2
$\begingroup$

Can you explain (or direct me to a link) the process by which a photon of $8.10^{14}$ Hz is absorbed by the skin without damage, whilst a slightly more energetic one (UVB) makes damage?

What area of skin is necessary to absorb one single photon? How many photons of visibile light per $\mathrm{cm^2}$ are required in order to produce burns to the skin?

$\endgroup$
  • 1
    $\begingroup$ I actually don't understand the question about the skin area, (but that might be just by my lack of physics background), could you explain that a bit further? $\endgroup$ – mpribis May 31 '17 at 6:35
  • $\begingroup$ You've put too many questions in there. Please narrow it down to, preferably, single question $\endgroup$ – another 'Homo sapien' May 31 '17 at 6:39
  • $\begingroup$ Sun rays burn more at noon and in summer, because the angle of incidence of the photons determines the quantity of photons per cm^2. That implies that the skin can absorb a limited amount of photons. If we concentrate sun rays with a lens we can get hundreds of Celsius degrees (550°, or even thousands. I'd like to know how many photons per square cm are tolerated. The questions are all related $\endgroup$ – user32816 May 31 '17 at 6:42
5
$\begingroup$

The reason that UV light causes sunburn, while normal light does not, is not founded in the energy carried by single photon. As stated on wikipedia:

Importantly, both sunburn and the increase in melanin production are triggered by direct DNA damage. When the skin cells' DNA is overly damaged by UV radiation, type I cell-death is triggered and the skin is replaced

The UV light can cause reactions in the DNA of the skin cells that directly damages the DNA. A low amount of this damage can usually be repaired by the cells, but a lot of DNA damage will lead the cells to die and cause inflammation reactions in the skin (the sum of which is then what we experience as sunburn). [Here is a citation also explaining this, the paper is behind a paywall though].

What area of skin is necessary to absorb one single photon? How many photons of visibile light per cm2 are required in order to produce burns to the skin?

A single UV photon has no measurable size and can therefore in principle be absorbed by any arbitrarily small patch of skin. In practice only those photons that interact with the DNA in your skin cells are relevant for sunburn (that will be a very small fraction of all UV photons). Additionally the melanin in your skin will also absorb photons and therefore protect the DNA from the UV radiation, which is why people with darker skin get sunburn less easily. This also means that the amount of radiation/skin area required to cause sunburn varies from person to person and can be influenced by the ability of your skin cells to repair the induced DNA damage.

$\endgroup$
5
$\begingroup$

Proteins and DNA absorb light in the UV spectrum (both UVA and UVB) because that is the energy level at which pi electrons in the aromatic rings get excited (nucleotides and aromatic amino acids have that), so exposure to UV will cause direct damage to our biomolecules causing an inflammatory response and even cell death. Also, different chemical modifications may occur at the DNA level upon UV irradiation that may result in mutations that accumulating over time can lead to cancer.

However, due to the different wavelength, UVB releases most of their energy on the superficial layers of the skin causing more damage at that level.

What area of skin is necessary to absorb one single photon? How many photons of visible light per cm2cm2 are required in order to produce burns to the skin?

It depends on the skin color! Melanin is the skin pigment responsible for the variation of color in the human skin and it acts as protectant by absorbing and thus shielding UV light (it also has aromatic rings). The more Melanin, the darker is the skin, the more resistant is to sunburn.

$\endgroup$
  • $\begingroup$ Linked pictures! $\endgroup$ – alec_djinn May 31 '17 at 18:01
  • 1
    $\begingroup$ and what about the pi electrons part! That's the core of it. $\endgroup$ – alec_djinn Jun 1 '17 at 8:09
  • $\begingroup$ @canadianer: Alright, that is fair enough ... I might have underestimated that part of the question slightly. The last edit also made this more clear to me. Thanks! Removed the -1 and deleted my comments. $\endgroup$ – AlexDeLarge Jun 2 '17 at 8:50
1
$\begingroup$

Technically- it does cause damage. The difference in wavelengths is utlimately the difference in potential energy that could be conveyed as damage; it just happens to be that the amount of damage visible light can do it less concerning than say ionizing radiation. Visible light, though weak by comparison, has been shown to accelerate the formation of free radicals (which spontaneously form on top layers of skin).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy