Thorium-229 has a famous isomer with an excitation energy of only about 7.8 eV. As I gather from the wikipedia page, this transition was discovered essentially by accident from gamma ray spectroscopy. It's gotten clock gurus all excited recently as a possible route to a very accurate nuclear clock.

This transition has two properties that make it attractive as a clock transition candidate: It is low energy for a nuclear transition, and it has a very narrow linewidth. I have a couple of questions about this transition:

  1. Do we understand theoretically why this transition exhibits these properties?
  2. With what sort of confidence do we believe that there are no other nuclear transitions that exhibit these properties?
  3. What are the next-lowest-energy known nuclear transitions?
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    $\begingroup$ Wow: 7.8 eV? That is a sooooooft gamma ray. $\endgroup$ – JEB Feb 3 '18 at 2:01

The existence of such a low-energy excited state is essentially random. This is an odd nucleus (even Z, odd N), so the low-lying states correspond to different states of the odd (unpaired) neutron. The energy spacing of single-particle states for a nucleus of this size is probably somewhere on the order of 10-100 keV, and when these single-particle levels differ in spin or parity (so that they can't interact), the spacing between them is essentially random. Because there is pairing, the energies of the excited states of the nucleus are the quasiparticle energies, not the single-particle energies, and for low-lying states, these energies all tend to be pretty close together (they're all roughly equal to the pair gap). Therefore you might expect the energy differences between successive low-lying states with different spin-parity to be on the order of 10 keV. If you go to the ENSDF database and look up 229Th, this seems to be pretty much the case on the average.

The excitation energy of this isomer is about 1000 times smaller than 10 keV. That's probably just pure luck. Out of a thousand odd nuclei in this mass region, you would probably expect to see roughly one with an excited state as low in energy as 10 eV.

The state seems to have a lifetime of hours, and its electromagnetic decay branch would be a dipole (probably magnetic dipole, M1). Atomic transitions are usually electric dipoles (E1). Ignoring the distinction between electric and magnetic dipole radiation, basically we expect dipole radiation to have a partial half-life that goes like $d^{-2}\omega^{-3}$, where $d$ is the (dynamic) dipole moment and $\omega$ is the frequency of the transition. Typical lifetimes for nuclei to decay by gamma emission are nanoseconds, but the very small $\omega$ of this transition would give it a very low transition rate for electromagnetic decay.

The narrow linewidth is presumably "narrow" in comparison with atomic decays of a similar (UV) energy. I assume the idea here is that the linewidth is $h/\tau$, where $\tau$ is the lifetime, and the lifetime of this state is very long compared to typical lifetimes of atomic states, due to the much smaller dipole moment for a nucleus compared to an atom. There are atomic states that are very long-lived due to selection rules. However, as pointed out in a comment by Akhmeteli, nuclear states have the advantage that they are easier to shield against external fields.

With what sort of confidence do we believe that there are no other nuclear transitions that exhibit these properties?

There are probably lots of others, but we just don't know that they exist. It takes considerable luck to detect states like this. The gamma ray transition rate is extremely low, so normally you would just not be able to detect the state by gamma-ray spectroscopy.

Probably an important property of this isotope for the technological purposes they're talking about is that it can be produced in macroscopic quantities in a reactor.

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    $\begingroup$ "it's not obvious to me why this nuclear states works so much better." According to the reference in my answer, "Atomic clocks ... accuracy depends on the line width of the states and on the accuracy to which the transition frequency can be shielded from external perturbations. Transitions between nuclear states could increase the accuracy of clocks by another two to three orders of magnitude.Nuclear transitions are extremely narrow and nuclei are about 100,000x smaller than atoms and thus less susceptible to external fields." (They have a reference for that) $\endgroup$ – akhmeteli Feb 3 '18 at 2:21

Looks like world’s second-lowest-energy isomer is uranium-235m, its energy is currently known as 76.8 ± 0.5 eV (https://ldrd-annual.llnl.gov/ldrd-annual-2015/nuclear/lowest).

EDIT (09/01/2018): However, according to L. von der Wense et al. (Measurement Techniques, Vol. 60, No. 12, March, 2018, p. 1178), this isomer is not a good candidate for use in a nuclear clock: "There is only one further nuclear state known with an energy of less than 1 keV above its nuclear ground state, which is $^{235m}$U, providing an energy of about 76 eV. This isomeric state, however, does not allow for the development of a nuclear clock, due to its extremely long radiative lifetime of 3.8·10$^{22}$ sec, which is about 100,000 times the age of the universe, not allowing for any significant laser coupling." By the way, the title of the relevant paragraph 2.3 there is "What makes $^{229m}$Th unique", pretty much the same as your question.

Let me also note that the excitation energy and lifetime of Th isomer (bare nucleus) was measured recently to be about 7.1 eV and 1880 s (https://arxiv.org/abs/1804.00299).

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    $\begingroup$ Note to flaggers/reviewers: This answer answers the third subquestion of the question, and does include the most relevant information from its link, so it is technically an answer. If you think this is answer is not useful, downvote it instead of flagging it. $\endgroup$ – ACuriousMind Feb 3 '18 at 12:52

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