A Electron capture in a molecule

[BillKet]

Hello! If I have a (diatomic for simplicity) molecule containing a nucleus that decays by electron capture, are there any theoretical calculations of how that would behave in practice? For example would the lifetime change? Would the resulting molecule still be bound? For example if I start with BeF and go to LiF, would LiF still be bound or would Li and F end up as separate atoms? I know these would require advanced theoretical numerical calculations, which I am by no means able to do, so I was wondering if anyone came across this in the literature or they know what groups would be able to do such calculations. Thank you!

 

[Vanadium 50]

BeF₂ is ionic solid, as is LiF. What does it even mean for this to be unbound?

Certainly you will have a crystal defect there - Be is not Li.

The ionization scale is eV and the beta energy scale is MeV. So you expect a million ionizations. Which million is probabilistic.

 

[BillKet]

BeF₂ is ionic solid, as is LiF. What does it even mean for this to be unbound?

Certainly you will have a crystal defect there - Be is not Li.

The ionization scale is eV and the beta energy scale is MeV. So you expect a million ionizations. Which million is probabilistic.

Actually my question was about BeF (monofluoride) and it was for the case in which both BeF and LiF are in gas phase. For example, say you trap $10^6$ BeF molecules (all containing $^7$Be, the details of how you achieve it in practice is not important for the question) in a MOT laser trap and one of Be decays. Will it turn into a LiF molecule, which will have some translational energy due to the decay energy (it will leave the MOT as the MOT works only for the BeF), or will I just end up with Li and F as separate atoms (also leaving the trap)?

 

[Vanadium 50]

Will it turn into a LiF molecule,

That's really more molecular physics, and arguably chemistry.

What happened to the other fluorine?

The question also needs to specify when - you have chemically active atoms near each other. Sooner or later something is going to happen.

 

[BillKet]

That's really more molecular physics, and arguably chemistry.

What happened to the other fluorine?

The question also needs to specify when - you have chemically active atoms near each other. Sooner or later something is going to happen.

What do you mean by other fluorine?

In this case we can ignore the other molecules, the distance between them will be too big for them to talk to each other and the resulting neutrino won't interact with them. Also, I said $10^6$ as an example, but my questions is about an ideally, isolated BeF molecule. Actually you can imagine that we have only one BeF molecule in ultra high vacuum (in practice this is how it would be, for example by creating a 2D molecular lattice with a single molecule at each lattice). What will be the final product once the Be turns to Li?

 

[Vanadium 50]

What do you mean by other fluorine?

You start with BeF₂, and the Be becomes Li. I now have two fluorine atoms, only one of which is needed to make LiF. So what happened to the other one?

 

[BillKet]

You start with BeF₂, and the Be becomes Li. I now have two fluorine atoms, only one of which is needed to make LiF. So what happened to the other one?

As I said before, you don't start with $BeF_2$ you start with BeF.

 

[InkTide]

EC is a bit unique in its susceptibility to chemical properties of the atom, especially for light elements like Beryllium, to the point that ionization can be used to extend the half life by reducing the number of electrons available for capture, up to the extreme of (for isotopes with insufficient decay energy for positron emission) shutting off the decay entirely with full ionization. Beryllium is so light that even non-ionization-related electron availability can have measurable impacts on Beryllium-7's half life, such as depending on whether the sample is surrounded by an electrically conductive or insulating material, but IIRC the difference there is a ~1% change. ⁷BeF₂ probably has a longer half-life than ⁷BeF, because of the strongly ionic nature of F bonds - F really likes to be F^-.

Following the capture, you're left with ⁷LiF unless the capture imparts enough energy to dislodge the LiF bond, in which case the bond may form again rather quickly unless you can get the Li atom very far away very quickly. EC usually involves an inner electron, so direct radiolysis of the BeF bond by capturing the electron bonded to the fluorine atom is less likely, the way I understand it, so even if the Li atom is pushed away, I suspect it's likely Li⁺, and the F atom is F^-, adding to the molecular-scale attractive forces.

I know it was just an illustrative simplification, but what's probably most impressive here is the isolation of large quantities of BeF without the formation of BeF₂. A better diatomic example would have probably been Beryllium Oxide, BeO.

 

[BillKet]

EC is a bit unique in its susceptibility to chemical properties of the atom, especially for light elements like Beryllium, to the point that ionization can be used to extend the half life by reducing the number of electrons available for capture, up to the extreme of (for isotopes with insufficient decay energy for positron emission) shutting off the decay entirely with full ionization. Beryllium is so light that even non-ionization-related electron availability can have measurable impacts on Beryllium-7's half life, such as depending on whether the sample is surrounded by an electrically conductive or insulating material, but IIRC the difference there is a ~1% change. ⁷BeF₂ probably has a longer half-life than ⁷BeF, because of the strongly ionic nature of F bonds - F really likes to be F^-.

Following the capture, you're left with ⁷LiF unless the capture imparts enough energy to dislodge the LiF bond, in which case the bond may form again rather quickly unless you can get the Li atom very far away very quickly. EC usually involves an inner electron, so direct radiolysis of the BeF bond by capturing the electron bonded to the fluorine atom is less likely, the way I understand it, so even if the Li atom is pushed away, I suspect it's likely Li⁺, and the F atom is F^-, adding to the molecular-scale attractive forces.

I know it was just an illustrative simplification, but what's probably most impressive here is the isolation of large quantities of BeF without the formation of BeF₂. A better diatomic example would have probably been Beryllium Oxide, BeO.

Thanks a lot for the reply! Actually I am curious about the last comment. Would $BeF_2$ actually form that easily, even at a relatively low density (which is the case in a MOT). I know there are efforts to MOT trap BaF and RaF so I would expect the same issue there, too, then, no? Or is the tendency to capture a fluorine much bigger for lower mass molecules.

 

[Vanadium 50]

So, as I understand it, you take an ionic solid, vaporize it, then put it in some kind of trap, and by the way, it's not the stable fluoride, it's an unstable one, and want to know if there is a calculation of the ionization probability after the nucleus undergoes EC.

I think the answer is "probably not" and "if there were an answer even under slightly different conditions, it would not apply."

 

[InkTide]

Thanks a lot for the reply! Actually I am curious about the last comment. Would $BeF_2$ actually form that easily, even at a relatively low density (which is the case in a MOT). I know there are efforts to MOT trap BaF and RaF so I would expect the same issue there, too, then, no? Or is the tendency to capture a fluorine much bigger for lower mass molecules.

I'm hardly an expert on it but most of what I've read about fluorides suggests that in systems with molar excess of the non-fluorine atom "X" that forms a difluoride or something with even more fluorines, XF_>1 + X tends to be more stable than XF (it's something that happens in transition metals, so I'd suspect alkali earths are even more susceptible, since they're already predisposed to +2 oxidation states). Some notable exceptions would be other halogens, which can sometimes form monofluorides. Lighter atoms may be more susceptible to momentum transfer from the decay, but they are also more susceptible to ionic attractive forces for the same reason - I doubt they'd behave all that differently.

It is partly a chemistry question, partly a nuclear question, but the nuclear part is just the scale of the decay energy, which has to be relatively low to favor EC in the first place. AFAIK, most of the energy released goes into the neutrino, not atomic recoil. It probably doesn't break the bond, but to theoretically model it you'd need to look at the decay energy and any potential trends in direction for the resulting momentum of the daughter atom - if they're all of insufficient magnitude to break the LiF bond, it won't get broken (outside, I suppose, the infinitesimally likely possibility of the neutrino being emitted directly into the fluorine nucleus), and if some are able to, the likelihood of it happening will be the likelihood of those particular recoil vectors.

 

[TeethWhitener]

(NB, BeF in the gas phase is extremely well-studied, and has been since at least 1931 by Mulliken (NIST). I imagine, given their post history, that @BillKet is already aware of this.)

Beryllium is an interesting case. Its electrons only occupy s subshells, which have nonzero electron density at the nucleus. This means that electron capture could feasibly come from one of the 1s core electrons or from one of the 2s valence electrons. Naively, this will be true even in a diatomic like BeF. If the capture comes from a core electron, then you need to consider Auger and/or shakeup-type processes where emitted electrons or photons leave behind a highly vibrationally excited molecule and could lead to dissociation. But if the capture comes from a valence electron, I’m not sure any of that is true, and you would almost certainly have a stable LiF molecule.

Ohtsuki et al showed that Be-7 trapped inside fullerenes (observed to decay at a faster rate because of the increased electron density at the Be nucleus) did not escape from the fullerene cages appreciably IIRC, which means that the total recoil energy imparted to the Be was likely less than a few hundred kJ/mol. Compare that the the $\Delta H_f =-340kJ/mol$ of gas phase LiF, and I’m guessing the molecule will stay intact.

To answer the OP’s question from post #1 succinctly, a lot of studies have been done on calculating the effects of chemistry on electron capture, but somewhat less seems to have been done on the effect of electron capture on chemistry.

 

[BillKet]

So, as I understand it, you take an ionic solid, vaporize it, then put it in some kind of trap, and by the way, it's not the stable fluoride, it's an unstable one, and want to know if there is a calculation of the ionization probability after the nucleus undergoes EC.

I think the answer is "probably not" and "if there were an answer even under slightly different conditions, it would not apply."

Actually the way it is done in practice is a bit different (at least in our lab), but the result is yes, you end up with BeF. What do you mean by unstable fluoride?