I know that $1 \:\mathrm{cm}^{-1} = 8065.54429 \:\mathrm{eV}$.

I know that wave-number has unit $\:\mathrm{cm}^{-1}$ and that $E=hc\times \mathrm{wavenumber}$.

But how do I arrive to the number $8065.54429$?

I know how to make the conversion but how do I derive that $1 \:\mathrm{cm}^{-1} = 8065.54429 \:\mathrm{eV}$?

  • 1
    $\begingroup$ Find wavelength of a photon with an energy of 1 eV. Take the reciprocal. Done. $\endgroup$ – Jon Custer Aug 21 '18 at 13:26

I know that $1 \:\mathrm{cm}^{-1} = 8065.54429 \:\mathrm{eV}$

You 'know' wrong. The quantity on the left is a reciprocal length and the quantity on the right is an energy. It makes as much sense to say that they are equal as to ask whether the surface area of Europe is bigger or smaller than the mass of the Moon.

What you can say is that, if a beam of light has wavenumber $\kappa$, then

  • its wavelength is $\lambda=1/\kappa$,
  • its frequency is $\nu=c/\lambda = c\kappa$,

and therefore its photon energy is \begin{align} E_\mathrm{photon} & = h\nu \\ & = hc\,\kappa \\ & = 0.000123984 \:\mathrm{eV}\:\mathrm{cm}\times\kappa \\ & = \frac{\mathrm{eV}}{8065.544} \times \frac{\kappa}{\mathrm{cm}^{-1}\!\!} \\ & = \frac{1}{8065.544} \frac{\mathrm{eV}}{\mathrm{cm}^{-1}\!\!} \times\kappa \\ & = 1\:\mathrm{eV} \times \frac{\kappa}{8065.544\:\mathrm{cm}^{-1}\!\!} . \end{align}

In other words:

  • a wave with a wavenumber of $\kappa = 8065.544 \: \mathrm{cm}^{-1}$ will have a photon energy of $E_\mathrm{photon} = 1\:\rm eV$, and
  • a wave with a wavenumber of $\kappa = 1 \: \mathrm{cm}^{-1}$ will have a photon energy of $E_\mathrm{photon} = \frac{1}{8065.544}\:\rm eV$.

(Or, in other words: you got the conversion factor backwards. This is the kind of catastrophic mistake that you can easily get sucked into by being sloppy with your units and pretending that you can equate quantities with different physical dimension.)

The actual numerical value comes from the physical constant $$ \frac{1}{hc} = 8065.544 \frac{\mathrm{cm}^{-1}\!\!} {\mathrm{eV}} $$ when evaluated in units that are powers of $\rm eV$ and $\rm cm$.

For other examples of how to do these conversions carefully, see e.g. p. 553 in Herschel and Schulz or p. 287 in Inguscio and Fallani

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