Heat Flow

[studiot]

6 hours ago, exchemist said:

In the visible, photons are absorbed by electronic transitions, creating excited states that relax by non-radiative means (i.e. intermolecular collisions) to populate vibrational and rotational excited states  - and thence eventually translational motion.  In the IR, they are absorbed by vibrational states, which then similarly cascade down by collisional relaxation. None of these radiative processes, so far as I know, can directly stimulate excitation of  translational motion. The exceptions would be ionisation or bond breaking, where the absorption causes an electron to jump right out of an atom, or two part-molecules to fly apart because the bond strength has been exceeded.

It seems to me you need a transition dipole moment in order to absorb a photon and I can't see how you get that in translational motion. So I think, subject to correction by one of the physicists here, that translational excitation is always the consequence of collisions, rather than radiation directly. 

I was answering your question how EM (no particular frequency) radiation could impart translational energy to molecules.

Changing the state from solid to liquid or liquid to gas seems to me to be adding KE to the molecules by the bucketful, since they are then rushing about in a way they did not before the irradiation.

Perhaps I misunderstood you.

[swansont]

15 minutes ago, exchemist said:

Do you think Rayleigh scattering can produce a significant heating effect, then? 

I’m not sure how much, but I see blue light scattered by the atmosphere, and that light originally came from the sun.

[exchemist]

8 minutes ago, swansont said:

I’m not sure how much, but I see blue light scattered by the atmosphere, and that light originally came from the sun.

Sure, but that's just the redirected blue wavelength photons. I don't think the effect can be much because this is never discussed in spectroscopy. In principle what you are saying is that the absorbed photon has to do two things: excite a transition to an excited state AND contribute to an increase in translational k.e. But we ought to be able to calculate the k.e. increase from the momentum, shouldn't we? For the photon E=pc = hν.    

[sethoflagos]

On 2/18/2023 at 7:20 PM, exchemist said:

Can it in all cases, though? Rotational and vibrational excitation can change by absorption or emission of radiation, but translational motion? Doesn't that have to change by heat flow?  

Bear in mind that there is a dynamic equilibrium between translational, rotational and vibrational components. Compression for example is a direct input of translational kinetic energy, but it it gets partitioned equally between all the degrees of freedom as part of the equilibriation process. 

If you look at purely the translational modes, the monatomic gases always behave elastically, but other gases only appear elastic on average. 

[studiot]

Just now, sethoflagos said:

Bear in mind that there is a dynamic equilibrium between translational, rotational and vibrational components. Compression for example is a direct input of translational kinetic energy, but it it gets partitioned equally between all the degrees of freedom as part of the equilibriation process. 

If you look at purely the translational modes, the monatomic gases always behave elastically, but other gases only appear elastic on average. 

The equipartition theorem no less.  +1

[swansont]

24 minutes ago, exchemist said:

Sure, but that's just the redirected blue wavelength photons. I don't think the effect can be much because this is never discussed in spectroscopy. In principle what you are saying is that the absorbed photon has to do two things: excite a transition to an excited state AND contribute to an increase in translational k.e. But we ought to be able to calculate the k.e. increase from the momentum, shouldn't we? For the photon E=pc = hν.    

Yes. The speed change from a scatter is going to be a few cm/s. You’d need the scattering rate, too.

[exchemist]

9 hours ago, sethoflagos said:

Bear in mind that there is a dynamic equilibrium between translational, rotational and vibrational components. Compression for example is a direct input of translational kinetic energy, but it it gets partitioned equally between all the degrees of freedom as part of the equilibriation process. 

If you look at purely the translational modes, the monatomic gases always behave elastically, but other gases only appear elastic on average. 

Of course. That's what I meant when I referred to collisional relaxation. What I have been trying to get at is whether radiation can directly stimulate the translational degrees of freedom. I think the answer is no, save for this issue of momentum transfer when a photon is absorbed by another form of excitation, which @swansontput me onto.

(By the way, this whole area of relaxation from excited states, esp. the mechanisms for the redistribution of thermal energy among the available degrees of freedom, is something that seems to me to be quite complex and often rather glossed over, in particular how electronic excitation comes to be redistributed into vibration, rotation and translation.) 

 

[sethoflagos]

3 hours ago, exchemist said:

Of course. That's what I meant when I referred to collisional relaxation. What I have been trying to get at is whether radiation can directly stimulate the translational degrees of freedom. I think the answer is no, save for this issue of momentum transfer when a photon is absorbed by another form of excitation, which @swansontput me onto.

I see scattering as a perfectly elastic exchange of momentum with no energy exchange. ie. the scattering body changes direction without a nett change in speed and no heat transfer. Ditto the photon so there is no change in frequency and no spectroscopic impact (other than directional).

Photon absorption processes are quite different though since there is a transfer of energy into the absorber and unless the photon is reemitted very quickly, the increase in internal energy must have a thermal impact. So yes, I agree with @swansont on this point.

3 hours ago, exchemist said:

(By the way, this whole area of relaxation from excited states, esp. the mechanisms for the redistribution of thermal energy among the available degrees of freedom, is something that seems to me to be quite complex and often rather glossed over, in particular how electronic excitation comes to be redistributed into vibration, rotation and translation.) 

 Yes indeed. if the photon is energetic enough to start shunting electrons into different orbitals then the physical nature of the particle starts changing. But most of the gases I've worked with don't absorb in the optical range so it's not something I've ever had to worry about. 

 

 

[exchemist]

13 hours ago, swansont said:

Yes. The speed change from a scatter is going to be a few cm/s. You’d need the scattering rate, too.

OK I've had a go at this, taking gas phase sodium atoms as an example, since I see that has been used in laser cooling experiments. My arithmetic may need checking but if it's right it goes like this:

Take gas phase sodium atoms, absorbing at the D line.

Wavelength of D line, λ  = 590nm

Momentum, p = h/λ (de Broglie)  =>  p ~ 6.6 x 10⁻³⁴/5.9 x 10⁻7

~ 10⁻²⁷ kg-m/sec

= Δmv for a sodium atom absorbing a D line photon.

                  ------------------------------

Mass of a sodium atom is 23g/mol, i.e. 0.023/6 x 10⁻²³ ~ 4 x 10⁻²⁶ kg.

So for the absorbing atom,  Δv = 10⁻²⁷/4 x 10⁻²⁶ = 2.5 x 10⁻² m/sec.

Hence the increase in translational k.e. will be:

1/2 . 4 x 10⁻²⁶ . ( 2.5 x 10⁻²)²

= 0.5 . 4 x 10⁻²⁶ .  6.25 x 10⁻⁴ = 1.25 x 10⁻²⁹ J

                 ----------------------------

Translational k.e. = 3/2 . kT.

So ΔT due to extra momentum = 2/3 . ΔE/k = 2/3 . 1.25 x 10⁻²⁹/1.4 x 10⁻²³ 

~ 6 x 10⁻⁷ Deg. 

This is a “temperature” increase, for each atom that absorbs a photon, of under a millionth of a degree.

 

Expressing that in terms of the energy of the photon, that energy will be hν = pc = 10⁻²⁷ . 3 x 10⁸ 

= 3 x 10⁻¹⁹ J.

So the proportion of its energy that contributes to thermal k.e. is 2.4 x 10⁻¹⁰ , so almost infinitesimal. Now I understand why this is never mentioned in spectroscopy. 

 

Edited February 20, 2023 by exchemist

[swansont]

2 hours ago, sethoflagos said:

Photon absorption processes are quite different though since there is a transfer of energy into the absorber and unless the photon is reemitted very quickly, the increase in internal energy must have a thermal impact. So yes, I agree with @swansont on this point.

The impact is from the recoil, which changes the KE, causing heating (or cooling under the right conditions)

2 hours ago, exchemist said:

This is a “temperature” increase, for each atom that absorbs a photon, of under a millionth of a degree.

Now multiply by the scattering rate.

The reason laser cooling works is that you can pretty easily scatter millions of photons/sec with the D2 transition.

Even for scattering rates that are significantly smaller, it will have an effect when integrated over the course of ~10,000 sec

 

Quote

 

Hence the increase in translational k.e. will be:

1/2 . 4 x 10⁻²⁶ . ( 2.5 x 10⁻²)²

= 0.5 . 4 x 10⁻²⁶ .  6.25 x 10⁻⁴ = 1.25 x 10⁻²⁹ J

 

Not quite. That’s the KE of a Na atom moving 2.5 cm/s

If the atom were moving 250 m/s, you could stop it or double its speed with 10^5 scatters (if the atom were moving toward/away from the source. That doesn’t match up with your numbers.

[exchemist]

1 hour ago, swansont said:

The impact is from the recoil, which changes the KE, causing heating (or cooling under the right conditions)

Now multiply by the scattering rate.

The reason laser cooling works is that you can pretty easily scatter millions of photons/sec with the D2 transition.

Even for scattering rates that are significantly smaller, it will have an effect when integrated over the course of ~10,000 sec

 

 

Not quite. That’s the KE of a Na atom moving 2.5 cm/s

If the atom were moving 250 m/s, you could stop it or double its speed with 10^5 scatters (if the atom were moving toward/away from the source. That doesn’t match up with your numbers.

The figure derived from 2.5cm/secs is only (if my arithmetic is good) the increase in k.e. from a single absorption event, so yes, obviously to explore what happens in laser cooling is a lot more involved. But I was simply intrigued to see how significant the translational energy input is, compared to the energy of the photon, and what the effect on the temperature of a spectroscopic sample might be.

I suppose since k.e. goes with the square of the speed, the kinetic energy transfer from the same amount of momentum to a lighter molecule, say hydrogen, would be somewhat greater.   

  

Edited February 20, 2023 by exchemist