radio - What is the lowest energy atomic transition ever detected and identified?

In this concise and insightful answer it was pointed out that atoms can have an extremely large (possibly infinite) number of bound excited states.

The Rydberg formula represents a simple model for the energy of transitions in a hydrogen-like (one-electron) atom. Originally written for the measured vacuum wavelength of the photon released by the transition:

$${1\over\lambda_{vac}}=\text{R}_{\infty}Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$

because the original work was done with grating spectrographs. Z is the atomic number of the nucleus. It is often written in terms of energy as well:

$$E=h\nu = \text{Ry}Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$

where the Rydberg energy Ry is roughly 13.6 eV.

As was pointed out, as $n$ increases, the number of energy levels for this simple model increases (mathematically, at least) without limit. So possible transitions can have arbitrarily small energy, and therefore long wavelength.

It's often the low energy transitions measured in radio astronomy that allow mapping gas temperatures and distribution, rather than the optical transitions.

My Question: What is the lowest energy atomic transition ever detected and clearly identified? I've added the discussion of the Rydberg model just to help explain why there can be extremely low energy transition. If there is a really compelling case for a lowest energy record that was achieved in - for example - a diatomic molecule, that would be OK here - as long as it is a well defined and unambiguously identified transition.

Naturally occurring atomic hyperfine transitions are fine (no pun intended) but I want to rule out things like NMR, where the transitions and their energy are caused and defined by (respectively) an externally applied fields. This should be about transitions of the atom itself.