nuclear force and atomic weight) which is beyond me.

My hunch is that the latter is true; searched some sites which had an answer to the question, but no acceptable results gained.

Answer

If the ratio of neutrons to protons in the nucleus is too high, the nucleus is unstable to beta decay. A neutron emits an electon and becomes a proton.

If the ratio of neutrons to proton is too low, the nucleus is unstable to positron decay and/or electron capture.

But how high/low is too high/low, and why does this occur?

We need to look at the Semi-empirical mass formula.

$$E = vA -sA^{2/3} - c\frac{Z^2}{A^{1/3}} - a\frac{N-Z}{A} - \delta (A,Z) $$

Where $E$ is the binding energy of the nucleus, $N$ is number of neutrons, $Z$ is number of protons, $A$ is $N + Z$, and $v$, $c$, $s$ and $a$ are empirical constants.

Only the $c$ (coulomb) and $a$ (asymmetry) terms involve the relative number of protons and neutrons.

The asymmetry term is zero when the number of protons and neutrons are equal, and relates to the Pauli exclusion princple (neutrons and protons being fermions).

Except for the coulomb term (that protons electrostatically repel each other), the ideal ratio of protons to neutrons would be 1 to 1. Electrostatic repulsion of protons explains why somewhat more neutrons than protons is favorable.