How Do Suncreams Absorb UV Radiation at the Atomic Level?

  • Thread starter Thread starter Air
  • Start date Start date
  • Tags Tags
    Radiation Uv
AI Thread Summary
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.
Air
Messages
202
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top