How Do Suncreams Absorb UV Radiation at the Atomic Level?

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Suncreams work by absorbing ultraviolet (UV) photons, which excites atomic electrons to higher energy levels. When these electrons return to their ground state, they release energy in the form of photons that are less energetic than the initial UV photon. This process helps to mitigate the damaging effects of UV radiation on skin tissue. The discussion emphasizes the importance of understanding electron transitions and the energy differences involved in these processes. Overall, the absorption and subsequent emission of lower-energy photons is a key mechanism in how suncreams protect against UV damage.
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Homework Statement


Scientists put a lot of effort into developing suncreams which absorb the damaging wavelength of ultraviolet radiation. Suggest what might be happening to the atoms within the suncream when they absorb ultraviolet photons.


Homework Equations


None. This is an explain question.


The Attempt at a Solution


I believe the photons are excited to higher energy levels.

^What more could I write to explain this?
 
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It sounds to me like you're supposed to be doing some research. Have you tried to dig up anything online?
 
No, it's a past exam question. One mark was obtained with what I had written about energy levels and the second mark says 'related to electron energy levels' which I don't know how to explain.

Surely, what I have written that photons are excited to different energy level is the same concept?
 
No, not really. Photons aren't excited to different energy levels, atomic electrons are. And you can be more detailed. Remember that UV photons are more energetic than visible photons, which is why they cause damage. So you've got a powerful UV photon coming into penetrate deep into your skin tissue. But you've protected yourself with this cream that somehow softens the blow. Let's think about how that could happen.

The UV photon slams into you and gets absorbed by an atom in the sunscreen. When this happens, an electron gets excited from the ground state to a high energy level (skipping some intermediate energy levels).

Can you briefly describe a deexcitation process in which the emitted photons are less energetic than the initial UV photon?
 
Thanks for the great explanation. It was very clear. Erm, I didn't quite understand the question, but I think you wanted me to explain the process when electrons fall back levels.

When the electrons de-excite, they emit photons corresponding to the change in energy levels. These photons are by definition traveling at 3.00 \times 10^8 \ \mathrm{ms^{-1}} and release light.

Is that correct? :confused:
 
Air said:
When the electrons de-excite, they emit photons corresponding to the change in energy levels.

Right, and if the UV photon promotes an atomic electron up several energy levels, can you think of a way that it could come back down by (perhaps more than one, hint hint) photons that are all less energetic than UV photons?
 
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