OK, so $\ce{Fe}$ is the most 'stable element'. As such, why do all elements above it not decay into $\ce{Fe}$? In all cases, would it not lead to an increase in binding energy and therefore energy been released, meaning it is energetically feasible, and should happen spontaneously (given enough time)?

  • $\begingroup$ How would you preserve charge in the decay of an element above iron into only iron? $\endgroup$
    – DJohnM
    Jun 15, 2014 at 17:54
  • 2
    $\begingroup$ @User58220 alpha emission. Or positron emission. $\endgroup$ Jun 15, 2014 at 18:08
  • 3
    $\begingroup$ Though much of the cosmology is now desperately obsolete, the section of Freeman Dyson's classic paper "Time without end: Physics and biology in an open universe" (the full text is available here and there around the web but it is not clear that these sources are copyright kosher) entitled "All matter decays to iron" discusses the gob-stopping time-scales ($\approx 10^{1500}\,\mathrm{yr}$) on which this is expected to occur. $\endgroup$ Jun 15, 2014 at 20:11
  • $\begingroup$ @BrandonEnright: but then you're talking about the decay of something into iron and helium, and the overall binding energy change of the total reaction is not necessarily positive... $\endgroup$
    – DJohnM
    Jun 16, 2014 at 0:16

3 Answers 3


There are two separate issues to consider.

Firstly there is usually an energy barrier to decay. Radioactive decay occurs due to quantum tunnelling through the barrier, and the rate therefore depends on the barrier height. One of the very first studies of this was by George Gamow back in 1928, who studied the alpha decay of uranium-238. Even though alpha decay produces about 5Mev of energy (nearly 500 gigajoules per mole!!) the half life of uranium-238 is about the same as the age of the Solar System. Gamow's calculation is discussed in this PDF, or Google for many similar articles. The decay is slow because there is a barrier of around 25Mev that prevents the decay.

So while it may be energetically favourable for a nucleus to decay to iron a kinetic barrier may reduce the rate to a negligably small value.

Secondly, although for example nickel-60 may have a lower binding energy per nucleon than iron-56 $^1$ this does not mean the reaction:

$$ \mathrm{^{60}Ni \rightarrow {}^{56}Fe + \alpha }$$

is exothermic because the $\alpha$ particle also has a lower binding energy per nucleon than iron $^2$. If you took 56 nickel nuclei, disassembled them into individual nucleons then reassembled them into 60 iron nuclei you might get an overall energy decrease, but this route isn't available. Decay pathways are limited to $\alpha$, $\beta$ and fission, and if any step is not energetically favourable the decay process will stop at that step.

$^1$ Actually, according to Wikipedia nickel-62 is the most stable nucleus not iron-56

$^2$ I have no idea whether this reaction is exothermic or not

  • $\begingroup$ +1, nice, take into account some specific "magic numbers" also, which are related to this as well $\endgroup$
    – Nikos M.
    Jun 15, 2014 at 18:31
  • $\begingroup$ Ni-62 has the lowest binding energy per nucleon, Fe-56 the lowest mass. I've never really gotten my head around which measurement should matter more, and in fact that's related to an unanswered question on this site. $\endgroup$
    – user10851
    Jun 16, 2014 at 4:08
  • $\begingroup$ @ChrisWhite: The relevant quantity is mass per baryon, because baryon number is conserved. The additional mass due to having too many neutrons is associated with weak isospin, which is not conserved. Thus, that mass can be shed (via weak interactions) in the arbitrarily long time limit. $\endgroup$
    – Xerxes
    Jun 16, 2014 at 6:55
  • $\begingroup$ @JohnRennie It lickly just my misunderstanding but in your answer you say "nickle-60 may have a higer binding energy per nucleoun then iron-56 this does not mean the reaction:..." surly if Ni-60 has a higher binding energy per nucleon then Fe-56, Ni-60 is more stable then Fe-56 already and this reaction would not happen any way as it would require energy?? $\endgroup$
    – user43487
    Jun 16, 2014 at 14:31
  • $\begingroup$ @Joseph: oops, that should be lower binding energy, i.e. less stable. $\endgroup$ Jun 16, 2014 at 15:42

There are a number of nuclei that can theoretically decay (based on conservation laws and energy) for which no decay has been observed. A list is in Wikipedia. There are more nuclei on it (164) than nuclei that are energetically prohibited from decay (90). The lifetimes are long enough that the decays are not observed.


In general, you would need to consider cluster decay processes, which are extremely rare. You can estimate the decay probabilities of such processes using a formula given in this article.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.