# Two-Photon Absorption

It takes about 6.2 eV of energy to ionize an air molecule. So any laser beam that has an energy-per-photon of 6.2 eV or more will almost immediately be absorbed by air. This corresponds to a wavelength of 0.2 μm.

But if you have a beam of light with a wavelength between 0.2 μm and 0.4 μm, then two photons acting together will have enough energy to ionize a molecule in the air. This is much harder than just ionizing with one photon, but if the beam is very intense it can happen. 0.4 μm happens to be just on the threshold between violet and ultraviolet light, and the 0.2 μm to 0.4 μm range spans the near ultraviolet between vacuum ultraviolet and the visible part of the spectrum, If you have an ultraviolet laser and are trying to focus it to a very intense spot at your target, two photon absorption can end up removing most of the power from your beam before it gets to your target!

If you start with a power of $P_{0}$ in your beam, emitted from a focusing aperture of diameter $D$ and focused to a spot of size $S$ on the target (which can be as small as the diffraction-limited spot size, but can also be larger), and the target is at a range of $R$ , and if the two-photon absorption coefficient is $\alpha _{2}$ , the power that hits the target will be 

$P={\frac {P_{0}}{1+(4/\pi )\,(\alpha _{2}\,P_{0}\,R)/(S\,D)}}$ So if $(4/\pi )\,(\alpha _{2}\,P_{0}\,R)/(S\,D)$ is much less than 1, two photon absorption will be negligible. If it is much larger than 1, two photon absorption will mean that most of your beam never gets where you want it to go.

The actual two-photon absorption cross sections of oxygen and nitrogen are difficult to find. However, based on typical two-photon cross sections of other atoms and molecules, the quantity $(4/\pi )\,\alpha _{2}$ for sea level air on earth can be expected to be somewhere in the range of $10^{-7}{\mbox{cm}}/{\mbox{W}}$ to $10^{-10}{\mbox{cm}}/{\mbox{W}}$ . If you are building a real-life ultraviolet laser death ray, you will need to get the correct value for the wavelength you are using. However, for the purposes of fiction just choosing $(4/\pi )\,\alpha _{2}=10^{-9}{\mbox{cm}}/{\mbox{W}}$ won’t be too wrong. This $(4/\pi )\,\alpha _{2}$ will be proportional to the atmospheric density.

## Credit

Author: Luke Campbell