Why do photons emit light?

[raid517]

OK I'm going to try to put this question the best way I can. But as far as I understand it, an atom will emit a photon when a photon of just the right energy is absorbed, thereby raising an electron from a lower energy state to a higher energy state. More specifically, the photon is emitted when the electron drops down an energy level from it's new excited state, to a lower energy level. From say n=4 to n=2 for example. But my question is what causes the electron to drop an energy level and thereby emit the photon? An example of this might be hydrogen whereby the electron dropping from the n=4 to the n=2 level results is a blue spectral line. However if just the right frequency of blue light is being shone on a sample to allow for an emission of exactly that frequency of light from the hydrogen sample, then why doesn't the electron stay in a continually 'excited' state?

 

What I mean is if work is continually being done on the electron as photons strike it, then why doesn't it just stay continuously in the n=4 energy state? Why would it drop down to a lower energy state an emit a photon of the same frequency as the source of light being directed at it?

 

The only (probably crude) explanations I can think of is that clearly no system is capable of continuously absorbing energy, without at some point releasing some of that energy. Therefore work done on a system, must equate to work produced by that system at some point or other. The only analogy I can think of is perhaps a sealed copper boiler containing water. If this boiler is continuously heated, this will equate to work done, but as the boiler is sealed there is no way for the energy supplied to escape. Therefore the energy of the boiler increases, until inevitably eventually there is an explosion and the energy of the system is transferred by means of work done in a variety of ways.

 

Therefore the electron cannot continue indefinitely to absorb photons, because to do so would result in it becoming increasingly unstable and would presumably cause it to break free of the atom at some point or other.

 

I'm not sure this explains everything though. Just because it would become unstable and would eventually break free of the atom, still doesn't explain (to me) why it may drop from one energy level to another.

 

Therefore the second (again probably erroneous) assumption I can make from this is that in some sense light isn't continuous, but instead comes in packets, rather like a series of tennis balls being thrown at a target, thereby giving the electron sufficient 'time' to absorb the photon, drop from one energy level to another and thereby emit a photon, before being struck by another photon (which is then absorbed), thereby raising it to a higher energy level again - and so on. If this was true (which it probably isn't), it would seem like a very simple mechanical process. So how close, or far off am I?

 

No doubt someone will 'hand me my ass' for trying to work out what's going on here though lol. I just don't feel I have a clear enough grasp of why electrons change energy levels and emit photons to be confident about this.

 

All input would be welcomed.

[swansont]

Systems tend to their lowest potential energy. Classically speaking, an energy gradient means there is a force which pushes in the direction which allows the most efficient reduction of potential energy. An atom just has quantized energy levels. Another concept is that any interaction that is not forbidden is mandatory. Dropping in energy and the creation of a photon are both allowed, so it will happen.

[raid517]

I'm not sure if this is still really explaining it. I think we are (roughly!) on the same page that an electron will tend to emit a photon and not contentiously store energy from succeeding photons, because there is 'enough time' between photons striking it for it to lose energy (although maybe this is too simplistic?), but the part where it gets a little fuzzy for me is why it emits this energy in the form of a photon and not in some other way? However I also suspect that the answer to this might be so complex that it's beyond me right now lol.

 

Just saying because something can happen it will happen is interesting, but maybe a little too philosophical for me at this point.

 

Systems tend to their lowest potential energy. Classically speaking, an energy gradient means there is a force which pushes in the direction which allows the most efficient reduction of potential energy.

 

So what is this force in terms of the atom (say hydrogen) and it's electron? If it was a ball the force pushing (or in this case pulling) the ball towards it's state of lowest potential energy would of course be gravity.

Edited August 15, 2011 by raid517

[mississippichem]

The force is from the charge of the nucleus. Lower energy orbitals are closer to the nucleus.

[raid517]

The force is from the charge of the nucleus. Lower energy orbitals are closer to the nucleus.

 

I'm still trying to visualise the full actual sequence in my head. The photon strikes the electron, the electron becomes excited and is raised to a higher energy level. A 'period of time' elapses (given that this will be extremely short), which is sufficient for the electron to lose energy. The electron (now having lost energy), drops to a lower energy state, as the force supplied by the photon to raise it to it's excited state is gradually overcome by the force of attraction of the nucleus, and finally the electron releases a photon.

 

A number of questions emerge from this. Firstly is this an accurate model of the process? Secondly if so is the photon that's emitted always exactly the same energy and wavelength of the photon that was absorbed? Thirdly, if so is the photon that was emitted the 'same' photon that was absorbed? Or for example is the photon that is emitted simply a result of energy transfer, where the kinetic energy of the excited photon is gradually lost and transferred to light energy, thereby causing a photon to be emitted? If so where did this 'new' photon come from? If it is the same photon that was absorbed, then why is the electron able to hold on to it in it's excited state and why does it release it in it's non-excited state? Is there some kind of force of attraction between a photon and an electron in it's excited state? Also in relation to the third point, are photons ever destroyed, or are they simply absorbed and re-emitted later?

 

Sorry for all the questions, it just seems like an interesting topic and it would probably be useful if I could get a firmer grasp of it.

[Greg Boyles]

OK I'm going to try to put this question the best way I can. But as far as I understand it, an atom will emit a photon when a photon of just the right energy is absorbed, thereby raising an electron from a lower energy state to a higher energy state. More specifically, the photon is emitted when the electron drops down an energy level from it's new excited state, to a lower energy level. From say n=4 to n=2 for example. But my question is what causes the electron to drop an energy level and thereby emit the photon? An example of this might be hydrogen whereby the electron dropping from the n=4 to the n=2 level results is a blue spectral line. However if just the right frequency of blue light is being shone on a sample to allow for an emission of exactly that frequency of light from the hydrogen sample, then why doesn't the electron stay in a continually 'excited' state?

 

What I mean is if work is continually being done on the electron as photons strike it, then why doesn't it just stay continuously in the n=4 energy state? Why would it drop down to a lower energy state an emit a photon of the same frequency as the source of light being directed at it?

 

The only (probably crude) explanations I can think of is that clearly no system is capable of continuously absorbing energy, without at some point releasing some of that energy. Therefore work done on a system, must equate to work produced by that system at some point or other. The only analogy I can think of is perhaps a sealed copper boiler containing water. If this boiler is continuously heated, this will equate to work done, but as the boiler is sealed there is no way for the energy supplied to escape. Therefore the energy of the boiler increases, until inevitably eventually there is an explosion and the energy of the system is transferred by means of work done in a variety of ways.

 

Therefore the electron cannot continue indefinitely to absorb photons, because to do so would result in it becoming increasingly unstable and would presumably cause it to break free of the atom at some point or other.

 

I'm not sure this explains everything though. Just because it would become unstable and would eventually break free of the atom, still doesn't explain (to me) why it may drop from one energy level to another.

 

Therefore the second (again probably erroneous) assumption I can make from this is that in some sense light isn't continuous, but instead comes in packets, rather like a series of tennis balls being thrown at a target, thereby giving the electron sufficient 'time' to absorb the photon, drop from one energy level to another and thereby emit a photon, before being struck by another photon (which is then absorbed), thereby raising it to a higher energy level again - and so on. If this was true (which it probably isn't), it would seem like a very simple mechanical process. So how close, or far off am I?

 

No doubt someone will 'hand me my ass' for trying to work out what's going on here though lol. I just don't feel I have a clear enough grasp of why electrons change energy levels and emit photons to be confident about this.

 

All input would be welcomed.

 

Surely not every photon in the light you shine at the collection of atoms will strike the right electron and not every electron in an excited state will receive another photon at the right time to keep it in that excited state. So you always have a mixture of electrons entering an excited state and droping back from an excited state.

[raid517]

Surely not every photon in the light you shine at the collection of atoms will strike the right electron and not every electron in an excited state will receive another photon at the right time to keep it in that excited state. So you always have a mixture of electrons entering an excited state and droping back from an excited state.

 

Yes that's rather what I said previously. But the rest of the questions are still interesting to me if anyone has an answer?

[swansont]

I'm not sure if this is still really explaining it. I think we are (roughly!) on the same page that an electron will tend to emit a photon and not contentiously store energy from succeeding photons, because there is 'enough time' between photons striking it for it to lose energy (although maybe this is too simplistic?), but the part where it gets a little fuzzy for me is why it emits this energy in the form of a photon and not in some other way? However I also suspect that the answer to this might be so complex that it's beyond me right now lol.

The interaction is electromagnetic, so photons are one way. You can also have this "relaxation" occur through nonradiative means in a molecule, e.g. exciting vibrational states, but in an isolated atom, there's really no other option.

 

I'm still trying to visualise the full actual sequence in my head. The photon strikes the electron, the electron becomes excited and is raised to a higher energy level. A 'period of time' elapses (given that this will be extremely short), which is sufficient for the electron to lose energy. The electron (now having lost energy), drops to a lower energy state, as the force supplied by the photon to raise it to it's excited state is gradually overcome by the force of attraction of the nucleus, and finally the electron releases a photon.

 

A number of questions emerge from this. Firstly is this an accurate model of the process? Secondly if so is the photon that's emitted always exactly the same energy and wavelength of the photon that was absorbed? Thirdly, if so is the photon that was emitted the 'same' photon that was absorbed? Or for example is the photon that is emitted simply a result of energy transfer, where the kinetic energy of the excited photon is gradually lost and transferred to light energy, thereby causing a photon to be emitted? If so where did this 'new' photon come from? If it is the same photon that was absorbed, then why is the electron able to hold on to it in it's excited state and why does it release it in it's non-excited state? Is there some kind of force of attraction between a photon and an electron in it's excited state? Also in relation to the third point, are photons ever destroyed, or are they simply absorbed and re-emitted later?

 

Sorry for all the questions, it just seems like an interesting topic and it would probably be useful if I could get a firmer grasp of it.

 

The loss of energy happens discretely — there's no "gradual overcoming"of the force.

 

Same energy to within the lineshape of the transition (dictated by the Heisenberg uncertainty principle)

 

You can't tell one photon from another (of the same energy) but the photon itself is not "stored" so it's not the same photon. Photon number is not a conserved quantity — photons are created and destroyed.

[raid517]

The interaction is electromagnetic, so photons are one way. You can also have this "relaxation" occur through nonradiative means in a molecule, e.g. exciting vibrational states, but in an isolated atom, there's really no other option.

 

 

 

The loss of energy happens discretely — there's no "gradual overcoming"of the force.

 

Same energy to within the lineshape of the transition (dictated by the Heisenberg uncertainty principle)

 

You can't tell one photon from another (of the same energy) but the photon itself is not "stored" so it's not the same photon. Photon number is not a conserved quantity — photons are created and destroyed.

 

This is all very vague. What does 'discretely' mean in this context? Also what does 'you can't tell one photon from another' mean? Do you mean that the energy of the photon transmitted will be the same as the energy of the photon that is absorbed, so you can't tell the difference between them? The last part too is a little unclear, how are photons created and how are they destroyed?

 

I think it's very difficult to lean on an internet forum, lol.

[swansont]

This is all very vague. What does 'discretely' mean in this context? Also what does 'you can't tell one photon from another' mean? Do you mean that the energy of the photon transmitted will be the same as the energy of the photon that is absorbed, so you can't tell the difference between them? The last part too is a little unclear, how are photons created and how are they destroyed?

 

I think it's very difficult to lean on an internet forum, lol.

Me, too. This is a place to fill in a few gaps. It's not a replacement for several semesters of physics.

 

Discrete means not continuous — the de-excitation is not a gradual process.

 

If by "how are photons created and how are they destroyed?" you want to know the details of a mechanism, I don't know. I don't even know that it's possible to know that. It's not part of any model we can test.

[raid517]

Me, too. This is a place to fill in a few gaps. It's not a replacement for several semesters of physics.

 

Discrete means not continuous — the de-excitation is not a gradual process.

 

If by "how are photons created and how are they destroyed?" you want to know the details of a mechanism, I don't know. I don't even know that it's possible to know that. It's not part of any model we can test.

 

I guess we are going around in circles a little here, lol. If the de-excitation of electrons is truly instantaneous, then clearly (or seemingly), there would be no effectively measurable time period over which this would occur? It would either be in a high energy state, or a lower energy state and nothing in between. Actually this part kind of makes sense given that the principal quantum numbers of the electron shells can only have 'discrete' values. So one would not expect values in between n=4, n=3, n=2 and n=1 and so on. But the question still remains (for me) if the energies for the electron shells n=4, n=3, n=2 and n=1 etc are always specifically the same? That is (for example) can an electron lose (or convert) any of it's energy at all while it occupies the n=4 energy shell, before it drops to n=3, or n=2 etc? Or is the amount of energy the electron has while it occupies these shells always the same?

 

If it cannot lose energy before dropping to subsequent shells, then what causes it to drop between one energy shell and the next, and for it to release a photon? I already accept that the attraction of the nucleus can pull the electron towards a lower state of potential energy, but surely if the photon was powerful enough to cause it to enter an excited state, either it must remain in that excited state, or lose some energy first before it can drop to a lower energy shell?

 

I wonder if the answer to this question might simply be that it does lose energy - in the form of a photon, lol.

 

Also it was you I believe who said that photons are both created and destroyed, but then when I ask how they are created and destroyed, you say that you know of no mechanism for this and that it cannot be tested. Very confusing...

 

I do know that photons can last for a very long time, in some cases (in terms of the cosmic background radiation), for as long as the Universe has existed. So they may not always be destroyed. But surely if some mechanism exists for their creation and destruction, we should at least be capable of knowing what it is? Just saying that a photon is released when an electron drops from one energy state to another and is absorbed when a photon of just the right energy excites an electron to enter a higher energy orbital, does not explain how a photon is absorbed, or emitted, it just says that it is.

 

I agree. I don't expect I'm going to get any vastly useful answers here. Small simple hints perhaps. It's just that as I read my textbooks a billion questions like this spring to mind, but perhaps I will have to simply shelve many of them until I have finished some of my reading.

[swansont]

It's not instantaneous, it's limited by the Heisenberg Uncertainty Principle [math]\Delta{E}\Delta{t} >\hbar[/math]

 

The mechanism is the electromagnetic interaction. If you want some more detailed, step-by-step explanation of what goes on, I don't have one.

[raid517]

It's not instantaneous, it's limited by the Heisenberg Uncertainty Principle [math]\Delta{E}\Delta{t} >\hbar[/math]

 

The mechanism is the electromagnetic interaction. If you want some more detailed, step-by-step explanation of what goes on, I don't have one.

 

"De-excitation is not a gradual process" yet "It's not instantaneous." Lol, I think I probably just need to keep reading my textbooks.

 

Thanks for trying anyway dude.

[spin-1/2-nuclei]

"De-excitation is not a gradual process" yet "It's not instantaneous." Lol, I think I probably just need to keep reading my textbooks.

 

Thanks for trying anyway dude.

 

Hello,

 

Sorry, this post is going to be long:

 

Regarding de-excitation..

the quick answer is - it depends on the environment. That is to say not all photos and electrons are created equal.

That being said:

 

Electrons which have "vibrational energy" - i.e. they can vibrate without colliding with other atoms undergo both excitation and de-excitation.

I've explained why excitation occurs in more detail in the previous post - but quickly - when an electron absorbs a photon excitation occurs. Once that happens the electron is now occupying a more UNSTABLE energy level. This means that the electron will not want to remain "there" - because this has a destablizing effect on the entire system to which it belongs - i.e. - all systems universally try to attain the lowest energy possible, that is to say - all systems are trying to achieve maximum stability. Moreover equilibrium isn't a process reserved just for chemical reactions, so not all pathways to the lowest energy "configuration" are created equal for all systems.

 

To go more in depth:

In thermal equilibrium the energy density ρ(ν) of photons with frequency ν is - at least in most examples - constant with respect to time - that is to say that the rate of absorption = the rate of emission.

 

Thus, in the Einstein model the rate of transition from a lower energy level to a higher energy level is proportional to the # of atoms with energy at that lower energy level & the amount of photons with that frequency - i.e. the energy density.

 

So, for de-excitation - i.e. emission, there are 2 options.

 

1. spontaneous emission

 

- in spontaneous emission the atom in the higher energy state decays to the lower energy state via the *spontaneous* emission of a photon.. The photon released had the energy equal to the difference in energy between the high energy and low energy levels, but in this case both the phase and the direction of the photon are random.

 

phase = in waves is the fraction of a wave cycle which has elapsed relative to an arbitrary point. [the relative to the arbitrary point will be important later]

 

To describe this, quantum mechanics has to be extended (because traditionally atomic levels are quantized but electromagnetic fields are not) So, we need what is called quantum field theory to explain all of this. In quantum field theory the electromagnetic field is quantized at all points in space.

 

I will get to this more later, but for now, you can think of it like this for a *qualitative view* - the decision to leave is made by the entire system if you will, because systems do whatever is best for the system as a whole (as a sum of whatever is best for all the sub-systems in the macrosystem) to attain maximum stability.

 

2. Stimulated emission - comes from interacting with an electromagnetic wave of the appropriate frequency, etc.

Photos emitted in this way will have the same phase, polarization, frequency, and direction when they are emitted as they did when they were absorbed.

 

Now there is no standard pathway to de-excitation for all electrons because in reality things do not exist in a vacuum...

 

Photons are qualitatively speaking - "balls" of radiant energy - and are basically the particle form of light. The amount of energy a photon has is calculated via E=hf, where f is the frequency of the light wave and h is 6.64x10-34j sec (Planck's constant). The energy of the photon that is released during de-excitation exactly matches the difference between the initial energy state of the electron and the electron energy states, i.e. no energy is lost.

 

Moreover, not all interactions are created equal. The relationship between the energy state of the photon and the fundamental energy states of the atom will greatly determine how many interactions take place. That is why for example, photons travel at different speeds through different mediums. Since no energy is lost or gained the photon exits with the frequency it entered with, however the travel time - i.e. the speed of the photon - has been altered. You can also look at this from the perspective of the atom - i.e. - orbital energies are unique for each atom of a different element.

 

sigh...

All of that being said:

 

In spontaneous emission the stationary state of the atom no longer meets the definition of a true eigenstate describing the co-system of the atom and the electromagnetic field. Most notably, when the electron moves from the excited state to the electronic ground state, these states mix with movement of the electromagnetic field from the ground state (the vacuum) to it's excited state (a single photon field state).

 

Thus, spontaneous emission in free space cannot occur without disturbances in the vacuum. Moreover, despite having only one electronic transition between the excited and ground state of the magnetic field mapping the process is not that straight forward because there are many mechanistic pathways the electromagnetic field can take when moving from the ground state to the excited state and back. Basically, with respect to the trajectories along which the photon can emitted the EM field has many many many more degrees of freedom than the atom - i.e. the EM's phase space is > the atoms.

 

Which is where the concept of electronic excitation and photonic excitation comes in to play - i.e. - the atom decays via spontaneous emission as a requirement of the system parameters. Which is basically saying that time spent in the excited state depends on the light source and the environment, as mentioned earlier in this insanely long post.

 

So, what does all this have to do with the uncertainty principle?

 

The uncertainty principle doesn't explain the how of spontaneous emission so much as the why it is difficult to empirically observe in the lab.

 

Basically, de Broglie suggested that - just like light - electrons and protons will have both particle and wave characteristics. That is to say that the behavior of electrons can be treated like wave functions. If we are looking at electrons in either their wave or particle form, the uncertainty principle basically states that the more we know about the current position of a particle in quantum mechanics the less we know about it's future momentum. In the case of wave functions: wave systems have non-zero amplitude - i.e. - their positions are unknown. You have to compress the waves to accurately obtain their position. The momentum is proportional the wavenumber of ONLY one, but, ANY one of the waves in the system/packet - i.e. - the more you know about the momentum the less you know about the position and the reverse is also true. So the uncertainty principle deals more with how we quantify quantum particles and less with how/why they actually behave the way they do. I like to think of the uncertainty principle as one of the bridges (albeit imperfect) connecting the actual reality of the universe with our observable reality.

 

Sorry, I didn't have as much time as I needed go into anymore detail.. my carbon is done so I have to return to the lab..

If you want more information on this I would suggest hitting up Sci-Finder and getting some review articles on photon emission.

Hopefully this helps,

Best of luck with your studies..

Cheers

Edited August 18, 2011 by spin-1/2-nuclei

[raid517]

I'm afraid it may take me the rest of my life to absorb what you just said. Nonetheless, I hope I get the opportunity to try. Thanks!