Interactions between e-m radiation and matter.

In summary: If there is dispersion, then the photon will be absorbed and the material will become opaque.Hope that helps.
  • #1
giuliopepe
3
0
Hi,
I'm an undergraduate physics student and I'd like to know more about interactions of light with matter.

Since I haven't studied Quantum Physics yet, I'd like to know what happen on a microscopic scale, when e-m radiation passes through a material (transmission), when it's reflected/scattered and when it's absorbed.

For example, glass is in general transparent to visible wavelengths, but it's opaque to ultraviolet or infrared. Does that mean that visible wavelengths, unlike ultraviolet and infrared, are not in the specific range to excite electrons and bring them to another orbital?
Hope this picture helps:
http://img135.imageshack.us/img135/4940/unexcited.jpg
This is just what I thought can happen.
Furthermore, is there any relation between polarization and transmission?

About reflection, I supposed e-m is absorbed by atoms in the material and then are re-emitted at the same angle of incidence. Then why not all materials reflect radiation with the same angle of incidence (scattering)?

About absorption, why a material should absorb radiation and transform it in internal energy instead of reflecting it?


I'm sorry for the poor language, but I've done the best I could do, since English is my second language.
I thank so much anyone that try to clarify what's going on when radiation encounters matter. :smile:
 
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  • #2
I must say that in solids such as glass, it is not individual atoms which participate in the photon interactions, but rather the solid as a whole. The individual levels of the atoms will "smear" when they are put close together and form a band.
 
  • #3
For example, glass is in general transparent to visible wavelengths, but it's opaque to ultraviolet or infrared. Does that mean that visible wavelengths, unlike ultraviolet and infrared, are not in the specific range to excite electrons and bring them to another orbital?

Yes. But electronic excitations only account for the part of the spectrum.

Furthermore, is there any relation between polarization and transmission?

There are many many kinds of 'polarization', so it depends. If you mean the circular polarization of the incident light, then yes, that can matter depending on the structure of the material, a fact which is exploited in a kind of spectroscopy called circular dichroism.

About reflection, I supposed e-m is absorbed by atoms in the material and then are re-emitted at the same angle of incidence.

Reflection is purely a bulk property of materials and can't really be explained in simple terms of individual atoms.

About absorption, why a material should absorb radiation and transform it in internal energy instead of reflecting it?

I'll just assume you mean 're-emitting it' rather than reflecting it, as reflection is a more specific process.

Where does that absorbed energy go? Heat. What's heat? Atomic/molecular motion/vibrations. (and the equivalent radiation, usually in the IR range). That energy can be transmitted through simple collisions ('non-radiatively')

So that's essentially your answer right there: If the absorption of light corresponds to a change in motion, either directly (absorbing heat radiation) or indirectly (e.g. changing the electronic state such that the vibrational state changes - 'vibronic coupling' being an example) then the re-emitted photon (if any) will have less energy.

Actually there's a third option, which is that the light is absorbed in a photochemical reaction (which is essentially another, but stable, electronic state). But that's a lot less common.
 
  • #4
It would be useful to examine why sky is blue, because the phenomenon is based on excitation and re-emission by individual molecules, with nearly complete polarization at 90 degrees, and with very little absorption of the incident sunlight. Here is a very brief discussion:
http://en.wikipedia.org/wiki/Rayleigh_scattering
 
  • #5
alxm said:
I'll just assume you mean 're-emitting it' rather than reflecting it, as reflection is a more specific process.

That's what I meant, sorry for that!

However, summing up, I've understood that to understand reactions between radiation and solid matter is very important to consider the "crystal lattice" formed by atoms in a solid and not atoms singularly, while in low pressure gases they can be considered singularly.

That's why is complicated to explain perfect reflection and why each material has its own optical properties.
For example diamond and graphite are both made of carbon atoms, but have different structure and indeed they have completely different properties.

Anyway can anybody suggest me in general what happens when visible photon passes through a transparent material (solid or liquid)? What are the conditions that permit the photon not to be absorbed? Or I need some more "quantum knowledge" to understand? :)

Thanks to all for the replies.
 
  • #6
A visible photon can pass through a "truly transparent material" ONLY IF there is NO dispersion. Review the Kramers Kronig relations (alias the dispersion relations) for proof.
 

FAQ: Interactions between e-m radiation and matter.

What is the definition of "Interactions between e-m radiation and matter"?

Interactions between e-m radiation and matter refer to the ways in which electromagnetic (e-m) radiation, such as light or radio waves, interacts with matter, such as atoms or molecules. This includes processes like absorption, scattering, and emission.

How does e-m radiation interact with matter?

The interaction between e-m radiation and matter depends on the properties of both the radiation and the matter. For example, e-m radiation can be absorbed by matter, causing the atoms or molecules to gain energy and potentially emit light of a different wavelength. It can also be scattered, changing the direction of the radiation, or transmitted, passing through the matter without being absorbed or scattered.

What factors influence the interactions between e-m radiation and matter?

The strength and type of interaction between e-m radiation and matter can be influenced by factors such as the frequency or wavelength of the radiation, the energy levels of the atoms or molecules in the matter, and the distance between the radiation and the matter.

What are some real-world examples of interactions between e-m radiation and matter?

Interactions between e-m radiation and matter are present in many everyday phenomena. For example, sunlight interacting with the Earth's atmosphere causes the sky to appear blue. Microwave radiation interacting with food in a microwave oven causes the food to heat up. X-rays interacting with body tissues allow for medical imaging. These are all examples of different types of e-m radiation interacting with different types of matter.

Why is it important to study interactions between e-m radiation and matter?

Understanding how e-m radiation interacts with matter is crucial in many fields, including physics, chemistry, and biology. It allows us to explain and predict various natural phenomena, develop technology such as lasers and solar cells, and use e-m radiation for various applications like medical imaging and communication. Additionally, studying these interactions can help us gain a deeper understanding of the fundamental laws of nature.

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