Thorium series:

Thorium decay chain

Uranium series:

enter image description here

Actinium series:

enter image description here

Why do all the radioactive decay series terminate at lead isotope? Why not hydrogen?

(All pictures were taken from Wikipedia.)

  • 3
    $\begingroup$ I'm no expert, but I thought they just kept on decaying until they reached a stable isotope, and lead just happens to be one $\endgroup$
    – NotEvans.
    Commented Dec 29, 2015 at 16:34
  • 13
    $\begingroup$ Well, terminating at hydrogen would imply that no elements beyond hydrogen are stable, which would imply a very different universe. Lead just happens to be the first reached stable nuclei. $\endgroup$
    – Jon Custer
    Commented Dec 29, 2015 at 16:43
  • $\begingroup$ You might benefit from studying the first diagram here while noting that most of the nuclides above and to the right of the intersection $Z = 82$, $N \approx 126$ decay towards lead, but lighter unstable isotopes generally do not. $\endgroup$ Commented Dec 30, 2015 at 5:51
  • $\begingroup$ I believe I read somewhere that at sufficiently high energies, lighter elements tend to fuse toward iron. $\endgroup$
    – hBy2Py
    Commented Dec 31, 2015 at 3:00

2 Answers 2


There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of this division (0, 1, 2 or 3, corresponding to the $4n$, $4n+1$, $4n+2$ and $4n+3$ mass number decay chains respectively) stays constant under alpha, beta or gamma decay. There are decays which change the remainder (neutron emission, proton emission, spontaneous fission, etc) and therefore allow hopping between the main decay chains, but for simplicity these can be approximately ignored for the isotopes which are not too unstable or too heavy.

Only three out of these four main decay chains "stop" at lead, so already the statement in the question is incorrect. The $4n$, $4n+2$ and $4n+3$ chains "stop" at $\ce{^208_82Pb}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$, respectively. The $4n+1$ decay chain reaches $\ce{^209_82Pb}$, but this lead isotope is quite short-lived ($t_{1/2}=3.25\ \mathrm{h})$ and decays further to $\ce{^209_83Bi}$ where it "stops" (or at least seemed to until 2003).

Note that I said the chains "stopped" at these isotopes. That's because in reality they don't. It just happens that $\ce{^208_82Pb}$, $\ce{^209_83Bi}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$ are all very long-lived isotopes, with half-lives billions of times greater than the current age of the Universe. This means that any continued decay is severely bottlenecked at these points, so for almost all practical purposes, the decay chains stop there. However, rigorously speaking, if you were really patient there would be further decays in the sequence. In fact, theoretically the heaviest isotope which is not susceptible to any known mode of spontaneous radioactive decay (except proton decay) is an isotope of zirconium, $\ce{^92_40Zr}$, so the decay chains actually go on well past lead or bismuth. It just takes an immense amount of time for the decays to happen.

  • $\begingroup$ Seems that proton decay has not yet been proved. Overall +1. $\endgroup$ Commented Feb 20, 2020 at 13:36

Short Answer: They don't.

As the comments above indicate, the decay chains will terminate at a stable nuclei. For the series you mentioned in your question, lead has the first stable isotope.

A really useful tool for looking at different decay chains is available from Oak Ridge National Laboratory's Risk Assessment Information System.

You can pick from lots of different isotopes and see the decay chains. For example, $\ce{^142_56Ba}$ gives $\ce{^142_58Ce}$ as the final stable product. $\ce{^14_6C}$ goes to $\ce{^14_7N}$, etc. Many terminate with lead, but many don't.

As for hydrogen? Keep in mind that for large nuclei, fission and other nuclear processes that decrease the number of protons are usually favored, but for small nuclei, processes that increase the number of protons are often favored (e.g., beta decay) - changing the neutron-proton ratio and generating more stable nuclides.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.