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In the paper

Characteristics of 2-photon ultraviolet laser etching of diamond. RP Mildren et al. Opt. Mater. Express 1, 576 (2011),

I found out that UV light cause the atom ejection from the surface of diamond. According to the article, the process is caused by the presence of UV light and there's no threshold meaning that the loss of mass will happen even under sunlight. My question is: if there's no threshold, does that mean that the process will happen even under visible light in standard irradiance or there's definitely necessary to have UV rays? Thank you.

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  • $\begingroup$ This was asked already here on in Chemistry and I short commented basically as in the more articulated answer by Floris. $\endgroup$ – Alchimista Mar 14 '18 at 19:14
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By analogy with the photoelectric effect, you can have "no threshold" of intensity, yet have a threshold of wavelength. If an individual photon does not have sufficient energy to kick an atom off the surface, then the effect will not occur.

If they say "you need UV", then you need UV. Of course there is a small component of UV in sunlight (some of it is filtered by air, so it's a function of altitude).

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  • $\begingroup$ And it seems to need 2 UV photons (for their particular laser). Sunlight seems unlikely to do much... $\endgroup$ – Jon Custer Mar 14 '18 at 19:55
  • $\begingroup$ Good point. “Two photon” usually implies very intense (focused, pulsed) light - not sure what the “no threshold” comment refers to then. $\endgroup$ – Floris Mar 14 '18 at 20:20
  • $\begingroup$ I presume that they needed to transfer enough energy in one shot and didn't have a laser wavelength short enough on its own. But, the two-photon process managed to be able to kick an atom off. $\endgroup$ – Jon Custer Mar 14 '18 at 20:55

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