Thread: Why do elements emit photons?

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.

raid517, Once an electron goes from n=2 to n=4, it will no longer absorb the same wavelength of light because the energy levels above it are not the same difference in energy above it (i.e. the energy difference between n=2 and n=4 is different than the energy difference between n=6 and n=4). However, another process called stimulated emission can occur. If light of the energy difference between n=2 and n=4 strikes an electron in n=4 and there is an electron hole in n=2, then the electron can be stimulated back to n=2, with the emission of another photon. So if you turn up the intensity of light on a sample you start by converting n=2 electrons to n=4 electrons, but once you build up a population of n=4 electrons you also have an appreciable amount of stimulated emission. When there are the same number of electrons in the n=4 level and the n=2 level there is no absorbance, because the same number of photons are emitted as are absorbed.

Secondly, an electron doesn't stay in the excited state of n=4 if there is an electron hole at n=2 because the n=2 position is lower in energy (there is a more complicated explanation using transition dipole moments but I don't think you were looking for that). So if you turn off the light there are nonradiative processes that can occur (coupling of the energy difference between n=4 and n=2 with the environment), or the electron can move down to n=2 while emitting a photon as fluorescence.

Thirdly, all four of these processes are going on at the same time (absorbance, stimulated emission, fluorescence, and non-radiative decay) when you're shining light on a sample. The rates of absorbance and stimulated emission depend on the concentration of electrons in the upper and lower states, and the intensity of light. The rate of fluorescence and non-radiative decay depend only on the concentration of electrons in the higher energy state.

Is that what you were looking for?

I think it might be. It may take me a while to digest it though! Thanks.

Hello,

What kpkent is saying is:

Electrons exist in what is typically called the electron cloud - but quantum mechanically speaking that's not really what is going on - so the best way to look at it is simply to say electron exist at energy levels which are directly related to how far they are away from the nucleus.

Furthermore, electrons prefer the lowest energy state - i.e. the groudn state - but cannot always exist there if it is being occupied by other electrons (which is typically the case for at least some electrons in every atom). In the ground state the electron is a close to the nucleus as it can get. So moving to a different distance from the nucleus = moving to a different energy state, etc.

Electrons that are in the ground state don't randomly move to different energy states. In order for them to do this they must absorb some energy. This energy typically comes in the form of a photo "floating" by - i.e. the electron absorbs the photo and moves to a higher energy state.. Then via decay - and subsequent release of said photon, *moves* the electron back to the ground state...

If you think of a photon as energy and less like a tiny ball, this will make more sense... This is a result of the universal concept of all systems strive to exist in the lowest energy state possible, as in they try to become the most stable.

When an electron abosorbs a photon it's called: absorbtion
When an electron decays back to the ground state it is called: spontaneous emission

This doesn't happen regularly, because if it did that would be bad.. most things would be in an excited (i.e. a reactive state) so what actually happens on the ground is a process called stimulated emission... which is basically a fancy way of saying the system compensates to prevent continuous existence in an excited state, because the overall stability of the system can be looked at as an average of all the energy states from T=0 till the end of time.. thus the best way to achieve the lowest energy state for the majority of that time is to do something to ensure that excited states are avoided whenever possible.. thus..
In this process.. you get a 2 for 1 deal, wherein a photon crashes in to an atom in an excited state and kicks the excited electron back to the ground state - and then emits two *equivalent/identical* photons.

hopefully this was helpful
Cheers

I think you also asked about de-excitation somewhere else.. in case you check this place more often, I will post my answer here as well.

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

It becomes easier if you conceive the photon as energy instead of a particle.

As stated before, electrons don't spontaneously jump to higher energy states, but they spontaneously jump to lower energy states. Spontaneous phenomena in nature tend to go towards lowering the system's potential energy.

If an electron is at n = 1, as it receives a photon, if that photon has enough energy, it will be excited. That means he will occupy a higher energy state (for instance n = 2).
Now the electron is not on the lowest possible energy state, that means it will spontaneously reduce the system's energy. So, when the electron goes back to n = 1, that surplus energy has to go somewhere, since energy cannot be destroyed. That means a photon (energy) will be release with the exact amount of difference between n=1 and n=2.

Hope it helps!

Originally Posted by 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.

Planck's model of the atom was based, to a high degree, upon thermodynamics. He reasoned that electrons act much like molecules in a gas. In a closed system, the atoms interact and either gain energy or lose energy. Each atom has stable states, and the electrons move from one state to another, depending on the amount of received or lost energy. His model also covers the absorption or emission of a photon. White light (broad spectrum) from an external source imparts energy to the system. Similarly, s system can emit white light, and his formula fits measurements to a high degree. In his state model, the atom absorbs and electron and jumps to a higher energy level. When an atom loses energy, it moves to a lower energy level. So you are quite right in that light is not continuous, which is the concept of the photon. However, not many would call it a "simple mechanical process". I have studied Planck's lectures that he presented at an American university some 15 years after he revealed his quantum theory of stable states. It is based to a high degree on probability theory and steady states. This is one of the most impressive presentations that I have ever read. You may be surprised to find that he called atoms "oscillaors".