I am reading more carefully about subatomic particles, and am on the Muon Wiki page, where it says:

The muon is an unstable subatomic particle with a mean lifetime of 2.2 μs, much longer than many other subatomic particles.

The pion page said it was unstable too, on the order of nano-seconds.

Then we have on Radionuclides:

A radionuclide (radioactive nuclide, radioisotope or radioactive isotope) is a nuclide that has excess nuclear energy, making it unstable.

Then we have the Stable nuclide page:

Stable nuclides are nuclides that are not radioactive and so (unlike radionuclides) do not spontaneously undergo radioactive decay. When such nuclides are referred to in relation to specific elements, they are usually termed stable isotopes.... Most naturally occurring nuclides are stable (about 251; see list at the end of this article), and about 34 more (total of 286) are known to be radioactive with sufficiently long half-lives (also known) to occur primordially.... Some isotopes that are classed as stable (i.e. no radioactivity has been observed for them) are predicted to have extremely long half-lives (sometimes as high as 1018 years or more).[2] If the predicted half-life falls into an experimentally accessible range, such isotopes have a chance to move from the list of stable nuclides to the radioactive category, once their activity is observed. For example, 209Bi and 180W were formerly classed as stable, but were found to be alpha-active in 2003.

So what is exactly meant by stable vs. unstable? Is it basically a spectrum (microscopic perspective vs. macroscopic perspective), sort of like how we perceive starfish to be still but if you speed up a video of them they are quite active and mobile? In the same way, all particles are at some level "unstable", depending on the time frame you are observing them at? Or are there some particles which are 100% stable permanently? If so, how does that work, how are they never unstable?

I get confused because (having done molecular biology as a major), I always thought all atoms/particles gained and lost electrons and could otherwise change into an unstable form through interaction with cosmic rays or other forms of radiation (sorry if I am confusing terms here, it has been a while). I thought all particles had radioactive decay, but maybe I fundamentally misunderstood that part of the class.

If you could help clarify what exactly is meant by stable vs. unstable, and when things are stable and how they can absolutely never become unstable, then that would help clarify this confusion I am currently facing. To me being stable means you can't interact with the environment in the grand extremes of things, so to me everything is inherently unstable. But I'm not sure exactly what is meant.

  • $\begingroup$ Strictly from an energy point of view, all nuclides other than Ni-62 are unstable, because they all will eventually form Ni-62. But your question is a bit unambiguous in its scope, because you mix elementary particles and nuclides. $\endgroup$
    – tobalt
    Feb 12 at 9:05

3 Answers 3


I start with the most general rule: if for a particular system a lower energy state is available then as long as a path exists to move to that lower energy state the system will proceed down that path.

(In sub-atomic processes: often the energy difference transforms to kinetic energy; velocity of resulting particles (if there are resulting particles). Some of the energy may transform to a rotational state with corresponding units of angular momentum.)

Example of proceeding to lower energy state from macroscopic physics:
The Earth's rotation has a shorter period than the orbit of the Moon around the Earth. As a consequence there are the tides caused by the tidal effect of the Moon's gravity. Over time this is slowing down the rotation of the Earth. (In addition there is a secondary effect arising from tidal deformation that over time increases the radius of the Moon orbit.) Given enough time eventually the Earth will be in tidal lock with the Moon. After that the Earth-Moon system still has rotational kinetic energy, but in the state of tidal lock there is no longer opportunity to dissipate rotational kinetic energy.

We observe that for a muon a path exists that releases energy. The associated probabitity of going down that path is what gives rise to the muon half-life.

We infer that for the isolated electron no path exists that leads to a lower energy state. (We have that for an electron-positron pair an energy releasing path does exist, demonstrating the energy equivalent of the inertial mass, but no such path exists, we infer, for electrons and positrons on their own.)

For completeness I point out explicitly: among the electron, muon, and tau particle the one for which no decay mode is observed is the one with the lowest inertial mass.

So: particles for which no lower energy state is available to decay to receive the designation: 'stable'.

See also:
Video about nuclear decay by the science youtuber Scott Manley.

Two things discused by Scott Manley:

Xenon 124 and Xenon 136 both have an extremely long half life. For Xenon 136 the decay mode is double beta decay. For that decay mode the energy pay-off is quite worth it, but: in order for the overall decay to occur the two beta decays must happen within a very, very narrow time window, which has an extremely low probability of occurring.

There is a state of Tantalum called Tantalum 180m. This state has an excess of nuclear energy, but in order to lose that excess energy (by emitting gamma radiation) that nucleus must lose 8 units of angular momentum in the process.

Article available on Arxiv:
Search for the decay of nature's rarest isotope 180mTa

I think the overall intention is that the word 'stable' is used when intrinsically no lower energy state is available.

Whether a lower energy energy state is available is inferred from observation. If for a particular isotope no decay is ever observed then it is reasonable to infer that no lower energy state exists for that configuration. But detection methods become ever more sensitive, so in some cases the status 'stable' has turned out to have been assigned prematurely.

When a lower energy state is available it may be that circumstances conspire to make the decay mode improbable. The lower the probability the longer the half life.

Now that I've written the above I realise I find the words 'stable' and 'unstable' very unsatisfactory. It feels silly to refer to an isotope with a half life trillions of times the age of the Universe as 'unstable'.

-There is the question of whether a lower energy state exists
-There is the question of the probability of actually going down the decay path.

  • $\begingroup$ "if for a particular system a lower energy state is available then as long as a path exists to move to that lower energy state the system will proceed down that path". In view of conservation of energy you might want to qualify that. $\endgroup$ Feb 11 at 12:36
  • 1
    $\begingroup$ @AndrewSteane This is a judgement call. It seems to me that the level of the question is such that it can be reasonably assumed the OP takes conservation of energy as a given. (The OP mentions having done molecular biology as a major). If I would cram too much information in the answer the answer would become unreadable. I must rely on some basis of physics knowledge, and take it from there. For a judgement of how wide a basis that is I take the question as guidance. Yes, preferably the answer should cater to a wider audience, but taking that too far is counterproductive. I added a paragraph. $\endgroup$
    – Cleonis
    Feb 11 at 13:35

There is a strict definition of stability covered in Cleonis' answer, which is based on possible modes of decay. In some cases, it is important for fundamental reasons (see proton decay for perhaps the most notable example).

There is also a more casual, "engineering" usage of the word. I would argue that with stability, scientific rigor usually gives way to practicality, and it is one of the examples where the word would be more commonly used as a shorthand for $t_{1/2}\gg\tau \Rightarrow \mathrm{decays\ can\ be\ ignored\ in\ this\ problem}$.

A great example here would be iodine: for a particle physicist working on an accelerator, $^{135}\mathrm{I}$ ($t_{1/2}=6.57 h$) is practically stable, as relevant reactions happen on a micro- and nanosecond scale. For a nuclear physicist dealing with a reactor, its half-life is of paramount concern. And for most other people, it is an extremely unstable isotope. For most people and in common parlance, $^{129}\mathrm{I}$ ($t_{1/2}=1.57×10^7 yr $) is stable, but this is not so from the point of view of theoretical physics.

Finally, there are fundamental particles which have no decay modes whatsoever, but as history has shown time and again, the definition may change later on. This is a whole separate can of metaphysical worms, however. For the purposes of natural sciences and physics in particular, we work with whatever is the most useful at the time. Both fundamental (has no lower energy states) and applied (does not decay while we care about it) definitions are useful in different scenarios.


A "stable" system is one whose state will not change spontaneously in an interesting period of time. An "unstable" system is one where the state change is too likely to be ignored in your calculations. Sometimes people use "metastable" to describe systems whose incipient state change is much slower than another interesting process.

These are plain English words. If we say that a building is stable, we mean that it is likely to remain a building for the foreseeable future, rather than spontaneously collapsing into a pile of rubble. If you have ever visited a construction site, you have been told "don't walk on that section there yet, it is still unstable." But "stable" has an implicit timescale. A house might be stable for decades or centuries without active maintenance, but rare processes like termites or tropical storms or urban redevelopment or warfare mean that the house you are currently in, right now, is unlikely to still look like a house in a hundred years. Last week the Gazientep Castle in Turkey was destroyed in an earthquake, after standing for 2000 years.

In a quantum-mechanical system, there is one "ground state" whose energy is lower than any other "excited" state. If the system finds itself in an excited state, it can release energy and increase entropy by transitioning to the ground state. But in some cases, there is some symmetry which forbids the transition. If the transition time is just very long, we call the excited state "metastable"; if the symmetry is exact and we have reason to believe the transition will never happen, we call the system stable.

For example, tantalum-180 is beta-unstable with a half-life of a few hours. There is a nuclear excited state, Ta-180m, which could in principle decay to the ground state by emitting a photon — but that photon would have very low energy (which generally makes transitions slower) and would also have to carry away an angular momentum of $\hbar\Delta J=8\hbar$. There is a small term in the transition probability which goes to the $\Delta J$-th power, and $(\text{small})^8$ is very small. Based on searches for decays which came up empty, the lifetime is at least $10^{15}$ years. If you dig tantalum out of the ground, some of it is Ta-180m. If you had made some Ta-180m at the beginning of the universe, at least 99.999% of it would still remain un-decayed. "Unstable" or "metastable" just don't feel like the right descriptions of this material.

Two other symmetries which limit decays are the conservation of "baryon number" and "lepton number." A baryon is a proton or a neutron, or an excitation which decays eventually into a proton or a neutron; a lepton is a particle, like the electron or its neutrino, which is blind to the strong force. We know of no process that changes baryon number or lepton number. (Antibaryons and antileptons with negative numbers, so pair production doesn't count.) We suspect that such a process must exist, because the universe contains more baryons than antibaryons. We also suspect that such a $B$-violating process might still take place down in the bowels of the proton, allowing processes like $\rm p^+\to \pi^0 e^+$. The Super Kamioka experiment was a skyscraper-sized vessel of water buried in a mine in Japan. After years of searching unsuccessfully for a proton decay event, the claim now is that the proton's lifetime is no shorter than $10^{33}$ years. If every minute since the beginning of the universe were stretched to the current age of the universe, that would still be shorter than the mean proton lifetime. Unfathomable time. The question of whether proton is absolutely stable is important for understanding how our universe came to be filled with enough protons that they came together and formed introspective humans. But at the same time ... the proton is stable. The proton is the ground state of the QCD vacuum, when the baryon number is nonzero. Protons aren't going anywhere.

The question of whether any particles are absolutely stable is a profound and tricky one, because more and more rare processes come into play as you look further into the future. There is currently a comment under your question which claims that all nuclei will eventually evolve into nickel-56. That's wrong. Nickel-56 (or whichever isotope, writing on a phone stinks) is the most tightly bound nucleus, but, as for the tantalum ground state, there isn't any pathway to get there. The processes which form nickel-56 are more likely to form iron, as evidenced by the fact that Earth's core is more iron than nickel. But the stellar processes which form iron and nickel also require a large amount of non-iron and non-nickel to function. Too much iron in a star, and the iron transitions via supernova to degenerate neutron matter, or to a black hole.

There is good reason to suspect that, on very long timescales, the matter sector of our universe will evolve towards being entirely black holes. It is also believed that, once its environment is cold enough, a black hole becomes unstable against photon emission, and begins to radiate away more thermal energy than it absorbs — which makes the black hole hotter, which makes it radiate faster, until it evaporates completely. (Its final instant also includes some massive-particle radiation; very exciting.) So a universe containing only black holes would evolve towards a universe containing only photons. We believe that the photon is a stable particle, but it is not clear whether a universe which contains only photons is stable; the accelerating expansion of our universe suggests that such a spacetime may be dominated by dark energy, whatever that is, and may undergo inflationary expansion.

Shortly after the discovery of the accelerating expansion of the universe (so, probably 1999 or 2000), I remember reading an article in Scientific American about the possibility of a universe which would undergo not a "big crunch" or a "big rip" but instead remain a flat spacetime basically forever. One point that the authors made was that, if our universe is currently about $10^{17}$ seconds old, then in the very distant future it would make sense to consider one $10^{17}$-th of the age of the universe as a "brief" interval. The article had a logarithmic timeline with events like "all matter is black holes" and "all black holes have evaporated." Somewhere around $10^{100}$ or $10^{150}$ years into this timeline was the phrase "quantum tunneling liquifies matter." Stability in this eternal sense is really an open question.

If the lack of a sharp dividing line between "stable" versus "unstable" bothers you, as it bothers the Wikipedia editors who wrote your quoted sentence about tungsten-180, I suggest you read an essay by Dawkins about "the tyranny of the discontinuous mind." It is the chapter in his The Ancestors' Tale where he discusses "ring species."

  • $\begingroup$ a) my comment didn't claim Ni56, but Ni62 b) nuclides remaining stable is only true if they are alone in the universe. If there is matter in general or even just some nuclides, there is a finite chance for nucleon exchanges. $\endgroup$
    – tobalt
    Feb 12 at 17:56
  • 1
    $\begingroup$ Yes, I got the isotope number wrong. But you have a quantitative question of whether nucleon exchange is faster or slower than the other rare processes I discussed. $\endgroup$
    – rob
    Feb 12 at 19:04

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