6

I think this must be a silly question but I would like the opinion of some experts.

A nucleus in excited states usually decay very quickly by isomeric transition, with a half-life of around $10^{-12}$ to $10^{-9}$ seconds. But there are many cases where a nuclide has an isomer which is not known to undergo isomeric transition, especially for odd-odd nuclides:

$\bullet$ $^{180\rm{m}}\rm{Ta}$ is not known to undergo any decay mode;

$\bullet$ $^{176\rm{m}}\rm{Lu}$ has a half-life of $3.664\,\rm{h}$ and a spin of $1^-$ compared to the $3.704\times 10^{10}$-years half-life and the $7^-$ spin of $^{176}\rm{Lu}$, but $^{176\rm{m}}\rm{Lu}$ is not known to undergo isomeric transition to convert to $^{176}\rm{Lu}$;

$\bullet$ There are two states of $^{248}\rm{Bk}$ known, one has a half-life of $23.7\,\rm{h}$ and a spin of $1^-$, the other is only known to have a half-life of $>9\,\rm{y}$ and a spin of possibly $6^+$. We don't know which one has lower energy due to lack of precise measurements, but neither is known to undergo isomeric transition to become the other one.

$\bullet$ There are many odd-odd isomers in the region just above the closed shell $Z=82$, $N=126$, whose isomeric transitions are not known due to alpha decays being too fast.

In conclusion, a nucleus in ground state cannot undergo isomeric transition; a nucleus in excited states is energetically possible to undergo isomeric transition (by definition), but such a process can be very hard to observe, possibly because the spin change would be too high, and/or other decay modes are too fast.

So I would like to know: How do we identify the ground states of nuclides? In other words, a nucleus in its state with the lowest known energy is possibly in its ground state, but perhaps it is also possible that the nucleus can undergo isomeric transition in theory and we haven't yet observe it? So how can we be sure that a nuclide (imagine the familiar ones that appear around us on a large scale) would not turn to be an isomer some day in the future?

Edit. I asked this question because I believed that we could only measure data like mass, half-life, spin, ... of a nucleus, so I wanted to know we could experimentally know if a nucleus is in the ground state.

Let me pick a concrete nuclide. Take the nuclide $^{150}\rm{Nd}$ as an example. It is classfied as being beta-stable since $^{150}\rm{Pm}$ is $83\,\mathrm{keV}$ higher in energy than $^{150}\rm{Nd}$, so beta decay of $^{150}\rm{Nd}$ is impossible. But wait! Perhaps it is not, because there may be an undiscovered state of $^{150}\rm{Pm}$ that has lower energy than $^{150}\rm{Nd}$, and we just don't know it? Perhaps $^{150}\rm{Nd}$ behaves as another nuclide like $^{48}\rm{Ca}$ and $^{96}\rm{Zr}$, its single beta decay being actually energetically possible but largely hindered?

Jianing Song
  • 217

2 Answers2

3

Generally speaking we cannot. A metastable state with very long time is indistinguishable from a true stable state until the transition occurs. For one thing, you don't even have to go for the complicated examples to see that. We don't even know whether the simplest nucleus, i.e. an individual proton is stable or not (https://en.wikipedia.org/wiki/Proton_decay). The only way to tell is to wait and hope to see the actual decay process (https://en.wikipedia.org/wiki/Irvine%E2%80%93Michigan%E2%80%93Brookhaven_(detector)). Therefore, very crudely speaking, physicists abide by the "presumption of innocence", that is, a state is considered to be stable ("ground state") until it is known otherwise (by means of explicit observation of decay).

John
  • 4,287
2

Well it might be the wrong approach to answer this question but the ground state is defined when the neutrons and the protons are occupying the lowest energy levels in the shell model (s,p,d,...). This would allow to define the excited states by removing one neutron/proton from the occupied states to one empty more energetic state.

One thing I should also mention is that you're talking about alpha decays while discussing the notion of a ground state of a certain nucleus. When there's an alpha decay, the nucleus lost two protons and two neutrons (a different element is now at play). Talking about a ground state in this context doesn't make any sense for me because we're considering two different elements.

For some very heavy elements, the alpha decay is the considered decay mode but that doesn't mean that they're not in their ground state before the decay, it might be that the ground state of the initial isotope is unstable. The ground state of 12C is stable but the ground state of 14C is not for example.

Edit:

I might have something but I'm not sure that it would work. You essentially have 2 states in your description (the real ground state and the meta stable state with a very long half-life). Now if you take a sample from the considered element, some nuclei should be in the real ground state because they underwent the decay we're trying to study. By irradiation the sample with an X-ray gun or a continuous photon beam and directly detecting the spectrum of the photon after the irradiation, you could see which photons are being absorbed by the nuclei at the ground state bringing them to an upper excited state.

By doing this you'll be able to see what possible excitations do you have with this material and then start comparing to the existing information that you have about the element. If you find that a certain photon is being absorbed by the element and you are absolutely sure that this doesn't correspond to any of the transitions between the "known" states. Then you might have found an unknown state that could be the real ground state.

I'm not sure about any of this but this is how I would go about it.

Hitman01
  • 76