Question on Excitation

[calbiterol]

Go easy on me, it's been a while since chemistry, and our defuncto Physics "teacher" didn't teach much of anything involving quantum mechanics...

 

Given the hydrogen energy level transitions:

n=3 to n=4: 0.66eV

n=2 to n=3: 1.89eV

n=1 to n=2: 10.2eV

Could one excite hydrogen gas from ground state to n=4 with two lasers; one of energy 10.2eV and one of 2.55eV, or would this violate conservation of angular momentum?

 

I ask because I want to do an experiment involving excitation of a gas in the nonvisible spectrum leading to visible emission. Essentially, I need a (n "easily" available) gas that can be excited with two lasers in the non-visible spectrum and will emit photons in the visible spectrum during de-excitation. I also need the frequencies / eV of the lasers I would need.

 

The problem is, I don't understand the process well enough to calculate this myself, and I seem to have misplaced my copy of the CRC Handbook to see if it has anything of help (and [rue the day] school hasn't started back up yet).

 

Hopefully that made sense. Cheers,

Calbit

[Klaynos]

IIRC, yes you can excite up to one level and then up again using a differnt souce. But the lifetime of the first excitement has to be long enough for it to have time to be excited up again else you'll be putting in ALOT of energy and not getting much of the correct frequency out.

 

To work out the frequency of the lasers you would require use:

 

E=hf

 

With E in joules, h is planks constant, and f is the frequency in Hz.

[swansont]

You can do two-photon transitions, but the angular momentum will change by either 0 or 2 units. Since you start in an S state, that means you will end up in an S or D state, but not a P state. The subsequent de-excitation will have to go through the P state, and if you are going to N=4, you could end up in the 2S state, which is metastable (lifetime is 1/7 sec, which is long for an atomic state), since there is no allowed transition directly out of it. (however, it can "mix" with the close-by 2P state, which decays pretty rapidly. The lifetime is actually an indication of the natural mixing of the states)

 

You don't even need a real state to be the endpoint of one of the transitions in two-photon excitation. In fact, to get to the 2S state, people use two-photon excitation, at ~243 nm, since a single photon won't work. To get to n=4 you might want to try and find a source of ~195 nm light. It'll be easier than finding the ~122 nm for the first transition, and then you only need 1 source.

[swansont]

A thought just occurred to me. In Rubidium, the D2 transition (IIRC the ground state is 5S, so this is the 5S_1/2-6P_3/2) transition is at ~780 nm, and the 6P-7D transition is at ~776 nm, so a two-photon transition at ~778 nm gets you there. There is a decay that goes through another state gives you a nice blue, at ~420 nm.

 

I saw this once with a single laser at 780 nm in a hot sample of the gas. My hypothesis was that two excited atoms collided, putting one in the 7D state. It's also possible that it was two-photon and the Doppler shift and local fields shifted it into resonance. (The high temperature damaged the vapor cell, however, so I wasn't able to investigate further. But that wouldn't be an issue at 778 nm)

 

Near-IR lasers are probably a little easier to obtain than UV ones.

[calbiterol]

I'm sorry, swansont. You are describing what I think is exactly what I am looking for, but I have absolutely NO idea what you mean by it. In other words, explain the transitions you were talking about, and the concept of two-photon excitation, and it'd be much appreciated.

 

Thanks much!

Calbit

[swansont]

You were right that you have to worry about conservation of angular momentum; each photon has 1 unit, but that won't necessarily stop you from doing what you want. Two photons together wil have either 2 units or 0 units.

 

An atom can absorb two photons at once, though the probability of doing so is generally lower than for any single absorpton, but the advantage of this is that you don't have excite the transitions with each beam — the intermediate state doesn't have to exist, as long as the total energy is right, and angular momentum is conserved.

 

So for H you need 12.75 eV, and any combination of two sources will work. If there's enough power, 195 nm will work if the atom absorbs two photons at once, even though there is no state there (enough power means the photon density is high enough that two photons can strike an atom essentially at the same time). But a 778 nm source might be easier to come by, so the Rb experiment might be easier.

 

BTW I checked a diagram; the Rb levels are 5S, 5P and 5D, with the 420 nm light coming through a de-excitation through the 6P state. S refers to a state with zero angular momentum, P has one unit, D has two. (So for a single photon, you can't go from S to S or S to D, since it has to change by one unit.)

[calbiterol]

Wait, you saw this in rubidium GAS? That must have been one HOT vapor cell, rubidium's boiling point is 688 C / 961 K. That's right about aluminum's melting point!

 

I still don't think I get the math.

 

How did you get 195 nm for hydrogen? Is that for a single source? If so, I think it makes sense. However, it isn't that I thought I would need two sources, it's that I want to have two sources. I'm still not sure of the math though, so could you explain it one more time for someone who's probably in over his head? Thanks again,

Calbit

[swansont]

Wait' date=' you saw this in rubidium GAS? That must have been one HOT vapor cell, rubidium's boiling point is 688 C / 961 K. That's right about aluminum's melting point!

 

I still don't think I get the math.

 

How did you get 195 nm for hydrogen? Is that for a single source? If so, I think it makes sense. However, it isn't that I thought I would need two sources, it's that I [i']want[/i] to have two sources. I'm still not sure of the math though, so could you explain it one more time for someone who's probably in over his head? Thanks again,

Calbit

 

It was vapor; all material have a vapor pressure, so if you pop some Rb (or watever) into an evacuated cell and seal it up, you will have some small amount of gas in there. And the vapor pressure is temperature dependent. I was looking at making the vapor "optically thick," i.e. dense enough that it would absorb all the light before it got through the cell.

 

[math]E = h\nu = \frac{hc}{\lambda} [/math]

 

so [math]\lambda = \frac{hc}{E} [/math] = 1243 eV nm/12.75 eV = 97.5 nm

 

But you want to do this with two photons, so they have half the energy, or twice the wavelength, or 195 nm

 

But you should be able to pick and choose, so find what sources you have available to you. As long as they add to the toal energy, it can work.

[calbiterol]

It was vapor; all material have a vapor pressure, so if you pop some Rb (or watever) into an evacuated cell and seal it up, you will have some small amount of gas in there. And the vapor pressure is temperature dependent. I was looking at making the vapor "optically thick," i.e. dense enough that it would absorb all the light before it got through the cell.

Makes sense. Was the Rb liquid beforehand (about 37 C IIRC)?

 

But you want to do this with two photons, so they have half the energy, or twice the wavelength, or 195 nm.

AAH! The lightbulb just turned on. You halve the energy, not the wavelength. Doh.

 

Thanks much, I think that will suffice! The lasers can be perpendicular to each other, correct? In other words, are there limits in orientation?

[swansont]

Makes sense. Was the Rb liquid beforehand (about 37 C IIRC)?

 

It would have been warmer, because I was heating it, but there wasn't enough to form a pool or anything.

 

Thanks much' date=' I think that will suffice! The lasers can be perpendicular to each other, correct? In other words, are there limits in orientation?[/quote']

 

That I'm not sure of. I can imagine that there would be, since the direction of the field from one laser may polarize the atom, and there are polarization restrictions on some interactions. It may be that they have to be co- or counter-propagating (the blue-light Rb experiment was like that), or if they are perpendicular, that they might have to be polarized along the same axis. I'm not sure.

[calbiterol]

I guess that's what experimentation is for, now isn't it?

 

I'll have to get back to you about that when I start trying things out.

[5614]

but the advantage of this is that you don't have excite the transitions with each beam — the intermediate state doesn't have to exist' date=' as long as the total energy is right, and angular momentum is conserved.

 

So for H you need 12.75 eV, and any combination of two sources will work. If there's enough power, 195 nm will work if the atom absorbs two photons at once, even though there is no state there[/quote']So does that mean that rather than even needing two beams you could just use one big 12.75eV beam? Or does that cause issues with angular momentum? Because you would have spin 1 (from one photon), and not 0 or 2 (from two photons).

 

And, maybe it's apparent, but I don't really understand how angular momentum and the conservation of it fits into this. This is something I have not yet been taught. I know what angular momentum is and the conservation of it. How does this fit into single and double-photon excitations?

[swansont]

So does that mean that rather than even needing two beams you could just use one big 12.75eV beam? Or does that cause issues with angular momentum? Because you would have spin 1 (from one photon)' date=' and not 0 or 2 (from two photons).

 

And, maybe it's apparent, but I don't really understand how angular momentum and the conservation of it fits into this. This is something I have not yet been taught. I know what angular momentum is and the conservation of it. How does this fit into single and double-photon excitations?[/quote']

 

You'd have to go to a P state* if it's one photon, since L (angular momentum eigenvalue) must change by 1 unit. You'd go from 0 to 1. With two photons, you go from 0 to 0 or 0 to 2. (These are known as "selection rules" in QM. Since the values are quantized, you can drop the [math]\hbar[/math] and go with the characteristic integer or fraction associated with it) Then you have the sign, depending on the orientation of the angular momentum vector.

 

With two photons they can be aligned; 1 + 1 = 2; or anti-aligned; 1 + (-1) = 0

 

*edit: (we won't go into P_1/2 vs P_3/2 here)