# Photoelectric Effect versus Laser Stimulated Emission [Confused]

• Bardia Sahami
In summary, the electron of the second layer (n=2) would use the energy of the photon we radiated to leave the atom and become a photoelectric electron.

#### Bardia Sahami

Our teacher taught us "Laser" today and it made me confused. So Electron s energy in n=1 is around 13 eV (-13eV) and in n=2 is around -3.4 eV. Our teacher told us if we radiate a photon which has 9.6 eV energy (the difference energy of the first and second layer, n1 and n2, which is 13-3.4=9.6 eV), the electron of n=2 goes back to n=1 and it s how lasers works.

But my question is here. Why doesn t the electron of the second layer (n=2) use the energy of the photon we radiated to leave the atom and become a photo electric electron? Because the energy of the second layer is -3.4 eV and the electron can use this energy to leave the atom as speed up its speed. (If we radiate a photon with 9.6 eV energy, 3.4 eV would be used for the electron to leave the atom and the other 6.2 eV energy would be used to speed up its speed [Kinetic energy).

So this is my question. Why does electron goes back to the previous layer (laser), which is n=2 to n=1 in my example, instead of using that energy to get out of atom (Photo Electric Atom)?

Regards

PS - Answer in easy words, thanks!

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I don't know what kind of laser you were modelling, because the numbers look like the energy levels of hydrogen, and I don't know about any laser working with that atom as a lasing medium.

Usually, what happens in lasers is that ionisation is not possible. The incoming photon would be only resonant with the n=2 → n=1 transition.

In the case you mention, there are indeed two possible outcomes: stimulated emission (n=2 → n=1) and ionisation. Since the photon is resonant with two discrete levels while ionisation is from a discrete level to a continuum, I would expect stimulated emission to still dominate.

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Demystifier, berkeman and sophiecentaur
Bardia Sahami said:
Our teacher taught us "Laser" today and it made me confused. So Electron s energy in n=1 is around 13 eV (-13eV) and in n=2 is around -3.4 eV. Our teacher told us if we radiate a photon which has 9.6 eV energy (the difference energy of the first and second layer, n1 and n2, which is 13-3.4=9.6 eV), the electron of n=2 goes back to n=1 and it s how lasers works.

But my question is here. Why doesn t the electron of the second layer (n=2) use the energy of the photon we radiated to leave the atom and become a photo electric electron? Because the energy of the second layer is -3.4 eV and the electron can use this energy to leave the atom as speed up its speed. (If we radiate a photon with 9.6 eV energy, 3.4 eV would be used for the electron to leave the atom and the other 6.2 eV energy would be used to speed up its speed [Kinetic energy).

So this is my question. Why does electron goes back to the previous layer (laser), which is n=2 to n=1 in my example, instead of using that energy to get out of atom (Photo Electric Atom)?

Regards

PS - Answer in easy words, thanks!
Isn't the idea of a laser that we somehow pump the electrons to a higher layer, then, rather than waiting for spontaneous decay, we give a small trigger to initiate decay in many atoms at once?

tech99 said:
Isn't the idea of a laser that we somehow pump the electrons to a higher layer, then, rather than waiting for spontaneous decay, we give a small trigger to initiate decay in many atoms at once?
That's the initial description of the laser process. There is, however, an important constraint for stimulated emission to happen in a significant amount. That is the excited state needs to be 'metastable' and have a relatively long average delay before it decays naturally. So, when a suitable photon happens to come along, it will find many excited atoms to be stimulated to emit. This is where the idea of Pumping comes in - which uses input Power to raise significant atoms into the metastable level (population inversion) that, once the system 'goes' the emitted photons are all in phase.

When a photon arrives 'near' an already excited atom, I think that the probability of stimulated emission is much higher than the probability of ionisation because the stimulated case is a resonance and it requires coupling of a (very) small amount of energy from the passing photon is much greater than the whole of the photon energy producing actual ionisation. It's Probability at work, I think.

tech99
DrClaude said:
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and
sophiecentaur said:
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According to my teacher, if we have many atoms and most of them are not in their primary layer (n=2 for example), if we radiate a photon (the energy of photon should be the difference of the first and second layer, n=1 and n=2), the electron would go back to n=1 from n=2 and radiates the second photon. These two photons go to other electrons and the same thing happens again. So the number of photons will be 2 then 4 then 8 then 16, etc..

My question is something else. According to the teacher, the first photon we radiate should has the exact energy difference between n=2 and n=1. What I'm saying is why the electron doesn't use this photon to get out of the atom and become a photo electron. Because the energy of the first photon we radiate (which its energy is the difference of n=2 and n=1) is higher than the Ion energy and electron can use this energy to for ionization instead of going back to the previous layer.

Bardia Sahami said:
My question is something else. According to the teacher, the first photon we radiate should has the exact energy difference between n=2 and n=1. What I'm saying is why the electron doesn't use this photon to get out of the atom and become a photo electron. Because the energy of the first photon we radiate (which its energy is the difference of n=2 and n=1) is higher than the Ion energy and electron can use this energy to for ionization instead of going back to the previous layer.
As @sophiecentaur and I said, it is because the probability of a transition between two discrete energy levels is greater than the probability of a transition from a discrete energy level to the ionisation continuum.

DrClaude said:
As @sophiecentaur and I said, it is because the probability of a transition between two discrete energy levels is greater than the probability of a transition from a discrete energy level to the ionisation continuum.
So you mean if for an atom the different energy of two layers is LOWER than the ionisation energy, it can NOT be used as a laser, right?

Bardia Sahami said:
So you mean if for an atom the different energy of two layers is LOWER than the ionisation energy, it can NOT be used as a laser, right?
No, that's not what I said. It is actually the contrary: the vast majority of lasers are based on transitions that are not sufficiently energetic to cause ionisation, therefore the possibility of ionisation is not a problem.

Bardia Sahami said:
What I'm saying is why the electron doesn't use this photon to get out of the atom and become a photo electron.
I just read this again and it looks to me as if you imply that it could use the energy of a photon that it just emitted(?) to give it enough energy to escape completely. That's just a 'pull yourself up by your won bootstraps' proposal and could be taken to a nonsense level. If the electron has been taken to the first level of excitement by (say) an absorbed photon from outside, there is only enough energy to have done just that. There is either a surplus of 'potential energy' for the electron at a high level or EM energy in the form of a re-emitted electron. There is no 'extra energy' for the electron to escape.
I think you need to do an Energy Budget for various stages of the process and it will show you that the limits to what can happen must be imposed by what's actually available

DrClaude said:
No, that's not what I said. It is actually the contrary: the vast majority of lasers are based on transitions that are not sufficiently energetic to cause ionisation, therefore the possibility of ionisation is not a problem.
But if it cause ionisation, there will be no second photon. (and the first one will be used)

Bardia Sahami said:
But if it cause ionisation, there will be no second photon. (and the first one will be used)
Why not take all this up with your teacher? You haven't answered my point about the total energy budget being what counts. One low energy input photon cannot produce ionisation. A second photon, passing near an excited atom could produce ionisation but that would require the second photon to have a (in classical terms) a direct hit. Possible but unlikely. If the second photon is only 'close', all that it can do is 'shake' the electron and release it (stimulated emission) back to its ground state, producing a photon in phase with the passing wave (=laser action). I am deliberately using the most conventional, non quantum, terms possible and it's a slope argument. However it doesn't have to be 'wrong' and gives a credible description of how it works.
Remember that significant laser action is really rare and needs special energy conditions to make it happen (it was only just invented when I was at Uni!). Otoh, mechanisms for producing ionisation have been studied for a hundred years or more and the two phenomena are not usually discussed in the same breath. If you think you have found a crossover then I don't think your idea is particularly worth pursuing. Find out much more about lasers and you will probably see what I mean.

Bardia Sahami said:
Our teacher taught us "Laser" today and it made me confused. So Electron s energy in n=1 is around 13 eV (-13eV) and in n=2 is around -3.4 eV. Our teacher told us if we radiate a photon which has 9.6 eV energy (the difference energy of the first and second layer, n1 and n2, which is 13-3.4=9.6 eV), the electron of n=2 goes back to n=1 and it s how lasers works.

But my question is here. Why doesn t the electron of the second layer (n=2) use the energy of the photon we radiated to leave the atom and become a photo electric electron? Because the energy of the second layer is -3.4 eV and the electron can use this energy to leave the atom as speed up its speed. (If we radiate a photon with 9.6 eV energy, 3.4 eV would be used for the electron to leave the atom and the other 6.2 eV energy would be used to speed up its speed [Kinetic energy).

So this is my question. Why does electron goes back to the previous layer (laser), which is n=2 to n=1 in my example, instead of using that energy to get out of atom (Photo Electric Atom)?

Regards

PS - Answer in easy words, thanks!

Btw, instead of using battery, of course, to bring the electron to a higher level we can use other source of energy. But, hopefully you get the idea... I read the two other answers and actually they are explaining the details on the laser emission process, which you might understand one day in the future :)

Bardia Sahami said:
So this is my question. Why does electron goes back to the previous layer (laser), which is n=2 to n=1 in my example, instead of using that energy to get out of atom (Photo Electric Atom)?
The 9.6 eV is the energy difference between n=1 and n=2. If you are in n=1 and add a photon with 9.6 eV, the atom can absorb the energy and transition to n=2. The atom could absorb MORE energy, and have the electron leave n=2, creating a free electron. But the 9.6 eV is unavailable for THAT transition. It is already the energy that enables n=2.

Without additional energy, the atom has 2 paths. Remain at n=2, or transition to n=1, emitting 9.6 eV as a photon.

Your question seems to read as if the atom has a surplus has 9.6 eV of "free" energy to use in any state transition. It only has 9.6 eV available if it transits downward. Think of a ball on a stairs. You lift it up one more step. It now can fall, and energy is transformed in falling from potential to kinetic. It cannot take that gravitational potential energy to throw itself further up. External energy can lift the ball higher, but the ball can only use the energy it has by falling.

The atom CAN be ionized in a photo-electric process, but not with its own internal state energy. That internal state energy can only be part of things remaining the same, or moving to a less energetic state with an emission of a photon.

Bardia Sahami said:
But if it cause ionisation, there will be no second photon. (and the first one will be used)
You are counting the Energy TWICE, here. The excited atom will lose the photon's energy and the electron will return to the ground state. How can that energy turn up again and re-arrange the atom?

sophiecentaur said:
You are counting the Energy TWICE, here. The excited atom will lose the photon's energy and the electron will return to the ground state. How can that energy turn up again and re-arrange the atom?
As far as I know, lasers are not made from one atom alone, but from many many similar atoms, so the energy/photon released from one atom may cause a ionization to the next (excited) atom ...

DanMP said:
As far as I know, lasers are not made from one atom alone, but from many many similar atoms, so the energy/photon released from one atom may cause a ionization to the next (excited) atom ...
The question is vague about whether it is initial excitation energy, or adding additional energy to a maximally excited state, to cause ionization. I am not expert, but the answer I see in other comments is that you should not make a laser from a material that will photo-degrade at the energy you are las-ing. If it photo-degrades into ions, that would be a bad choice.

DrClaude said:
the vast majority of lasers are based on transitions that are not sufficiently energetic to cause ionisation, therefore the possibility of ionisation is not a problem.

So if a material has an energy transition of 5 eV to the maximum state, and then it needs 6 eV to achieve ionization (from that excited state), the laser frequency of 5 eV will never ionize anything. If you have a material that has 5 eV to the maximum state, and 4 eV to achieve ionization from that state ... don't use that material in a laser.

The question is vaguely worded though so there are two answers being given, and I have some confusion as to what is being asked.

votingmachine said:
don't use that material in a laser.
+1
That's the whole thing in a nutshell!

DrClaude said:
I don't know what kind of laser you were modelling, because the numbers look like the energy levels of hydrogen, and I don't know about any laser working with that atom as a lasing medium.
As the original question has been answered, I would like to add to @DrClaude's correct response. While not suitable for visible light spectrum, 'molecular hydrogen' was used in predecessor masers (microwave or molecular amplification by stimulated emission of radiation) described here: Hydrogen maser.

sophiecentaur said:
[snip]
Remember that significant laser action is really rare and needs special energy conditions to make it happen (it was only just invented when I was at Uni!). [snip]
The first practical maser was invented the same year 1953 my family moved to 'Silicon Valley' (Santa Clara Valley in Northern California). As a child at a science fair on the grounds of SRI International in the 1960's, I spent several hours with a docent scientist testing and tuning a Hydrogen maser (This article is more descriptive than in my previous post.).

I remember the unit gave off a purple glow similar to hydrogen discharge tubes I used later in the USAF. The docent explained the 'glow' in terms of the EM spectrum. Our eyes responded to the 'glow' but we needed instruments to measure the microwaves. Tuning altered dimensions of the microwave cavity as well as varying electronic feeds.