How does the electron come back down in energy level after absorbing a photon?

[jjschwartz1]

A common explanation for the origin of (some) color is that a material absorbs a, say, red photon and an electron moves from one energy level to a higher energy level, the difference in energy being the energy of the red photon. The material then appears bluish having absorbed the red light.

What happens to that excited electron? If it just falls down to the original energy level it will emit red light and the material won't have color. There must be other processes.

Thank you in advance for any insights.

 

[Vanadium 50]

Where did you get that explanation? It is certainly not the case for solids.

 

[jjschwartz1]

Among other more valid references, this youtube video produced by the American Chemical Society, beginning at 1:07,

 

[PeroK]

A common explanation for the origin of (some) color is that a material absorbs a, say, red photon and an electron moves from one energy level to a higher energy level, the difference in energy being the energy of the red photon. The material then appears bluish having absorbed the red light.

What happens to that excited electron? If it just falls down to the original energy level it will emit red light and the material won't have color. There must be other processes.

Thank you in advance for any insights.

I suspect the full answer to this question may be quite complicated and that different types of objects have colour for different reasons. One key point to note is this:

The light from the Sun or a lamp is largely in the visible (and infrared) spectrum. This determines the spectrum incident upon an object. Some of that spectrum may be absorbed. Note that visible light is relatively energetic, so the absorbtion will excite a molecule from the ground state to a relatively high excited state. Typically, a molecule has a large number of possible excited energy levels.

When the molecule reemits the light, however, it may do this in several stages, resulting in several emissions each of a lower energy (longer wavelength) than the original incident light. The object may emit light generally in the invisible, infrared part of the spectrum. It's possible that statistically very little light is re-emitted in the visible spectrum.

This would tie in with the thermodynamic picture of the incident light heating the body, as the light is absorbed, then the energy is radiated according to the blackbody spectrum, which would be largely invisible at normal temperatures.

 

[Vanadium 50]

Solids don't have orbitals in the way you are imagining. They have energy bands. The energy to move an electron from band to band varies with the size of the band. Further, electrons can gain or lose energy as they travel in a band.

 

[snorkack]

A common explanation for the origin of (some) color is that a material absorbs a, say, red photon and an electron moves from one energy level to a higher energy level, the difference in energy being the energy of the red photon. The material then appears bluish having absorbed the red light.

What happens to that excited electron? If it just falls down to the original energy level it will emit red light and the material won't have color. There must be other processes.

If the deexcitation takes place by emitting same frequency light (in the case, red), then the photons would be spontaneously emitted in random directions. Absorption and reemission would thus work as scattering, and the substance would be blue in transmitted light, but red in scattered light, including backscattered light.
A good example of a substance which is similar colour as gas and as liquid is bromine. Simple homoatomic diatomic gas too. Does a Br₂ molecule in gas have orbitals?
How much do the spectra of gaseous, liquid and solid bromine differ? And does liquid bromine have bands instead of orbitals?
For water, there is a nice comparison of spectra:
https://en.wikipedia.org/wiki/File:Water_infrared_absorption_coefficient_large.gif
but that´s somewhat more complex, infrared absorption, compared to a simple diatomic molecule.

 

[sophiecentaur]

Does a Br2 molecule in gas have orbitals?

Bromine is one of the few gasses with very visible colour. Could that be because it is dense and has high van der Waal's forces - meaning it is more likely to absorb light? That would tie in with the fact that solids and liquids tend to exhibit colour, where most gases do not.
There is a great temptation to discuss what goes on in solids and liquids by referring to a simple Hydrogen Atom model; in fact it's nothing like that because there are many more possible energy levels involved (bands not lines)

 

[DrClaude]

The origin of the color of the halogens stems from the excitation between the highest occupied π* molecular orbital and the lowest unoccupied σ* molecular orbital. The energy gap between the HOMO and LUMO decreases according to F2 > Cl2 > Br2 > I2. The amount of energy required for excitation depends upon the size of the atom. Fluorine is the smallest element in the group and the force of attraction between the nucleus and the outer electrons is very large. As a result, it requires a large excitation energy and absorbs violet light (high energy) and so appears pale yellow. On the other hand, iodine needs significantly less excitation energy and absorbs yellow light of low energy. Thus it appears dark violet. Using similar arguments, it is possible to explain the greenish yellow color of chlorine and the reddish brown color of bromine.

https://chem.libretexts.org/Bookshe.../8.13.01:_Physical_Properties_of_the_Halogens

 

[snorkack]

Bromine is one of the few gasses with very visible colour.
There is a great temptation to discuss what goes on in solids and liquids by referring to a simple Hydrogen Atom model; in fact it's nothing like that because there are many more possible energy levels involved (bands not lines)

And solids and liquids are too complex compared to hydrogen atom.
Halogens are a good next step to look at - just two atoms and a symmetry because the atoms are identical. Just a few extra degrees of freedom compared to hydrogen atom.
One obvious degree of freedom to look at is the internuclear distance.
Something I recalled in pondering it is Franck-Condon principle.
Movement of nuclei is slow compared to excitation of electrons.
Therefore, when molecular orbitals are excited, they are often excited to vibrationally excited states of the electronically excited state of the system.
I said "system" because I am not sure whether to call it "molecule". When a halogen molecule has one electron excited to antibonding orbital, is it excited to a bound (and therefore quantized) state of the molecule, or an unbound continuum state of two dissociated atoms propelled away from each other by repulsion of their opposite wavefunction orbitals?

 

[sophiecentaur]

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/08:_Chemistry_of_the_Main_Group_Elements/8.13:_The_Halogens/8.13.01:_Physical_Properties_of_the_Halogens

What about the attenuation of the light with those strong colours? Could there be something to do with the probability of photon/electron interaction (i.e. do other gasses just absorb EM of other frequencies just as much?)
PS or is it to do with the fact that the absorption is broader band so more electrons are stimulated by photons of a range of frequencies? That could make sense to me.

 

[sophiecentaur]

Solids don't have orbitals in the way you are imagining.

That's a popular misconception.

 

[Vanadium 50]

Can you suggest a better one-sentence explanation?

 

[sophiecentaur]

That's a popular misconception.

The word "orbital" is hardly accurate to describe the way an electron 'moves' anywhere but in a simple molecule. What would it 'go round' in the middle of a block of plastic or a crystal? Or through a piece of copper.

 

[snorkack]

The word "orbital" is hardly accurate to describe the way an electron 'moves' anywhere but in a simple molecule. What would it 'go round' in the middle of a block of plastic or a crystal? Or through a piece of copper.

But you seem to agree with Vanadium 50 on the issue.
When Vanadium 50 captures an electron and turns into Titanium 50 in a block of solid metal vanadium, is the electron taken from 1s (or even 2s or 3s) orbital of the specific Vanadium 50 atom? Or is it taken from a conduction band of metal vanadium?
When Vanadium 50 captures an electron and turns into Titanium 50 in a molecule of VCl₄ in frozen VCl₄ (melts at -25 degrees), is the electron taken from 1s (or even 2s or 3s) orbital of the specific Vanadium 50 atom? Or is it taken from a molecular orbital, shared with the 4 Cl atoms but not the other VCl₄ molecules, bound by dispersion forces alone? Or from a band of the solid (and nonconductive) VCl₄?
VCl₄ incidentally can also be excited by visible light - it is bright red. It can also be observed as a gas, because it boils at 148 degrees.

 

[sophiecentaur]

But you seem to agree with

There is no "but" involved. I agree with his (@Vanadium 50 's) and your ideas but I question why the term "orbital" is ever used to describe anything but the very basics of the Hydrogen Atom - particularly when QM stresses that the shapes that you see in the associated graphics are probability densities.
Having learned my Physics without an A Level Chemistry past I was just confused by the term Orbital when nothing goes 'around' anywhere. I'd be surprised if no one else would have been helped by the use of a better term.
Too late to change now, though. (The term AND me!)

 

[DaTario]

When the molecule reemits the light, however, it may do this in several stages, resulting in several emissions each of a lower energy (longer wavelength) than the original incident light. The object may emit light generally in the invisible, infrared part of the spectrum. It's possible that statistically very little light is re-emitted in the visible spectrum.

A process that is possibly important in explaining the phenomenon of color by subtractive synthesis is the fact that once excited, the electron decays and emits a photon in a superposition of momentum states (Weisskopf-Wigner theory for spontaneous decay). In practice, this means that the red photon is not generally directed toward the same retina that collected the other incident and reflected photons. In many cases, this photon must head toward the sample of matter itself, giving rise to reabsorption or thermal processes.

 

[snorkack]

A process that is possibly important in explaining the phenomenon of color by subtractive synthesis is the fact that once excited, the electron decays and emits a photon in a superposition of momentum states (Weisskopf-Wigner theory for spontaneous decay). In practice, this means that the red photon is not generally directed toward the same retina that collected the other incident and reflected photons. In many cases, this photon must head toward the sample of matter itself, giving rise to reabsorption or thermal processes.

If the only effect is reemitting the photon in a random and new direction, at the same frequency, over time small relative to reaction time and with 100% efficiency, it looks like scattering. And selective scatterer of e. g. red light should be red in reflected light, blue in transmitted light.