Is there a simple account of why technetium is unstable?

From the Isotopes section of Wikipedia's article on Technetium:

Technetium, with atomic number (denoted Z) 43, is the lowest-numbered element in the periodic table that is exclusively radioactive. The second-lightest, exclusively radioactive element, promethium, has an atomic number of 61. Atomic nuclei with an odd number of protons are less stable than those with even numbers, even when the total number of nucleons (protons + neutrons) are even. Odd numbered elements therefore have fewer stable isotopes.

It would seem that simply its atomic number is part of the reason why it is unstable, though this just pushes back the mystery back one step for me: why are nuclei with even atomic number more stable? And why then are all of the elements from 45 through 59 stable — notably including silver (Z=47) and iodine (Z=53) — not to mention higher odd-proton nuclei such as gold (Z=79)?

Even the most stable isotope of technetium has a half-life less than a hundredth that of uranium-235, which has a half life of 703.8Ma:

The most stable radioactive isotopes are technetium-98 with a half-life of 4.2 million years (Ma), technetium-97 (half-life: 2.6 Ma) and technetium-99 (half-life: 211,000 years) [...] Technetium-99 (99Tc) is a major product of the fission of uranium-235 (235U), making it the most common and most readily available isotope of technetium.

It's perhaps an unfair comparison, as uranium has an even atomic number (however that is suppose to help mitigate its instability); but it also has nearly twice the number of protons. This deepens the mystery for me. Even granted that Tc has no stable isotopes, how does it come to be so unstable that all of its isotopes are essentially absent naturally, compared for instance to uranium-235?

(This question is a specific case of an earlier question on synthetic isotopes.)

  • $\begingroup$ Isobars are more convenient to study nuclide stability than isotopes. $\endgroup$ – voix Oct 17 '12 at 17:51
  • $\begingroup$ There is no exclusively beta-unstable isobars. $\endgroup$ – voix Oct 17 '12 at 18:02
  • $\begingroup$ @voix: If there are no beta-unstable isobars, is this really a good way to examine why there are no beta-stable isotopes of Z=43? Or is there a good account of how Tc falls short of stability for atomic weights 96-100? $\endgroup$ – Niel de Beaudrap Oct 17 '12 at 18:12
  • $\begingroup$ "If there are no beta-unstable isobars, is this really a good way to examine why there are no beta-stable isotopes of Z=43?" - Yes, because beta-stability depends of proton-neutron ratio. Look at the table of 154-isobars in my linked answer. $\endgroup$ – voix Oct 17 '12 at 18:39
  • $\begingroup$ Except that (for example) 98Tc and 99Tc both have a Z/N ratio much closer to that of 154Gd than does 154Sm, so that simply comparing the magnitudes of the ratios is insufficient. Should I be looking for something subtler? $\endgroup$ – Niel de Beaudrap Oct 17 '12 at 18:55

This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.

If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.

Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.

Response to comment:

The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).

However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:

Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a quali­tative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements.

So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.

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    $\begingroup$ Even a good description of the boundaries of knowledge would be good. "Even Z nuclei are generally more stable than odd, and nobody knows why" could be a partial answer; "Nucleons beyond a core of Z=10 form ring structures which tend to fill up with 8, 12, 16, ... nucleons and Tc is too far between complete shells to be stable" is a (totally made up) possible partial answer. I've heard enough echos of what we know of nuclear structure to know that we know something beyond charge/volume. What are the pertinent factors, however mysterious, and what roles do they play in Tc? $\endgroup$ – Niel de Beaudrap Oct 17 '12 at 10:19
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    $\begingroup$ I've added some links to interesting articles, but note that the general rules fail for Tc, so that just emphasises how limited the genereal rules are. $\endgroup$ – John Rennie Oct 17 '12 at 14:51
  • $\begingroup$ The quoted material about the Weizsäcker semi-empirical mass formula is relevant as to the ability of that mass formula to predict something like this. However, that mass formula is the crudest classical approximation you could possibly make, so its failure in this case doesn't indicate that nuclear physicists in general can't retrodict beta-decay energies with the necessary precision. $\endgroup$ – user4552 Aug 28 '13 at 23:44

this just pushes back the mystery back one step for me: why are nuclei with even atomic number more stable?

There is no big mystery about this. There is a pairing interaction in nuclei. It's loosely analogous to the Cooper pairs in a superconductor.

It's easy to construct an argument as to why, if you're looking for an element with no stable isotopes, Tc is a good candidate.

Any even atomic number is guaranteed to have an even-even isotope that's stable against beta decay, because even-even isotopes are more stable than odd-odd ones, due to pairing.

In general we expect a richer variety of stable isotopes for elements whose atomic numbers are near a magic proton number, or for elements such that for that element, the line of stability comes close to a magic neutron number.

Based on these considerations, if we were hoping to find a light element with no stable isotopes, it would be one that was an odd $Z$, one whose $Z$ was far from any magic proton number, and one for which the line of stability is not close to a magic neutron number. Technetium fulfills these requirements.

To retrodict this theoretically, the simplest thing we could try would be to use the semi-empirical mass formula, which is essentially the classical energy of a charged liquid drop, with a couple of terms thrown in to approximate quantum-mechanical effects. As John Rennie's answer describes, this works surprisingly well, in the sense that although it's incredibly crude, it correctly predicts that, e.g., 97Tc is very, very close to the dividing line between stability and instability.

The next step up in sophistication would be the Strutinsky smearing technique (Strutinsky 1968; also described in Salamon 2010). This method involves taking a classical liquid-drop energy and adding on a correction for quantum effects. It correctly reproduces the loss or gain in binding energy due to the higher- or lower-than-average density of single particle levels. E.g., it correctly predicts that magic numbers are far more bound. The technique has the advantage of being computationally cheap, and of working for both nuclei near closed shells and mid-shell nuclei.

If you want to know whether the Strutinsky technique is good enough to retrodict the beta-instability of 97Tc and 99Tc, I think the answer is basically that the question is ambiguous. These models have a fairly big number of adjustable parameters, maybe 30 or so, that need to be fitted to the experimental data. There are also qualitative choices involved, such as the use of a Woods-Saxon potential as opposed to Nilsson. When you use these ~30 parameters to predict some very large number of experimental observables across the entire chart of the nuclei (tens of thousands of masses, electric quadrupole moments, ground-state spins, ...), you get points for honesty but you don't get the best precision. People who are interested in a certain region of the chart of the nuclei, e.g., superheavy elements, will fit their parameters to that small region and get much better precision. If you keep on narrowing your focus like this, and you're making retrodictions about well-studied regions of the chart of the nuclei, then eventually what you're doing is just a fancy exercise in interpolation. I'm sure that at this level, one could correctly calculate the very small difference in binding energy between 97Tc and 97Mo to the precision needed in order to show that 97Tc is beta-unstable, but all you'd really be doing would be interpolating.

Strutinsky, Nucl. Phys. A122 (1968) 1

P. Salamon, http://arxiv.org/abs/1004.0079

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    $\begingroup$ Not a big mystery, perhaps, but an explanation that uses magic numbers seems a little bit mysterious... $\endgroup$ – Art Brown Aug 29 '13 at 2:08
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    $\begingroup$ @ArtBrown: I don't know if it's the word "magic" that's putting you off, but the idea isn't really that mysterious. Maybe you should post a question, "What is a magic number in nuclear physics?" $\endgroup$ – user4552 Aug 29 '13 at 2:24
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    $\begingroup$ I looked it up. The comment was meant to be humorous but obviously failed. Perhaps you could include a wikipedia link to "magic number (physics)". Nice answer in any case. $\endgroup$ – Art Brown Aug 29 '13 at 2:31
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    $\begingroup$ @ArtBrown: Good suggestion, done. $\endgroup$ – user4552 Aug 29 '13 at 2:39

here's a view from a pure mathematician in an abstract ideal world, untroubled by real-world complexities. I hope it's helpful even if it is a gross simplification of reality.

Atomic shells feel evenly balanced when they are of a size equal to twice a square number, i.e. 2, 8, 18, 32, 50, 72, 98, 128 etc. Every period in the periodic table has a number of elements equal to one of these numbers, and elements which occupy the spaces at the end of these periods are so contentedly balanced that they have no desire to change or react at all. Elements on either side of these noble gases are reactively desperate to emulate the harmonious nirvana of their immediate neighbors, but elements in the middle of a period (such as carbon) are pulled in two different directions and can consequently end up wrapped up in very complex molecules.

Atomic nuclei are most stable when their number of protons is even, and close to one of the numbers listed above. Technetium (43) and Promethium (61) don't satisfy either of these criteria, and are unstable.

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    $\begingroup$ Welcome to Physics.SE! This answer is essentially correct, but neither the electronic closed shells (2, 10, 18, 36, 54, 86, 118(?)) nor the nuclear magic numbers (2, 8, 20, 28, 50, 82, 126) follow the pattern you have named — the physics is more complicated than $2n^2$. $\endgroup$ – rob Jan 29 '16 at 7:13
  • $\begingroup$ Thanks Rob! The numbers 2, 10, 18, 36, 54, 86, 118 are computed by cumulatively adding 2, 8, 8, 18, 18, 32, 32, which are all among the numbers I mentioned. The early magic numbers also approximately correspond with these numbers, though as you say, physics is more complicated than pure mathematics. But as a rough rule-of-thumb, it's not too bad :-) $\endgroup$ – Phil Jan 29 '16 at 9:24
  • $\begingroup$ The sequences actually come from the number of fermions you can fit into an orbital with angular momentum $L$, which is $2(2L+1)$. This is why the periodic table has columns in groups of two (electrons with $L=0$), six ($L=1$), ten ($L=2$), and fourteen ($L=3$). But yes, you've got the gist of it. $\endgroup$ – rob Jan 29 '16 at 19:07
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    $\begingroup$ Great, I didn't know that part of the story. So angular momentum gives us the sequence 2, 6, 10, 14, 18, 22..., and when we cumulatively add those numbers together we get 2, 8, 18, 32, 50, 72..., which are the numbers I mentioned in the original post. It all fits together beautifully now $\endgroup$ – Phil Feb 5 '16 at 8:13

I have long wondered about why certain elememts nuclei are more unstable the the others. Technetium is one example. It is odd atomic numbered but has no reason to be less stable then other fermionic-like atoms, some heavier but stable like gold. I shall explain further (as I am also unsure) but I will try to explain as far as I can without the vague use of 'uncertainty' and 'magic numbers'.

At first look, it seems like magic numbers are to blame. But no. When looked closer atoms do have nuclei structure. Look at Uranium class elements. More massive but are significantly more stable. Magic numbers are calculated through spin and these spinning particles should have a configuration where they rest in lowest energies. Yes I agree its also uncertainty, but uncertainty IS uncertainty due to the minute mass and size and unpredictablity of these particles trajectories. Atoms are bounded by the strong force, where colour charge interactions between neutrons and protons with spin exchange muons. They swap places. This usually leads to some physicists stating vaguely that atomic nuclei are just soups of protons and neutrons. Once again they are quite 'wrong' that subatomic particles do have a structure, not in that sense which it stays rigid but in one that has the lowest energy. Hydrogen 3 decays to Helium 3 due to energy(mass).

To summarise my topic, I suggest that Technetium is unstable due to nucleus shell instability (I apologise for my vagueness), where even some nuclei (metastable 'm') with stronger spin can be more stable then its lower spin state. The atomic nuclei is indeed a very complex structure which I hope future physicists can explain. I thank all previous answers for their ideas.

  • $\begingroup$ for the record, the atomic number has no impact on whether or not an atom is fermionic or bosonic. An atom, as opposed to an ion, has the same number of protons and electrons. This means that regardless of the atomic number, the spin of the atom due to the protons and electrons always is an integer. The neutron number is what determines if an atom is fermionic or not. $\endgroup$ – Jim Jun 12 '15 at 14:33

I'm not very well versed in nuclear models and many of the above answers are probably a lot more accurate and in-depth but from my basic understanding nuclei decay when they are heavier than the masses of their potential products which means they're more stable as they're products.

All the isotopes of technetium are heavier than their decay products and thus will decay into them to enter a more stable state, for example tc-97 which is the most stable isotope of technetium weighs more than its decay product mo-97 (96.9060215amu vs 96.9060215amu) and so will decay into it, however as for why there is no stable configuration of technetium with 43 isotopes, I do not know.


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