# What advantage is there in reporting the Lennard-Jones well depth in wavenumber units?

The Lennard-Jones well depth $$\epsilon$$ is typically given in energy units ($$\mathrm{kJ}$$ or $$\mathrm{kcal}$$; sometimes per mole). Jasper and Miller, however, reported $$\epsilon$$ values in reciprocal centimeters ($$\mathrm{cm^{-1}}$$).

Why use $$\mathrm{cm^{-1}}$$?

1. It seems unnatural using $$hc$$ as a conversion factor for a classical description.
2. Absorption peaks (e.g. in wavenumbers) are not related to the LJ potential well, which is for non-bonded interactions.

• Why not? It's just a conversion factor. – Todd Minehardt Nov 20 '16 at 22:41
• @ToddMinehardt it seems unnatural using hc as a conversion factor for a classical description. – Sparkler Nov 20 '16 at 22:44

I don't think there's any inherent advantage as one can always convert to another unit. It seems natural to me, however, because chemists generally have a good idea of the energy of vibrational and rotatial states in $cm^{-1}$, so being able to compare the depth of the well with some familiar chemical process is quite nice. For instance, if the well-depth were quite small, one could immediately identify that the system may not spend a lot of time there if it is comparable to the zero-point energy or something similar.
It's similar to how physicists report everything in electron-volts ($eV$). They do it because it's familiar and they have many physical processes they can use as a reference. For instance physicists know that the band gap of semiconductors is between $2-5\ eV$, so it's quite natural to report other energies in the same unit.
• They are related in some sense. Non-covalent interactions absorb light like covalent bonds do. For instance, a water dimer has intermolecular vibrational normal modes, and these normal modes have a certain energy which corresponds to a frequency of light that might be absorbed when using spectroscopy. These modes can only exist because the water molecules are bound in some potential energy well. If I observed one of the modes were at an energy of $1000\ cm^{-1}$, then I know the well must be deeper than that, and I would express the well-depth in $cm^{-1}$. – jheindel Nov 22 '16 at 2:59