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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.

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    $\begingroup$ Yes, it is true that nuclear physics can not be reduced to a brief explanation. This might be better over in Physics... $\endgroup$ – Jon Custer Dec 17 '14 at 21:28
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    $\begingroup$ Protons and neutrons are fermions like electrons, i.e. they obey the Pauli exclusion principle. If there's a sufficiently large excess of one type, the highest energy proton/neutron has to stay far above the highest energy other particle, which makes it favourable for some type of decay to happen and even the counts out more. $\endgroup$ – Nicolau Saker Neto Dec 18 '14 at 0:36
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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.

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The principle that "heavier nuclei are more stable" is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested this paper for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.

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  • $\begingroup$ Well, it's not exactly incorrect; there are no non-radioactive isotopes of any element after lead. It's just a very broad trend, and you can't use it to judge the relative stability of the metastable isotopes of the later elements accurately. $\endgroup$ – J. LS Dec 19 '14 at 8:42
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"The heavier the atom, the more unstable it gets, right?" Wrong. Nickel-62 is the "most stable" isotope, measured by binding energy per nucleon. Since in the core of massive stars, it is subject to photodissociation, Fe-56 is often considered to be the "most stable" isotope. This discrepancy shows that context matters. Take a look at the graph of binding energy/nucleon vs nucleon number https://en.wikipedia.org/wiki/Nuclear_binding_energy#/media/File:Binding_energy_curve_-_common_isotopes.svg You can see that the graph is nearly flat around ~±10 nucleons of Fe-56, so there's no value in making broad statements, imho. Even if unlike the OP, the statement was accurate, the devil is in the details (context, environment, what is meant by "stability" , etc.). Why did anyone bother to resurrect this old question, I wonder? Heavier nuclei are neither or both more and less stable than lighter nuclei. It isn't (imho) a "very crude approximation", it is simply wrong. A more accurate "crude" approximation is to say that low mass (not weight, for God's Sake!) elements tend to release less and less fusion energy up to approximately iron, and also past iron, the tendency is a slow increase in the fission energy released. (but details matter, especially the exact balance between the various forces). You simply can not predict the stability of an isotope by using only its mass (at least, not below ~ 250 or 300 amu).

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