What Causes Some Substances to Absorb Light?

[davo789]

This seems like a really simple question, and perhaps it is, but I just can't work it out so I'm asking for some help!

Let's say that we have a bucket of black paint. It is black, and so it absorbs pretty much all the visible light it receives (except for some on the glossy surface that is reflected back at us). It is also paint, which is fairly inert, there are no electrons swimming freely around, and it's not in a nice regular crystal structure. I suppose there must be a dye in the paint to make it black, but what is it about this dye which means it can absorb all light? It can't be just via electron level changes, because that'd only select a few distinct frequencies of light, so how does it work?

My guess would be that rather than interacting with the dye and shifting electrons around, the photons might interact as you'd expect IR light to, and start the molecules vibrating. But if this is right, when do the photons decide do stop vibrating and start changing electron levels, or not interact at all?

Thanks in advance!

 

[Edi]

and what about black, isolating rubber? It doesn't have, by definition, the so called "free electrons", witch could be accelerated by the incoming photon and then it would release its energy in other particles by hitting them and exchanging kinetic energy..

 

[HallsofIvy]

Atoms absorb photons if they have electrons that have the correct energy levels to absorb the energy of the photon.
Different types of atoms will absorb photons of different energy levels (frequencies). Black paint has enough atoms of all different types to absorb all different frequencies.

 

[davo789]

Thanks for your reply HallsofIvy! I just find it really difficult to believe that objects are able to absorb a continuum of light by purely using the absorption lines available to them. Also, when the excited electron drops back down, it will usually re-emit light at a similar wavelength; black things get hot, not bright, so I guess I'm asking how object turn the visible light into IR light?

And Edi, yes; that's a similar problem!

 

[haael]

Atoms can exchange energy. Visible light has low entropy, IR light has high entropy. So, random energy transfers between atoms converts visible light into the black body spectrum of certain temperature.

 

[davo789]

Thanks for your reply haael. What I want to know is by which mechanism is the light originally absorbed; I am struggling to believe that it is purely by the photons raising electrons up in their configuration, and that there is no other mechanism involved.

 

[davo789]

Unless, of course, if the photons are allowed to do a kind of 'low energy' type Compton scatter. This would mean that the photons can collide with the electrons without having enough energy to boost them to the next level, and the recoil photon is lower in energy (i.e. IR). The major floor with this idea is that what happens to this electron, who now has an un-quantized amount of energy...

I'm just thinking out loud so apologies if I'm spewing all sorts of rubbish!

 

[chrisbaird]

Thanks for your reply HallsofIvy! I just find it really difficult to believe that objects are able to absorb a continuum of light by purely using the absorption lines available to them. Also, when the excited electron drops back down, it will usually re-emit light at a similar wavelength; black things get hot, not bright, so I guess I'm asking how object turn the visible light into IR light?

And Edi, yes; that's a similar problem!

I think the problem is that you are picturing one atom with only a handful of possible states, say N, and therefore only a handful of frequencies (N²)that it can absorb. But when you bring two identical atoms close to each other, their energy levels split (some go slightly up, and some go slightly down) so that there are now (2N)² frequencies that the two-atom object absorbs. A bucket of paint has something like 10^26 atoms and therefore can absorb up to (10^26 N)² different frequencies. If you were to try plot this many frequencies on a spectrum graph, you would end up with something that looks essentially continuous. And this is just the regular atomic number transitions, there are also a broad range of vibrational and rotational states between which the atoms can transition by absorbing light. This explanation is a little simplistic, but it gives you the idea.

 

[Andy Resnick]

Thanks for your reply HallsofIvy! I just find it really difficult to believe that objects are able to absorb a continuum of light by purely using the absorption lines available to them. Also, when the excited electron drops back down, it will usually re-emit light at a similar wavelength; black things get hot, not bright, so I guess I'm asking how object turn the visible light into IR light?

In a bulk material, the absorption spectrum is smeared out- there are no longer discrete absorption lines. Furthermore, the light is not "converted" from (say) visible to IR- the absorbed energy is dissipated among all the possible decay modes, which is also a continuum.

 

[davo789]

Those are the two answers I was looking for; thanks a lot Chris & Andy!

 

[DaveC426913]

And remember that the paint is nothing more than suspended pigments - molecules that are chosen because they absorb certain frequencies. In the case of black paint, they have added several different pigments, designed to ensure that all frequencies are absorbed.

 

[Edi]

ok.. so far so good, but - if the electrons absorb photons by rising their energy levels, they should drop back to their lower energy levels and re-emit a equal frequency photon. Basically that is reflection. .. how can electrons drop back without releasing visible photon, but by giving the energy yo internal kinetic energy, aka, heat.. ?
or do they re-emmit the photo, but in a different direction.. in witch case, there would be a little bit of redshift, when the photon is emited.. repeted this millions of times and .. it wuld explain my question very well - but could it be true?

 

[Andy Resnick]

ok.. so far so good, but - if the electrons absorb photons by rising their energy levels, they should drop back to their lower energy levels and re-emit a equal frequency photon. Basically that is reflection. .. how can electrons drop back without releasing visible photon, but by giving the energy yo internal kinetic energy, aka, heat.. ?
or do they re-emmit the photo, but in a different direction.. in witch case, there would be a little bit of redshift, when the photon is emited.. repeted this millions of times and .. it wuld explain my question very well - but could it be true?

What is the energy scale of a chemical bond? Pigments (and most other things) are not made of isolated atoms.

And reflection is the the absorption and emission of a photon- reflection is an elastic scattering process.

 

[chrisbaird]

ok.. so far so good, but - if the electrons absorb photons by rising their energy levels, they should drop back to their lower energy levels and re-emit a equal frequency photon. Basically that is reflection. .. how can electrons drop back without releasing visible photon, but by giving the energy yo internal kinetic energy, aka, heat.. ?
or do they re-emmit the photo, but in a different direction.. in witch case, there would be a little bit of redshift, when the photon is emited.. repeted this millions of times and .. it wuld explain my question very well - but could it be true?

As stated, the key is that the molecules are not isolated. The energized electron bumps into other molecules or electrons before it has a chance to release a photon and its energy is converted to the kinetic energy of the thing it bumped into. Because the bumping (scattering) is random, the distribution of kinetic energy gained is random, so that any interesting quantum information or coherent states are typically lost and the material just heats up classically. In summary, optical loss in a material is a function of non-optical scattering. If the material is a crystal, you can describe this process using phonons (lattice vibrations) and you would say that the electron absorbs a photon, then emits a phonon. For amorphous materials/mixtures, it gets more complicated.

The difficult design of lasers is in a sense the art of suppressing scattering so that the electron has a chance to emit a photon.

 

[Edi]

As stated, the key is that the molecules are not isolated. The energized electron bumps into other molecules or electrons before it has a chance to release a photon and its energy is converted to the kinetic energy of the thing it bumped into. Because the bumping (scattering) is random, the distribution of kinetic energy gained is random, so that any interesting quantum information or coherent states are typically lost and the material just heats up classically. In summary, optical loss in a material is a function of non-optical scattering. If the material is a crystal, you can describe this process using phonons (lattice vibrations) and you would say that the electron absorbs a photon, then emits a phonon. For amorphous materials/mixtures, it gets more complicated.

The difficult design of lasers is in a sense the art of suppressing scattering so that the electron has a chance to emit a photon.

oh, cool. Thank you. This is, actually, a very good explanation to my, not so good, and made in a hurry, description of my question..
you could say that the atom/ absorbing system sort of expands.. and pushes on the surrounding atoms/ molecules.. and the energy to make that push comes from the incoming photon.

 

[Claude Bile]

In short, because there is an allowed transition in the energy levels of the atoms/molecules/lattice that corresponds to the energy of the incoming photon.

The allowed energy levels are determined by the atoms that are present and how they bond with other atoms.

The bond bit is important because it explains why graphite looks different to diamond, even though they both consist of carbon atoms.

Claude.