Tellurium-128 is the longest lived radioisotope so far with a half-life of 2.3 septillion ($2.3*10^{24}$) years. I know I've heard something about so called "magic numbers" that if you have the perfect amount of protons and neutrons in an atomic nucleus it usually is still unstable but has an absurdly long half-life (examples bismuth-209,xenon-124,). But why is tellurium-128 this stable? And why is this for the ones I have just said?

Wikipedia Link: https://en.wikipedia.org/wiki/Isotopes_of_tellurium

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    $\begingroup$ Somebody has to come out on top. $\endgroup$ Jan 11, 2022 at 3:16

2 Answers 2


Tellurium-128 is in no way extraordinary or unique.

Cape Chelyuskin in Russia is the northernmost extreme point of continental Eurasia. Tellurium-128 is not an extreme point of anything. There are hundreds of isotopes which are less stable than it and hundreds of those which are more stable. The latter kind is called simply "stable".

Now you might think there is a clearcut boundary between the stable isotopes and the radioisotopes. No, there is none. As our detection methods improve, more and more isotopes previously thought to be stable are found to have some decay path, usually with an absurdly long half-life (otherwise they would have been found earlier). When I was a kid, the heaviest stable element had the number 83. This is no longer the case.

Magic numbers of protons and neutrons have no role in this. Well, I mean, they are related to the same field: if a nucleus has either of these, or better both, it is expected to be more stable than its peers. $\rm^{128}Te$ doesn't and isn't.

If anything, $\rm^{128}Te$ is (or was; I didn't check) a temporary winner in a perpetual race. In due time, longer half-lifes will be detected, and then longer yet, and so on.

So it goes.

  • $\begingroup$ We have $_{180}^{73m}Ta$ waiting in the wings for a predicted, but highly spin-forbidden, decay, no? $\endgroup$ Jan 11, 2022 at 10:52
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    $\begingroup$ Apparently so. Then tellurium-128 is not even the record holder. $\endgroup$ Jan 11, 2022 at 11:20
  • $\begingroup$ When it comes to double beta decay, the mode by which Te-128 decays, there are many nuclei which are predicted to be unstable but no decay has been observed as yet - see en.wikipedia.org/wiki/Double_beta_decay $\endgroup$
    – Ian Bush
    Jan 11, 2022 at 16:07
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    $\begingroup$ And then there's always the projected decay of the proton, predicted half life > 10^30 years $\endgroup$
    – Ian Bush
    Jan 11, 2022 at 16:13

There are unstable nuclides more stable than tellurium-128.

Multiple nuclides are considered thermodynamically unstable with predicted decay path, but they are observationally, kinetically stable. Some of them are eventually observed to decay, directly or indirectly, as the detection techniques improve over time.

Two of interesting examples are

  1. calcium-48 ($\pu{6.4E19 y}$), the double-magic-number nuclide with the extra long half-life in spite of the highly disbalanced n/p ratio.
  2. Metastable tantalum-180m, the only (just observationally) stable odd-odd nuclide heavier then nitrogen-14. It is stabilized by its high spin state and "forbidden" transition. The lower energy tantalum-180 has the half-life about 8 hours, beta decaying to hafnium-180 or tungsten-180.

Some long-life nuclides like calcium-48 and tellurium-128 perform double beta decay, like

$$\ce{^{48}Ca -> ^{48}Ti + 2 e- + 2 \bar{\nu_\mathrm{e}}},$$


$$\ce{^{128}Te -> ^{128}Xe + 2 e- + 2 \bar{\nu_\mathrm{e}}}.$$

They are quantum tunnelling through less stable odd-odd nuclide ( scandium-48 or iodine-128, respectively ) with higher energy (even-even -> odd-odd -> even-even). These intermediate nuclides form an energetic barrier analogical to very high activation energy of chemical reactions (like kinetically stable diamonds).


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