How do Atomic Nuclei 'know' what the Temperature is?

[sethoflagos]

I was reading up on the equilibrium relationship of ortho and para (molecular) hydrogen (ortho having proton spins in parallel alignment, para ant-parallel). 

With para hydrogen being the lower energy state, relatively pure para hydrogen (~99.8%) can be obtained by chilling pure hydrogen down to around 20 K and keeping it there for a couple of days. 

If this is rapidly heated to a more moderate temperature and then kept fully insulated, the gas will slowly lose temperature as the the appropriate ortho/para equilibrium is established. Essentially an endothermic reaction. Though it can take several days to do this.

It's sort of a remarkable behaviour in itself but it got me wondering exactly how the protons manage to extract energy from the bulk internal energy of the gas. I'm guessing the surrounding electron orbitals aren't the insulating cushions I'd pictured them to be.

[Lorentz Jr]

Bulk motion of nuclei is simple enough. That's just electromagnetic interactions between the protons and electrons and other nuclei. The hard part would be internal excitations of the nucleus. The short answer may be that they don't know for a while. Maybe a long while. The nuclear forces are short-range, so I wouldn't expect much from them.

Edited March 1, 2023 by Lorentz Jr

[exchemist]

1 hour ago, sethoflagos said:

I was reading up on the equilibrium relationship of ortho and para (molecular) hydrogen (ortho having proton spins in parallel alignment, para ant-parallel). 

With para hydrogen being the lower energy state, relatively pure para hydrogen (~99.8%) can be obtained by chilling pure hydrogen down to around 20 K and keeping it there for a couple of days. 

If this is rapidly heated to a more moderate temperature and then kept fully insulated, the gas will slowly lose temperature as the the appropriate ortho/para equilibrium is established. Essentially an endothermic reaction. Though it can take several days to do this.

It's sort of a remarkable behaviour in itself but it got me wondering exactly how the protons manage to extract energy from the bulk internal energy of the gas. I'm guessing the surrounding electron orbitals aren't the insulating cushions I'd pictured them to be.

I would imagine it has to be via emission and absorption of microwave radiation corresponding to the energy of the transition between the two states, as in nmr. 

Later note: Sorry, no, it can't be. The probability of spontaneous emission at RF frequencies is negligible, because of the ν³ rule. It must be the presence of fluctuating external fields from neighbouring molecules that does it, cf. spin-lattice relaxation in nmr. (This was all a long time ago so my recollection is hazy, to say the least.)

 

Edited March 1, 2023 by exchemist

[swansont]

The proton and the electron spins can be aligned or anti-aligned, and you can have correlations of the electron spins. That’s probably how you get the nuclear spin alignment. The lower ground state is F=0, so that’s anti-aligned electron and proton spins. F=1 is higher in energy, corresponding to 1420 MHz (the hyperfine splitting)

Cold atoms tend to be in the lower hyperfine state.

[sethoflagos]

50 minutes ago, exchemist said:

I would imagine it has to be via emission and absorption of microwave radiation corresponding to the energy of the transition between the two states, as in nmr. 

Later note: Sorry, no, it can't be. The probability of spontaneous emission at RF frequencies is negligible, because of the ν³ rule. It must be the presence of fluctuating external fields from neighbouring molecules that does it, cf. spin-lattice relaxation in nmr. (This was all a long time ago so my recollection is hazy, to say the least.)

 

It may be relevant that the transition rate can be sped up by a factor of twenty or so in the presence of an iron catalyst. The immediate assumption would be that some sort of electrochemical mechanism was in play, but perhaps it's more magnetic than chemical.

[exchemist]

18 minutes ago, sethoflagos said:

It may be relevant that the transition rate can be sped up by a factor of twenty or so in the presence of an iron catalyst. The immediate assumption would be that some sort of electrochemical mechanism was in play, but perhaps it's more magnetic than chemical.

How interesting. Molecular hydrogen can get adsorbed onto iron compounds and may dissociate on the surface into atoms. If it does that, then I would expect it to prefer to pair up with an atom of opposed spin before desorbing again, since that has the lower of the two energy states. But I'm just guessing about the mechanism. 

[studiot]

Here is some more information on Ortho v Para from Greenwood and Earnshaw  Chemistry of the elements.

Sorry for the quality of the scan but it is difficult for this book as it has over one and a half thousand pages.

 

[exchemist]

17 minutes ago, studiot said:

Here is some more information on Ortho v Para from Greenwood and Earnshaw  Chemistry of the elements.

Sorry for the quality of the scan but it is difficult for this book as it has over one and a half thousand pages.

 

Thanks, that seems to support my bond breaking hypothesis then.

[sethoflagos]

45 minutes ago, exchemist said:

How interesting. Molecular hydrogen can get adsorbed onto iron compounds and may dissociate on the surface into atoms. If it does that, then I would expect it to prefer to pair up with an atom of opposed spin before desorbing again, since that has the lower of the two energy states. But I'm just guessing about the mechanism. 

So you go with electrochemical. Makes sense.

A smidgen more:

Wikipedia tells me that the energy transition from para to ortho is 1.455 kJ/mol. Dividing by Avogadro's number and Planck's constant gives a frequency of 3.646 THz which just about crawls into far infra-red, I think (if I've done it right. Not a calculation I often get asked to do!).

Also, I find it intriguing that it's impossible to obtain by heating an ortho/para ratio that's higher than 3:1. Is it purely coincidental that ortho is a triplet state and para a singlet?

PS : Apologies - @studiot's attachment is beyond the capabilities of my prescription lenses!

Edited March 1, 2023 by sethoflagos

[studiot]

7 minutes ago, sethoflagos said:

PS : Apologies - @studiot's attachment is beyond the capabilities of my prescription lenses!

No apology needed. If it is really inadequate the please do speak up.

But do you realise that ScienceForums offer the full sized scsn by clicking on the pictures ?

In my browser (firefox on and old XP Dell) this opens a new browser window whcih you can zoom to full size.

 

 

[sethoflagos]

12 minutes ago, studiot said:

No apology needed. If it is really inadequate the please do speak up.

But do you realise that ScienceForums offer the full sized scsn by clicking on the pictures ?

In my browser (firefox on and old XP Dell) this opens a new browser window whcih you can zoom to full size.

 

 

That's neat! Thanks!

[swansont]

2 hours ago, exchemist said:

I would imagine it has to be via emission and absorption of microwave radiation corresponding to the energy of the transition between the two states, as in nmr. 

Later note: Sorry, no, it can't be. The probability of spontaneous emission at RF frequencies is negligible, because of the ν³ rule. It must be the presence of fluctuating external fields from neighbouring molecules that does it, cf. spin-lattice relaxation in nmr. (This was all a long time ago so my recollection is hazy, to say the least.)

 

But such transitions could be induced in collisions

[sethoflagos]

1 hour ago, swansont said:

But such transitions could be induced in collisions

Getting back to the OP, how are the momentum and energy transfers in a collision actually transferred to the nucleus? 

I can picture electrostatic repulsion and possibly even the Pauli exclusion principle acting on the electrons, but I struggle to picture the corresponding forces of repulsion acting on the positively charged nucleus.

Okay, I admit it, I've not the faintest idea of what keeps a nucleus centralised within the atom.   

[swansont]

2 minutes ago, sethoflagos said:

Getting back to the OP, how are the momentum and energy transfers in a collision actually transferred to the nucleus? 

I can picture electrostatic repulsion and possibly even the Pauli exclusion principle acting on the electrons, but I struggle to picture the corresponding forces of repulsion acting on the positively charged nucleus.

Okay, I admit it, I've not the faintest idea of what keeps a nucleus centralised within the atom.   

The electrons attract nucleus. If you move the electron cloud around, it will drag the nucleus with it. 

 

[sethoflagos]

40 minutes ago, swansont said:

The electrons attract nucleus. If you move the electron cloud around, it will drag the nucleus with it. 

How does this work? It seems to suggest that the electrons further away from the nucleus must exert more attractive force than those closer. Or is it a case of 'more electrons behind than in front'? If so, I'm struggling a little with the inverse square aspect of Coulomb's Law.

Edited March 2, 2023 by sethoflagos

[exchemist]

8 hours ago, swansont said:

But such transitions could be induced in collisions

Yes....by creating the transitory fluctuations in magnetic field that I referred to.

6 hours ago, sethoflagos said:

Getting back to the OP, how are the momentum and energy transfers in a collision actually transferred to the nucleus? 

I can picture electrostatic repulsion and possibly even the Pauli exclusion principle acting on the electrons, but I struggle to picture the corresponding forces of repulsion acting on the positively charged nucleus.

Okay, I admit it, I've not the faintest idea of what keeps a nucleus centralised within the atom.   

The way I think of it (rightly or wrongly) is like this. If you consider one nucleus, it is experiencing a magnetic field from the other one in the molecule, so it partially aligns, either with against that field. Standard space quantisation. In a collision, a 3rd nucleus comes up, just as close as the one to which it is bonded, maybe closer. So what magnetic field does the first nucleus now see? Some sort of resultant, with different alignment and different field strength. So it will now try to align with that. But this state has only transitory existence, so its energy levels will be poorly defined (uncertainty principle). And then after the collision the situation reverts to what it was before. But as a result of this there is a probability that the nucleus does not come out of the interaction with the same orientation in which it entered.  

Regarding centralisation of the nucleus, the electron cloud is centred on the nucleus and if it moves, leaving the nucleus off-centre, the electron cloud becomes distorted, leading to a higher energy state, which is resolved by the nucleus re-centring itself.  

Edited March 2, 2023 by exchemist

[studiot]

11 hours ago, sethoflagos said:

That's neat! Thanks!

Glad it worked out for you!

It was late and I was rushed last night.

One of the very first books I bought at university back in 1968 was Denbigh's Chemical Equilibrium.

He gives a much wider treatment of these effects both from a fugacity/Chemical Potential  point of view (P1`22 - 132) and also form a statistical/calorimetric entropic point of view.

Here are some pages of the latter which may be of interest since they cover a much wider range of gases that just hydrogen.

The note at the bottom of p420 refers to Glasstone - an earlier work from 1940 - on page 588
My 1951 second ed has that material at the beginning pages 94 - 97 entitled  Ortho and Para States.
This may be more accessible for you as it is american.

In deference to your opthalmology I have increased the scan deoth, although it means a larger file size.

2 hours ago, exchemist said:

Regarding centralisation of the nucleus, the electron cloud is centred on the nucleus and if it moves, leaving the nucleus off-centre, the electron cloud becomes distorted, leading to a higher energy state, which is resolved by the nucleus re-centring itself.  

Yes I agree, the standard chemist's method of avoiding mechanical mechanisms is to go the least energy/ least action route.

 

 

Another good reference is

Wilson Thermodynamics and Statistical Mechanics 1966 

Chapter 6 Specific Heats

p 139ff

6.12 The Rotational Specific Heat of Hydrogen

 

Edited March 2, 2023 by studiot

[exchemist]

20 minutes ago, studiot said:

Glad it worked out for you!

It was late and I was rushed last night.

One of the very first books I bought at university back in 1968 was Denbigh's Chemical Equilibrium.

He gives a much wider treatment of these effects both from a fugacity/Chemical Potential  point of view (P1`22 - 132) and also form a statistical/calorimetric entropic point of view.

Here are some pages of the latter which may be of interest since they cover a much wider range of gases that just hydrogen.

The note at the bottom of p420 refers to Glasstone - an earlier work from 1940 - on page 588
My 1951 second ed has that material at the beginning pages 94 - 97 entitled  Ortho and Para States.
This may be more accessible for you as it is american.

In deference to your opthalmology I have increased the scan deoth, although it means a larger file size.

Yes I agree, the standard chemist's method of avoiding mechanical mechanisms is to go the least energy/ least action route.

 

 

Another good reference is

Wilson Thermodynamics and Statistical Mechanics 1966 

Chapter 6 Specific Heats

p 139ff

6.12 The Rotational Specific Heat of Hydrogen

 

The problem with trying to tackle it mechanically is that means quantum mechanically.

What has happened is the wave functions of the orbital have become distorted by the nucleus being off-centre. One can't really speak of nice neat forces, acting between the nucleus and electrons as particles, in this scenario. So the energy approach, which is what the Hamiltonian does in Schrödinger's equation, seems to be the only way to describe what happens, so far as I can see.

I'm not sure how the heat capacity stuff relates to what we have been discussing. Can you elucidate? 

[sethoflagos]

1 hour ago, exchemist said:

The problem with trying to tackle it mechanically is that means quantum mechanically.

What has happened is the wave functions of the orbital have become distorted by the nucleus being off-centre. One can't really speak of nice neat forces, acting between the nucleus and electrons as particles, in this scenario. So the energy approach, which is what the Hamiltonian does in Schrödinger's equation, seems to be the only way to describe what happens, so far as I can see.

I'm not sure how the heat capacity stuff relates to what we have been discussing. Can you elucidate? 

The mental picture I'm getting is that the electron fields recoil first from the collision due in part to their much lower inertia. The consequent relative shift of -Ve potential away from the point of impact then acts on the nuclei and as @swansont says, drags them along.

In the extreme case, this mechanism fails and the protons and electron seperate leading to a plasma, which we know happens so that's consistent. It's a classical picture, but there's room for a buch of vitual photons to be showering the nuclei to create some sort of Feynman diagram out of it.

As regards the isomeric transition perhaps there's some really narrow window in the collision spectrum where the colliding nuclei are sufficiently closely aligned for an exchange of spin states to occur? The energy transition is in the same ball park as the latent heat of vapourisation (1.445 kJ/mol vs 0.904 kJ/mol) so the kinetics seem credible even if the actual mechanism is opaque.

In passing, I find it intriguing that such a quantum oriented phenomenon can have such a marked impact on bulk thermodynamic properties. The anomalous behaviour of the specific heat of hydrogen at low temperatures has been known for a long time of course. It's a classic case of Gibb's paradox: the tiniest imaginable difference between particle species is sufficient to cause a significant step change in macroscopic entropy. 

[swansont]

10 hours ago, sethoflagos said:

How does this work? It seems to suggest that the electrons further away from the nucleus must exert more attractive force than those closer. Or is it a case of 'more electrons behind than in front'? If so, I'm struggling a little with the inverse square aspect of Coulomb's Law.

The electrons repel each other, too. Any perturbation of the electrons that distort the cloud leaves a charge imbalance. I don’t see why the inverse square law is a problem. 

 

4 minutes ago, sethoflagos said:

In the extreme case, this mechanism fails and the protons and electron seperate leading to a plasma

Yes, but thermal energy is typically much smaller than the ionization energy, so that’s not a problem unless the temperature is quite high.

[studiot]

1 hour ago, exchemist said:

I'm not sure how the heat capacity stuff relates to what we have been discussing. Can you elucidate?

Very quickly before I head for the bank

Part 3 of Callen   Thermodynamics and Thermostatistics

pages 455 - 471

Callen goes through modern interpretation of the quantum angular momentum rules via Noethers theorem (symmetries), Goldstone's theorem (broken symmetries) and gauge symmetries in relation to accessible and inaccessible microstates of a system and the state space of a system.

The quantum rules forbid certain transfers of angular momentum which leaves ortho and para hydrogen as essentially different gases in a mixture.

[John Cuthber]

"How do Atomic Nuclei 'know' what the Temperature is?"
They don't.
They get batted about by the laws of physics and we interpret the way in which they behave as meaning that they have a temperature.

 

[exchemist]

1 hour ago, studiot said:

Very quickly before I head for the bank

Part 3 of Callen   Thermodynamics and Thermostatistics

pages 455 - 471

Callen goes through modern interpretation of the quantum angular momentum rules via Noethers theorem (symmetries), Goldstone's theorem (broken symmetries) and gauge symmetries in relation to accessible and inaccessible microstates of a system and the state space of a system.

The quantum rules forbid certain transfers of angular momentum which leaves ortho and para hydrogen as essentially different gases in a mixture.

For a while at least. Though from what I read, the rotational J=1 state of ortho does eventually relax to J=0 as the ortho gradually converts to para, by the various non-radiative means of relaxation we have been discussing.  

[sethoflagos]

4 hours ago, exchemist said:

For a while at least. Though from what I read, the rotational J=1 state of ortho does eventually relax to J=0 as the ortho gradually converts to para, by the various non-radiative means of relaxation we have been discussing.  

I found the discussion of J state populations a bit hard to follow. Is the 75% ortho equilibrium limit simply due to equipartition amongst these extra rotational degrees of freedom, or is there more to it than that?

Edited March 2, 2023 by sethoflagos

[exchemist]

1 hour ago, sethoflagos said:

I found the discussion of J state populations a bit hard to follow. Is the 75% ortho equilibrium limit simply due to equipartition amongst these extra rotational degrees of freedom, or is there more to it than that?

Hmm, I see what you mean. There are no extra degrees of freedom, though. Diatomic molecules all have 2 rotational degrees of freedom. But ortho can only populate odd numbered rotational energy levels while para can populate only even levels. I had to look this up (it's badly explained or not explained in Wiki) but it appears the issue is that ortho hydrogen is a triplet state, in which the total nuclear spin of 1 can be orientated +1, 0 or -1 with respect to the axis of rotation, thereby multiplying the numbers of rotational states available by 3, i.e. each rotational level has 3-fold degeneracy, whereas the para states do not. So at RTP, with kT>> ε for rotation, you end up with a  3:1 ratio, just because there are more ways for ortho to have a certain amount of rotational energy.

I think that's it, at least.