Colorless Compounds and electromagnetic radiation

In summary, the molecule is transparent because it has a gap in energy between the HOMO and LUMO levels.
  • #1
pisluca99
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I was trying to understand why some compounds appear colorless (transparent) and tried to give an explanation.

I take benzene as an example: it is a chromophore group in which there is π-conjugation, so a certain energy gap is generated between HOMO and LUMO. This energy gap is such that in order for an electronic transition to take place, the molecule goes to forcibly absorb electromagnetic radiation that falls into the UV. As a result, the molecule will go on to reflect other electromagnetic radiation, again from the UV, which does not allow the same electronic transition to take place. This reflected radiation reaches our eye, but we are unable to perceive it.

Contextually, the molecule is not able to interact with visible light, that is, it is neither absorbed nor reflected, so it is transmitted, which allows us to observe "through" the solvent.

Can this be considered roughly correct reasoning?
 
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  • #2
pisluca99 said:
Can this be considered roughly correct reasoning?
We live in a universe of incompatible interfaces. We see the reflection from impedance mismatches, differences in refractive index, at the surface contact with air. The colour of a surface is due to variations in the RI with wavelength.

If the material is transparent, we do not see detail inside the object, because there are no changes in RI at the wavelengths we can see.
 
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  • #3
pisluca99 said:
As a result, the molecule will go on to reflect other electromagnetic radiation, again from the UV, which does not allow the same electronic transition to take place.
Basically I believe you are correct. In order to be visible at all the object must interact with the light (in a quantum way).
I would take issue with this characterization. If there is no appropriate state that to receive the energy there will be no scattering whatsoever. Reflection, in my vernacular, implies an elastic scattering which commonly involves absorption and re-emission often with a change in direction.
 
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  • #4
hutchphd said:
Basically I believe you are correct. In order to be visible at all the object must interact with the light (in a quantum way).
I would take issue with this characterization. If there is no appropriate state that to receive the energy there will be no scattering whatsoever. Reflection, in my vernacular, implies an elastic scattering which commonly involves absorption and re-emission often with a change in direction.
Reflection/refraction and scattering are different processes.
Consider liquid helium. Its absorption spectrum is nearly perfect blank bluewards of 60 nm where electronic excitations of He atom lie. In principle you could excite the translations of He atoms relative to one another but because of the symmetry, it´s forbidden unless by three atom processes, and I could not find their cross section - looks like the absorption of liquid He stays low in the far-IR/microwave region where these should happen.

Yet liquid He interacts with light and is visible. Its polarizability is much lower than that of water but much higher than that of air, so it refracts and therefore retards light.
What is the process of retarding light? Note that it is not "absorption and reemission", it does not change direction, just phase.
 
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  • #5
hutchphd said:
If there is no appropriate state that to receive the energy there will be no scattering whatsoever.
This isn’t really correct. Unless you’re on resonance, all elastic (Rayleigh) and inelastic (Raman) scattering proceeds through a “virtual” state, which is really just a mathematical convenience and not a true eigenstate of any operator.

Reflection/refraction is a collective phenomenon, caused by the coherence of scattering from lots and lots of resonators. For instance, glass has a band gap that puts absorbance squarely in the UV, but specular reflection from a glass windowpane is easily observable from outdoors on a sunny day when the room inside is darkened.
 
  • #6
But in the end the process gets put on the energy shell. Whether you wish to consider the intermediate states merely a "mathematical convenience" is up to you.
Of course refraction/reflection from a surface is a collective phenomenon. It is difficult to make a surface without a collection of something. Typically the scattering then is treated using the collective excitations (phonons, rotons, plasmon......) of the surface (or bulk)
 
  • #7
Refraction does not need a surface.
Consider clean air in visible light. Dinitrogen and dioxygen absorb far UV but are clean of excitations redwards of 200 nm.
Gradual changes of air density, like from less dense hot air on a road surface to cooler and denser air above, or from denser air near Earth surface to less dense air high above, will not cause coherent reflection but will cause refraction. And Rayleigh scattering goes on.
 
  • #8
snorkack said:
Gradual changes of air density, like from less dense hot air on a road surface to cooler and denser air above, or from denser air near Earth surface to less dense air high above, will not cause coherent reflection but will cause refraction.
The eikonal paths which (semiclassically) describe the refraction can be considered a coherent reflection. We have reached a semantic impasse far outside the original scope of the question.
 
  • #9
hutchphd said:
But in the end the process gets put on the energy shell. Whether you wish to consider the intermediate states merely a "mathematical convenience" is up to you.
Ok. But then this quote:
hutchphd said:
If there is no appropriate state that to receive the energy there will be no scattering whatsoever.
becomes tautological. You’re basically saying nothing happens unless something happens.

The “state” that scatters light (off-resonance) is not a state of the molecule/material. It’s a transient state formed by the interaction of light with matter. It’s different from the states that impart color to chromophores, which are energy eigenstates of the molecule/material.
 
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  • #10
Any state of the system is one that includes both the atom and the electromagnetic field. For instance EM field with no excitation and atom in ground state. As soon as there is interaction (in Hamiltonian) these simple states are no longer eigenstates of energy.
A usual method of solution is essentially time dependent perturbation theory where there is a sum over all intermediate states while the interaction is supposed to be acting. If there are coupled bound states then these will be predominant and produce strong resonances. Absent resonances there is no single state responsible
TeethWhitener said:
The “state” that scatters light (off-resonance) is not a state of the molecule/material. It’s a transient state formed by the interaction of light with matter.
The scattering eigenstate state which contains photon(s) and atom is knowable in principle. It is presumed to be an eigenstate of the complete Hamiltonian. We just don't know it usually and must do perturbation theory in one form or other.

I presume chromophores have strong resonances but don't really know.
 
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  • #11
Note the Sellmeier equation:
n2(λ)=1+∑i Bi22-Ci)
C has dimension of area. This is actually confusing... because Ci is actually square of an absorption wavelength. So better write it:
n2(λ)=1+Σi Bi22i2)
Now, note that for λ>>λi, λ2i2→λ2, so (λ22i2)→1 and n2(λ)→1+Bi
For λ<<λi, λ2i2→-λi2 and (λ22i2)→-λ2i2

If a substance has no excited states then it cannot retard or refract light either, because then it has no Sellmeier coefficient.
When an excitation actually happens, it happens to a specific excited state.
But when a molecule has several Sellmeier coefficient and light is merely retarded, is it then retarded by one Sellmeier coefficient/excited state, or all of them simultaneously?
 
  • #12
Baluncore said:
We live in a universe of incompatible interfaces. We see the reflection from impedance mismatches, differences in refractive index, at the surface contact with air. The colour of a surface is due to variations in the RI with wavelength.

If the material is transparent, we do not see detail inside the object, because there are no changes in RI at the wavelengths we can see.
To the final sentence, I would add somewhere "and no absorption inside the object".
 
  • #13
pisluca99 said:
I was trying to understand why some compounds appear colorless (transparent) and tried to give an explanation.

I take benzene as an example: it is a chromophore group in which there is π-conjugation, so a certain energy gap is generated between HOMO and LUMO. This energy gap is such that in order for an electronic transition to take place, the molecule goes to forcibly absorb electromagnetic radiation that falls into the UV. As a result, the molecule will go on to reflect other electromagnetic radiation, again from the UV, which does not allow the same electronic transition to take place. This reflected radiation reaches our eye, but we are unable to perceive it.

Contextually, the molecule is not able to interact with visible light, that is, it is neither absorbed nor reflected, so it is transmitted, which allows us to observe "through" the solvent.

Can this be considered roughly correct reasoning?
Your sentence on visible light is basically correct (Why use the word contextually?). However, I do not understand why you are mentioning UV light and excited molecules. Visible light will transmit without any UV involvement.
 
  • #14
snorkack said:
Note the Sellmeier equation:
...

If a substance has no excited states then it cannot retard or refract light either, because then it has no Sellmeier coefficient.
When an excitation actually happens, it happens to a specific excited state.
But when a molecule has several Sellmeier coefficient and light is merely retarded, is it then retarded by one Sellmeier coefficient/excited state, or all of them simultaneously?
I think your understanding is little bit off. The Sellmeier equation is an empirical fit to the tails of resonances outside the measured spectrum. Every material has excited electronic states.
 
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  • #15
Dr_Nate said:
To the final sentence, I would add somewhere "and no absorption inside the object".
We cannot see the inside unless there is a reflection, which requires a change of RI or chemistry in the material.
 
  • #16
hutchphd said:
Basically I believe you are correct. In order to be visible at all the object must interact with the light (in a quantum way).
That is one of the objections to "invisible man". A man who manages to eliminate all absorption and scattering in his body AND match his index of refraction to that of air might indeed be invisible... while in air; he would be immediately conspicuous in water (as a bubble/hole in water!).

For visibility of detail, consider trying to mix syrup in a colourless transparent glass of transparent nearly colourless (its blueness is too faint at that size) water.

You don´t need "reflection" to see details:
You may see schlieren lines. These are interfaces between water and syrup, where index of refraction changes. These are NOT reflection. Transition between solutions is too gradual compared to wavelength. But the light is refracted anyway.
And you may see absorption, when the syrup was absorptive in visible light.

Whether you see the unmixed syrup by light absorption, light refraction or both, it is a detail you see, and not through reflection.
 
  • #17
This thread has become problematic because of its generality: the scattering of light by matter. We have the blind men and the elephant problem. Many of the comments take the form "it is not this it iis that".
For instance (paraphrased) "It is not scattering but the change in refractive index" is a false dichotomy. Collective scattering is fundamentally what causes the change in refractive index.
So please do not assume that an explanation involving say, the index of refraction, prcludes a more involved explanation involving virtual scatt6ering processes.
Such is the nature of semantic physics.
 
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  • #18
Dr_Nate said:
Your sentence on visible light is basically correct (Why use the word contextually?). However, I do not understand why you are mentioning UV light and excited molecules. Visible light will transmit without any UV involvement.
Essentially because I wanted to highlight the fact that the absence of color is due not only to the fact that benzene transmits all visible radiation, but also because our eye cannot perceive the UV radiation that is reflected (or transmitted ?? This is another doubt I would like to clarify) by benzene, considering that benzene has absorbance peaks in the UV.

That is, benzene transmits all visible radiation and at the same time transmits/reflects (give me clarification as to which phenomenon is preponderant as far as UV is concerned) some UV radiation, i.e., that which is not absorbed (lambda max), and it is these two factors that determine the absence of color.

If our eye could have perceived UV radiation then we would be talking about something else.
 
  • #19
pisluca99 said:
Essentially because I wanted to highlight the fact that the absence of color is due not only to the fact that benzene transmits all visible radiation, but also because our eye cannot perceive the UV radiation that is reflected (or transmitted ?? This is another doubt I would like to clarify) by benzene, considering that benzene has absorbance peaks in the UV.

That is, benzene transmits all visible radiation and at the same time transmits/reflects (give me clarification as to which phenomenon is preponderant as far as UV is concerned) some UV radiation, i.e., that which is not absorbed (lambda max), and it is these two factors that determine the absence of color.
Benzene also reflects some visible radiation.
The reflectivity of a flat surface of a homogenous dielectric, at normal incidence, is
ρ=((n-1)/(n+1))2
If you solve it for ρ=0,5, you will get almost n=6. Which is not a common index of refraction for clear substances.
Actually, benzene has index of refraction just 1,50 at 590 nm. Thus reflecting just 4% of yellow light and transmitting the rest.
Mind you, this is still a lot. Contrast benzene with hexane. Hexane´s index of refraction is just 1,375. Benzene can be distinguished by its visibly higher index of refraction. Which is caused by the absorption band in UV.
As you go from 590 nm towards blue and then approach the actual absorption band in UV, the index of refraction increases due to dispersion. But how high does it get before the absorption gets heavy?
 
  • #20
snorkack said:
Benzene also reflects some visible radiation.
The reflectivity of a flat surface of a homogenous dielectric, at normal incidence, is
ρ=((n-1)/(n+1))2
If you solve it for ρ=0,5, you will get almost n=6. Which is not a common index of refraction for clear substances.
Actually, benzene has index of refraction just 1,50 at 590 nm. Thus reflecting just 4% of yellow light and transmitting the rest.
Mind you, this is still a lot. Contrast benzene with hexane. Hexane´s index of refraction is just 1,375. Benzene can be distinguished by its visibly higher index of refraction. Which is caused by the absorption band in UV.
As you go from 590 nm towards blue and then approach the actual absorption band in UV, the index of refraction increases due to dispersion. But how high does it get before the absorption gets heavy?
So essentially, every wavelength is absorbed, reflected and transmitted to some extent. For example, in the case of benzene, it does not absorb at all (or at least extremely little) between 400 and 700 nm, so all this radiation is largely transmitted and to a smaller extent reflected. Since transmission is prevalent it appears transparent.

Of course, this also applies to UV radiation: each is absorbed to some extent, as well as reflected and transmitted. Even at an absorbance peak there is still some degree of reflection and transmission. So this is not an 'all or nothing' kind of discussion.

This explains, for example, why ethanol is an excellent solvent in spectrophotometry: it transmits most of the UV-vis radiation and absorbs and reflects only a very small percentage of every wavelenght.

Hope this Is correct.
 

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