# What makes light?

1. ### rogerk8

220
Hi!

I wonder what makes light when it comes to excited and/or ionized atoms.

I have tried to study this phenomena both by looking up ionization and recombination on Wikipedia but the answer is missing.

Let's say that an atom is hit by a high energetic particle.

Two things, I believe, then might happen:

1) The outer electron escapes from the atom and the atom is thus ionized.
2) The outer electron is pushed into one of perhaps several outer shells but still resides in the atom and is thus excited.

Now,

In the case of 1)
The only reason for emitting light is if the state of energy is higher before it finally recombines with an electron.

In the case of 2)
Emission of light is made when the atom simply falls back to its original and lower energy state.

But all talk about ionizations making light makes me confused.

Which is true and how can an ionized (excited is ok due to "mgh") atom have a higher energy than a neutral one?

Much obliged for an answer on this one.

Best regards, Roger

2. ### DrClaude

2,327
I don't know where you got the idea that ionization of an atom is accompanied by emission of light, because it most cases it isn't. Exceptions to that might be if the electron being ionized is not the least-bound electron, but a core electron, in which case another electron from the atom will fill the gap, and hence emit a photon.

And why do you mention "mgh"? The binding of electrons in an atom is due to the Coulomb interaction, not gravity. Electrons can exist in different energy orbitals around the nucleus, and if there is a vacancy in a lower energy orbital, a higher energy electron can change state to occupy that lower energy orbital, and the difference in energy is compensated by the emission of a photon of the proper wavelenght.

3. ### rogerk8

220
This is a very good explanation. Thanks!

This is also a very good explanation. Thanks! I do however feel stupid now

Elaborating:

$$F=k\frac{q_1q_2}{r^2}[N]$$

$$W=\int_{r_1}^{r_2} Fdr=-kq_1q_2(\frac{1}{r_2}-\frac{1}{r_1})=-ke^2(\frac{1}{r_1}-\frac{1}{r_2})[Nm]$$

which means that for high r2 the work done by ionizing the atom is

$$W=-\frac{ke^2}{r_1}[Nm]$$

hence, the energy of the atom has increased(?)

Best regards, Roger

Last edited: Jan 18, 2014
4. ### DrClaude

2,327
It doesn't work like that. You have to use quantum mechanics, where electrons do not follow "orbits" at fixed ##r## around the nucleus.

5. ### rogerk8

220
Not even as a principle, you mean?

I do however know that

$$E=hf$$

where Planck's constant comes from quantum mechanics, right?

And E is the energy difference of the "orbits" or possible energy states, right?

Best regards, Roger

6. ### DrClaude

2,327
Not if you want the correct answer. It is possible to use the Bohr model to get a number that is correct for hydrogen, but this is not what an atom really looks like.

That equation is valid for photons. If ##E## is the energy lost by the electron due to the change in orbital, the formula will give you the frequency ##f## of the emitted photon.

7. ### rogerk8

220
A lazy question, what determines E?

Best regards, Roger

8. ### rogerk8

220

The electron's total energy at any radius:

$$E=\frac{1}{2}m_ev^2-\frac{Zk_ee^2}{r}=-\frac{Zk_ee^2}{2r}$$

This means that it takes energy to pull the orbiting electron away from the atom.

And when this energy is taken as in an ionization, the atom energy must lessen, right?

And an electron closer to the nucleus (and therefore higher energy) may fill that gap which then gives rise to light, right?

But what happens to that closer orbit and it's vacancy?

Best regards, Roger

9. ### DrClaude

2,327
You can see this basically as giving the energy to the electron to remove it from the atom. From that point of view, the energy of the atom left behind is unchanged. But we have to be careful, because you usually don't consider each independently, but take the ion + electron system as a whole.

That electron would have lower energy (there is a negative sign in the equation).

10. ### rogerk8

220
So you are giving energy to the electron (and not the atom) to remove it from the atom. And the energy of the atom is unchanged (and not increased while work is done to it or perhaps I should say the electron). Removing the electron far from the atom, according to the formula, however makes the electron energy "zero" as relative to the atom.

Recombination then would mean nothing to the atom. Yet, I think, the electron finding it's way back to the "shell" it came from would mean that the photon energy sent out is equivalent to the energy of that specific shell. Because that is the increasement of energy for that electron.

An electron closer to the nucleus obviously has higher (negative) energy.

I have a really hard time understanding this but let's look at it in layman terms. Positive energy gives "heat", negative energy then must give "cold". In other words, positive energy is an increasement, negative energy is a decreasement.

So if the negative energy is high, there is actually a decreasement of (electron) energy?

This would mean that an electron close to the nucleus would need a higher additional energy to escape the atom, right? All of this because it's energy is so negative which has to be overcome.

This actually sounds reasonable

Best regards, Roger

Last edited: Jan 19, 2014
11. ### DrClaude

2,327
I've been trying to keep it simple, which may have made things less clear than they should be!

We should take the system ion + electron and look at the energy of the system in total. Setting the zero of energy as the the energy when distance between the two is infinite, the energy is negative for the netral atom, as it is the more stable configuration.

Again, from the point of view of the ion + electron system: yes, if the ion captures the electron, the binding energy must be released, most probably through emission of a photon.

A bigger negative energy means lower energy. The zero of energy is arbitrary.

As I said above, where you put the zero is arbitrary, so there is no difference between negative and positive energy. All that is important is relative energy. We usually set the zero for the system at hand in order to easily make a distinction between different possiblities. In some cases, you set the zero for an infinite separation (as above). Sometimes, you will set the zero as the ground state of the atom.

Also, in this context you shouldn't think about "hot" or "cold". There is no temperature here, as we are working at fixed energy. Only when you have many particles does it make sense to use the language of thermodynamics.

Yes. The closer an eletron is to the nucleus, the greater the energy needed to remove it from the atom.

And just to be clear: the picture here is very classical. There are some subtleties due to quantum mechanics that need to be taken into account (such as the fact that an electron cannot be said to be at a given distance from the nucleus). Also, in many-electron atoms, electron-electron interactions also affect the picture.

12. ### rogerk8

220
You have obviously put lots of time and effort into describing this to me and I thank you for it!

Two things:

1) I didn't know that energy could be relative (and thus negative) but as you describe it it makes sense.
2) I do however still don't get how a higher negative energy could mean a lower energy because intuetively an electron closer to the atom ought to have higher energy simply to keep its distance from the atom's electromagnetic force.

This negative stuff obviously confuses me

Best regards, Roger

13. ### DrClaude

2,327
Most often, the exact total energy is not important, only the relative energy. For instance, when studying atoms, it is exceptional to have to take into account the "rest energy" ##mc^2## when calculating the energy of the atom. Where we chose to put to zero of energy is arbitrary.

It is the same as with gravity: the lowest energy is when the two bodies are closest together. If you want to move an electron away from the nucleus, or lift something off the ground, you have to put in energy into the system.

14. ### rogerk8

220
I totally understand now, thanks!

Best regards, Roger

15. ### rogerk8

220
I have understood that there are more ways to generate light. The mechanism I am thinking of is thermal radiation and reading about this in Wikipedia took me so far as to charge-acceleration and dipole oscillation.

While surfing around in Wikipedia I got to remember that light is a TEM-wave.

I now have an idea how thermal radiation makes light. Let's consider charge-acceleration first and please tell me if I'm wrong.

Being an electrical engineer, I began to think:

$$U=-L\frac{dI}{dt}=-jwLI$$

where

$$I=I_0e^{jwt}$$

which both means that jw is a derivation operator for any stationary signal and that the current and the voltage is 90 degrees out of phase in an inductor.

If we consider the Bohr Model and the Bohr Radius rB we have an orbiting moving charge (at a constant speed, preliminary) which by definition is equal to current.

Now, the magnetic flux through a coil of area S may be simplified to

$$\phi=BS=\mu_0HS=\mu_0NIS[Wb]$$

Where N=1 for our orbiting electron.

$$e=\oint Edl=-\frac{d\phi}{dt}=-jw\phi[V]$$

states the emf-induction due to magnetic flux change.

I have some stupid thoughts regarding how this emf may be measured. Perhaps simply on either side of the electron

Viewing these equations we do however have both B and E 90 degrees out of phase. And this is by definition a TEM-wave.

The speed of the (circulating) charge/current need however to be non-constant (otherwise no induction can be made) which means that we have to accelerate the charge by for instance heat.

For a while there I thought "what is acceleration of charge?". But then I stupidly realised that it just is change of velocity like any sinusoidal (f.i) signal.

My reasoning makes ordinary atoms radiate no light unless they are (being) heated up or excited. Strange. I must be wrong because almost everything has color. On the other hand, might vibration be the reason in that case?

The conclusion must be that every AC-driven coil (which might include dipole oscillation) or accelerated charge like above give rise to TEM (even though frequency differ enormously). Sounds a bit like radio

While the Bohr Model just is for simplification, a free accelerated or decelerated charge might suit even better for this reasoning. In this case it is more obvious that there are no spectral lines when it comes to thermal radiation and this is simply because of the "linear" speed states while adding kT.

How wrong am I?

Best regards, Roger

Last edited: Feb 1, 2014