Harmonics

[calbiterol]

I think this is appropriate in QM and not CM, since it deals with atoms and waves.

 

Can someone please explain to me how resonant frequencies work in atoms, and what happens when atoms encounter their resonant frequency?

Thanks.

[swansont]

Atoms have resonant frequencies because energy and angular momentum are quantized. If you solve the equations governing their behavior, you find that only certain energies are possible for the orbital electrons.

 

Atoms "encounter" their resonant frequencies when bombarded by photons at that frequency. These photons can be absorbed, which puts an electron into a higher energy state. The electron will eventually release energy in the form of another photon, and the atom will go to a lower energy state. (atoms in molecules can release this energy in other ways)

 

Photons not at a resonant frequency will not be absorbed.

[calbiterol]

So, it's possible (in theory) to bump an electron up to a higher energy state with a photon in the nonvisible spectrum, yes? Is it possible that the same electron could, in theory, emit a visible photon when lowering its energy state?

[Tom Mattson]

So, it's possible (in theory) to bump an electron up to a higher energy state with a photon in the nonvisible spectrum, yes? Is it possible that the same electron could, in theory, emit a visible photon when lowering its energy state?

 

 

Yes and yes.

 

The visible spectrum ranges in wavelength from about 400nm (violet) to about 700nm (red). You can compute their energies via the formula:

 

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

 

where hc=1240 ev-nm. That gives E_red=1.77eV and E_violet=3.1eV.

 

Now look at the hydrogen atom. In first approximation its energy levels are given by:

 

[math]E_n=\frac{-13.6eV}{n^2}[/math]

 

So the energy of a transition from E₁ to E₄ would require a photon whose energy is about 12.75eV, which is way more energetic than an optical photon.

 

Now this atom can de-excite one energy level at a time. Here are the energies (to the nearest 0.01eV) of the transitions. Note that those below 1.77eV and above 3.1eV result in invisible photons.

 

n=4 to n=3: |E_4-->3|=0.66eV

n=3 to n=2: |E_3-->2|=1.89eV

n=2 to n=1: |E_2-->1|=10.2eV

 

The second transition (and only the second transition) in the above list results in a visible photon.

[J.C.MacSwell]

Atoms have resonant frequencies because energy and angular momentum are quantized. If you solve the equations governing their behavior' date=' you find that only certain energies are possible for the orbital electrons.

 

Atoms "encounter" their resonant frequencies when bombarded by photons at that frequency. These photons can be absorbed, which puts an electron into a higher energy state. The electron will eventually release energy in the form of another photon, and the atom will go to a lower energy state. (atoms in molecules can release this energy in other ways)

 

[b']Photons not at a resonant frequency will not be absorbed[/b].

 

Is this absolute or decreasing probability?

[Tom Mattson]

Is this absolute or decreasing probability?

 

That would be an absolute probability of 0.

[calbiterol]

Ai'ight. I think I get it now. 3 questions though: first, how do you excite through multiple energy levels? Do you use more power, or just sustain what power you had?

 

Also, are there any other gasses that have the same effect that hydrogen does?

 

And is there any way for me to determine these kinds of things (like hydrogen electrons only being able to drop one level at a time, the spectra emitted during each drop, etc)?

[J.C.MacSwell]

That would be an absolute probability of 0.

 

That seems at odds with quantum uncertainty. I would have guessed that the chance of absorbtion would peak at certain wavelengths but not go immediately to zero either side.

[swansont]

Is this absolute or decreasing probability?

 

It drops off rapidly with the square of the frequency difference with a Lorentzian shape, and also depends on the width of the transition (which is the inverse of the lifetime; this stems from the Heisenberg Uncertainty Principle). A resonance is not infinitely narrow.

 

As an example, the transition from the first excited state in Rb is around 780 nm, putting the frequency at around 3.8 x 10¹⁴Hz. The transition is around 6 MHz wide, since its lifetime is ~26.5 ns (IIRC).

[swansont]

Ai'ight. I think I get it now. 3 questions though: first' date=' how do you excite through multiple energy levels? Do you use more power, or just sustain what power you had?

 

Also, are there any other gasses that have the same effect that hydrogen does?

 

And is there any way for me to determine these kinds of things (like hydrogen electrons only being able to drop one level at a time, the spectra emitted during each drop, etc)?[/quote']

 

There are restrictions. Photons carry one quantum of angular momentum, so all transitions involving a photon must change the atom's angular momentum bu one (this is known as a selection rule) So, for example, you will not see a pure S-state to S-state transition, since these both have zero angular momentum. Likewise, you cannot go from an S to a D state, because D has an angular momentum of 2. So you have S->P transitions. From the P state, you can the go to S, P or D (some P->P transitions are allowed, because it flips the spin of the electron, even though the overall state has the same angular momentum. QM is confusing of you aren't familiar with it. And even when you are. This isn't introductory stuff, even in a QM class)

 

You excite through multiple levels by having all of the frequencies available to the atom. A Google on "three step excitation" will yield many papers (mostly pdf).

 

There are many publications that show the spectra of elements. The CRC handbook has a decent list, and there are journals that publish this kind of result. And online resources as well.

[Tom Mattson]

That seems at odds with quantum uncertainty. I would have guessed that the chance of absorbtion would peak at certain wavelengths but not go immediately to zero either side.

 

Never mind, my mistake. My mind was stuck in the first approximation.

[calbiterol]

Can someone please explain this to me? I don't quite get it.

 

What you want to investigate is a field called "four wave mixing" If you shine two laser frequencies' date=' w[sub']1[/sub] and w₂, onto certain nonlinear materials, you can get the sum and difference frequencies out: w₃ = w₁+w₂ and w₄ = w₁-w₂ (assuming w₁>w₂) A special case, where w₁ = w₂ is called second-harmonic generation. You get twice the original frequency and there is no difference frequency.

 

 

If the materials were transparent to the frequencies involved, you would in principle have minimal absorption except at the overlap region. However, four wave mixing has certain alignment requirements; I'm not certain if the beams can be perpendicular, or have to be copropagating or have some other alignment/polarization relationships.

 

I get that you are shining two laser frequencies onto a material that is not a line, and that the two frequencies may combine to give the sum and the difference of the two frequencies, and that when the sum equals the difference, it is called second-harmonic generation. I don't really get the rest of it...

[swansont]

I get that you are shining two laser frequencies onto a material that is not a line, and that the two frequencies may combine to give the sum and the difference of the two frequencies, and that when the sum equals the difference, it is called second-harmonic generation. I don't really get the rest of it...

 

I meant: I'm not familiar enough with this to know if the beams can be perpendicular, or some specific angle between them, or if they have to be co-propagating, or what the polarization requirements are. Four-wave mixing also requires certain types of materials.

 

The lasers are typically off-resonance, so there is no absorption. The interaction only takes place at the overlap region.

[calbiterol]

Hypothetical situation:

 

Take two lasers. One is at the frequency that will excite an electron to its P state. The other is at the frequency that will excite the electron from P to D. Will this end up with the electron in the D state? I would think so... not that I claim to know too much about these things... because wouldn't the first laser bump the electron up to the P state, with the first laser having no interaction with the electron (because it was the wrong frequency to be absorbed), and then the second laser would have an effect but not the first, bumping the electron up to its D state?

[calbiterol]

Now look at the hydrogen atom. In first approximation its energy levels are given by:

 

[math]E_n=\frac{-13.6eV}{n^2}[/math]

 

So the energy of a transition from E₁ to E₄ would require a photon whose energy is about 12.75eV' date=' which is way more energetic than an optical photon.

 

[b']Now this atom can de-excite one energy level at a time.[/b] Here are the energies (to the nearest 0.01eV) of the transitions. Note that those below 1.77eV and above 3.1eV result in invisible photons.

 

n=4 to n=3: |E_4-->3|=0.66eV

n=3 to n=2: |E_3-->2|=1.89eV

n=2 to n=1: |E_2-->1|=10.2eV

 

The second transition (and only the second transition) in the above list results in a visible photon.

 

(Emphasis added)

 

How do you know that? Is it always only one level at a time (I'm talking other elements now, Neon in particular) or is there some rule or something? Sorry if this sounds like a really stupid question, but I'm trying to understand something that I have basically no teaching in ('cept for this ) and I need to know nitpicky details... Also, can an atom absorb photons that are at any of the harmonics (harmonics being the energies an electron jumps to, right?) IE, the ~12.75 eV to get to E₄, OR the ~12 eV to get to E₃, etc (in the H atom) could all be absorbed (at different times) to excite the electrons? And is this universal between all elements, or is there some rule/chart of acceptable jumps for this too?

[Tom Mattson]

How do you know that?

 

I know that the atom can deexcite one level at a time because transitions from state n to state n-1 are allowed transitions. So there is the possibility of it occuring that way.

 

Is it always only one level at a time (I'm talking other elements now' date=' Neon in particular) or is there some rule or something?

[/quote']

 

It is not always that way. Any allowed transition from the higher state to the lower state can occur. Furthermore there is no way to tell a priori which decay channel the system will take.

 

Also, can an atom absorb photons that are at any of the harmonics (harmonics being the energies an electron jumps to, right?) IE, the ~12.75 eV to get to E₄, OR the ~12 eV to get to E₃, etc (in the H atom) could all be absorbed (at different times) to excite the electrons?

 

Photons that have energies that are equal to that of any allowed transitions from a lower level to a higher level can be absorbed. If a photon does not have an energy corresponding to an allowed transition then it will not be absorbed.

 

And is this universal between all elements, or is there some rule/chart of acceptable jumps for this too?

 

The energies quoted in my post are unique to the hydrogen atom. The energy spectrum of each atomic species is different, and in principle it can be determined from quantum mechanics. I say "in principle" because only for hydrogenlike atoms does QM admit an exact solution. For more complex atoms one must invoke approximations. Energy spectra for the most complicated atoms are determined empirically.

[calbiterol]

Thanks Tom, that post was exactly what I needed to know.

 

One thing, though:

 

The energies quoted in my post are unique to the hydrogen atom. The energy spectrum of each atomic species is different, and in principle it can be determined from quantum mechanics. I say "in principle" because only for hydrogenlike atoms does QM admit an exact solution. For more complex atoms one must invoke approximations. Energy spectra for the most complicated atoms are determined empirically.

 

I knew that, but I guess it would have been more appropriate for me to say, "how do I ensure beforehand that it will only take the one path." So in reality, there's no way to know? Out of curiosity, how do neon lights work with different colors? ... Wait, different gasses. Duh. But how do they emit only one color of light?

[swansont]

I knew that, but I guess it would have been more appropriate for me to say, "how do I ensure beforehand that it will only take the one path." So in reality, there's no way to know? Out of curiosity, how do neon lights work with different colors? ... Wait, different gasses. Duh. But how do they emit only one color of light?

 

The only way to ensure they take only one path is to restrict the level to which you excite. e.g. if you only go to the first excited state, there's only one path for de-excitation.

 

Of course this gets immensely more complicated if you start to incorporate the hyperfine levels. One way of ensuring that is to polarize the light, so only a certain excitation/de-excitation is possible. Basically making the selection rules work to your advantage.

 

The various gases in "Neon" lights may give off more than one frequency, but the additional transitions are not in the visible part of the spectrum. I imagine the gases are chosen specifically because they only have one visible transition.

[calbiterol]

There are restrictions. Photons carry one quantum of angular momentum, so all transitions involving a photon must change the atom's angular momentum bu one (this is known as a selection rule) So, for example, you will not see a pure S-state to S-state transition, since these both have zero angular momentum. Likewise, you cannot go from an S to a D state, because D has an angular momentum of 2. So you have S->P transitions. From the P state, you can the go to S, P or D (some P->P transitions are allowed, because it flips the spin of the electron, even though the overall state has the same angular momentum. QM is confusing of you aren't familiar with it. And even when you are. This isn't introductory stuff, even in a QM class)

 

Can you explain that again? Something my chem teacher said a while ago seems to contradict this. Plus, Tom's example jumps from ground state to E=4 with one photon of energy ~12.75 eV. Or is there a difference between s/p/d/f and E=1,2,3, etc? Sorry if the question seems dumb, but this is something that I haven't been taught in at all, other than what I've taught myself or knowledge I've acquired through very specific questions.

[swansont]

Can you explain that again? Something my chem teacher said a while ago seems to contradict this. Plus, Tom's example jumps from ground state to E=4 with one photon of energy ~12.75 eV. Or is there a difference between s/p/d/f and E=1,2,3, etc? Sorry if the question seems dumb, but this is something that I haven't been taught in at all, other than what I've taught myself or knowledge I've acquired through very specific questions.

 

Tom's example only focused on the quantized energy - that's one thing that must be satisfied. That's the E value. But it's possible that you could have a state at the right energy, but the system in that state must have a certain value of angular momentum. And the original state + photon can not give you that value of angular momentum. That's the s/p/d/f value.

 

What did your chemistry teacher say that seems to contradict this?

[DQW]

Or is there a difference between s/p/d/f and E=1,2,3, etc?

Yes there is a big difference. For hydrogenlike atoms/ions, the energy depends only on the principal quantum number 'n'. This relationship is what Tom wrote out (following the Bohr picture). For any value 'n' of the principal quantum number, there are n allowed values of the angular momentum quantum number 'l' (from 0 to n-1). It is these numbers that are designated s (l=0), p (l=1), d (l=2), f (l=4), ... by spectroscopists.

 

The principal quantum number does not determine angular momentum. And in multielectron atoms (in the absense of applied fields), the energy depends both on 'n' and 'l'.

[DQW]

http://www.chemistrycoach.com/quantum.htm

 

http://hyperphysics.phy-astr.gsu.edu/hbase/qunoh.html

 

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

 

http://scienceworld.wolfram.com/physics/QuantumNumbers.html

[calbiterol]

The only thing my chem teacher said that seemed to contradict anything - and keep in mind he said this about 3 months ago at the end of school (and hence I have not seen him since) was something about how all transitions were allowed. I took this to mean a transition from any level to any level - I guess one of us (probably I was) wrong.

[swansont]

The only thing my chem teacher said that seemed to contradict anything - and keep in mind he said this about 3 months ago at the end of school (and hence I have not seen him since) was something about how all transitions were allowed. I took this to mean a transition from any level to any level - I guess one of us (probably I was) wrong.

 

 

No, that's wrong. Either your teacher misspoke, or you misinterpreted what was said. Some transitions are forbidden, though there will still be a way to get from state A to state B.

 

One of the reasons that there are things called metastable states is that an electron gets "shelved" in an excited state from which it cannot easily decay - the direct decay to a lower state is not allowed. Atoms and molecules are complex, though, so even if one transition is forbidden, it's possible that coupling through other states will eventually allow the transition. An example is the 2S-1S transition in Hydrogen. Since no change in angular momentum is present, you can't get between the two states with a single photon dipole transition. However, you can get there with two photons (since their angular momentum can add to zero) or by coupling through the weak nuclear interaction that mixes the 2S and 2P states.

[calbiterol]

What I was getting at was that I probably either misheard him or misinterpreted what he said. He could have just said that more transitions than 1-2, 2-3, 4-3, etc are possible. To be honest, it was so long ago that I can't remember exactly what was said.

 

Can you explain more in depth the rules with 2 photons, and the situatations under which these are valid? And what's the weak nuclear interatction mixing the 2s and 2p states? I know what 2s and 2p are, just what's the weak nuclear interaction? Is it just the weak force (like the strong force, but the weak force; not just a weak force... You get the idea)?