# Why is kelvin used for Energy Separation in atoms?

I have been reading a 1974 paper on the anisotropy of Sm$$^{3+}$$ ions, in which the authors say the following:

IN THE series of trivalent rare-earth ions, Sm$$^{3+}$$ is unique in the sense that the energy separation between the ground J 5/2 multiplet and the first excited multiplet is only 1400K, ...

I understand that K can be used as a unit of energy, but I don't really understand why people would choose it over something like eV. What reason is there for people to use K over something like eV?

• It could be to have rounder numbers but 1400 K -> 120 meV, so it is not that much better. Maybe to simplify some statistical physics calculation. Dec 20, 2021 at 13:03
• My best guess is that sometimes it's useful to express energy gap in terms of black body temperature shift. Dec 20, 2021 at 13:45
• Also Samarium is from Lanthanide group, which is suitable for using as active elements in a Laser systems, because it's easy to achieve population inversion with them, cause they decays to ground level slowly. $\text {K}$ units has somewhat finer resolution than $\text {MeV}$, may give a better insight how laser system energy levels / output will change given exact transition schema. Thus it may add to my explanation above. Dec 20, 2021 at 14:15

Sometimes this is done to give an idea of the temperature above which an excited state is likely to be found occupied just by thermal agitation. The environment can in fact provide some energy to the system by means of thermal fluctuations, whose amplitude is of the same order of magnitude as $$k_BT$$ ($$k_B$$ being the Boltzmann's constant and T the temperature).
For example in this case, if we imagine to sweep up gradually the temperature, we would observe that the first excited state of the system would be almost always empty for $$T\ll1400K$$, because the fluctuations would be too small to "kick" the electron from the ground to the excited state. Then for intermediate temperatures the probability for the electron to successfully jump would become higher and higher. Finally for $$T\gg1400K$$ the excited state would be almost certainly occupied.