From the Bohr formula the Rydberg energy of the muonic hydrogen would be some 200 times larger than regular hydrogen. Anyone know how that plays a role?
It can't be sqrt(spring/mass) for vibration since the proton is anyway already some 2000x heavier than the electron. Unless spring somehow depended on the Rydberg energy, which is possible since the P.E-K.E would depend on mass via the K.E.
The decrease in potential energy upon vdW bond formation in reactions with typical hydrogen isotopes dominates the negligible gains in vibrational zero-point energy from reactants to products. With the muonic hydrogen substitution, the authors claim instead that the driving force for "bond formation" results from a decrease in zero-point energy which compensates for the expected losses in potential energy.
I have a few published articles in quantum chemistry, and I barely can understand your explanation. It feels right, but I think I need to take 30 minutes to try to understand the details. Can you explain this like I'm a graduate student with only 3 years in the university, please?
The authors of the article in topic provide a more coherent explanation than I could ever articulate:
>> Conventionally, the formation of chemical bonds is due to a decrease in potential energy (PE), often accompanied by small increases in vibrational zero point energy (ZPE). In principle, this basic mechanism can be completely reversed, wherein chemical bonds may even be formed by an increase in PE if there is a sufficiently compensating decrease in vibrational ZPE, giving rise to what has been coined “vibrational bonding” of molecules stabilized at saddle-point barriers on a potential energy surface (PES), far away from potential minima.
Thanks. But I think your previous comment has an interesting point about why muons are different than electrons. I'm not sure because I hadn't made the calculations, so any confirmation or refutation is welcome. Let's try:
When two normal molecules, with electrons, are close, they can form different kind of bonds. The weakest bond is the "van der Waals" bond. It's caused because the electrons of the molecules change their position slightly due to the presence of the other molecule.
In this experiment they only replace the electron of a hydrogen atom by a muon. The muons have much more mass than the electrons, so the radius of the orbit is much smaller. (They are quantum particles, so they don't have orbits, but please forgive this technical detail.) As the orbits are smaller, the displacement caused by the other molecule is smaller, so the van der Walls force is smaller.
In the normal (electron) case the van der Walls force cause the formation of the intermediate molecule. In this case (muon) the van der Waals forcé is so weak that other effects are more important.
[I left out the part about zero point energy. It's also interesting but this explanation is becoming larger than the article :) .]
The role of muon instead of proton could be: (i) to make the bond detectable as the muon decays, or (ii) 10X lighter mass of muonium compared to regular hydrogen leads to a dynamical regime with saddle points in P.E vs K.E (as pointed out earlier).
> chemists experimenting at a nuclear accelerator in Vancouver observed that a reaction between bromine and muonium—a hydrogen isotope—slowed down when they increased the temperature
Isn't saying muonium is "a hydrogen isotope" sort of like saying a car is "a horse isotope"?
The antimuon is "proton-like" for a lot of purposes: it has a +1 charge and it's much heavier than an electron. So you've got a heavy bit with +1 charge and an electron orbiting it, you call it hydrogen.
It would have been very nice had the article said that, and described what "muonium" is. I took it to be a proton plus a muon, not an anti-muon plus an electron.
To describe it as "a hydrogen isotope" is just about the optimal mix of uninformative and misleading!
It's probably as close as you can get to accuracy while still keeping those who passed their high school chemistry classes in your audience.
For the purposes of explanation, the antimuon-electron pair is chemically similar enough to a proton-electron pair to call it an isotope. It has a +1 massive center, with a probability distribution for a less-massive -1 charge cloud.
Deuterium and tritium, two true isotopes of hydrogen, increase the mass of the center by adding neutrons. An antimuon is, I think, about 10% the mass of a proton. So muonium would act a lot like extra-light hydrogen for the instant before it decays.
It would have been clearer to call it an "exotic atom, chemically similar to hydrogen", though. That wouldn't conflict with the common definition of isotope as an atom with the same number of protons as the reference atom, but with a different number of neutrons.
I'm actually not aware of any exotic atoms that are not chemically similar to hydrogen or antihydrogen, as getting an exotic helium-like atom would require forming a +2 exotic center and making two -1 charges interact with it before it decays. That is likely beyond our current capabilities.
That's funny, because despite having no clue what muonium is, I instantly knew that line was typical science journalism bullshit. Am I too cynical or does dumbing down something to the point it makes no sense have actual value for some people?
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[ 2.9 ms ] story [ 66.3 ms ] threadFrom the Bohr formula the Rydberg energy of the muonic hydrogen would be some 200 times larger than regular hydrogen. Anyone know how that plays a role?
It can't be sqrt(spring/mass) for vibration since the proton is anyway already some 2000x heavier than the electron. Unless spring somehow depended on the Rydberg energy, which is possible since the P.E-K.E would depend on mass via the K.E.
>> Conventionally, the formation of chemical bonds is due to a decrease in potential energy (PE), often accompanied by small increases in vibrational zero point energy (ZPE). In principle, this basic mechanism can be completely reversed, wherein chemical bonds may even be formed by an increase in PE if there is a sufficiently compensating decrease in vibrational ZPE, giving rise to what has been coined “vibrational bonding” of molecules stabilized at saddle-point barriers on a potential energy surface (PES), far away from potential minima.
When two normal molecules, with electrons, are close, they can form different kind of bonds. The weakest bond is the "van der Waals" bond. It's caused because the electrons of the molecules change their position slightly due to the presence of the other molecule.
In this experiment they only replace the electron of a hydrogen atom by a muon. The muons have much more mass than the electrons, so the radius of the orbit is much smaller. (They are quantum particles, so they don't have orbits, but please forgive this technical detail.) As the orbits are smaller, the displacement caused by the other molecule is smaller, so the van der Walls force is smaller.
In the normal (electron) case the van der Walls force cause the formation of the intermediate molecule. In this case (muon) the van der Waals forcé is so weak that other effects are more important.
[I left out the part about zero point energy. It's also interesting but this explanation is becoming larger than the article :) .]
The role of muon instead of proton could be: (i) to make the bond detectable as the muon decays, or (ii) 10X lighter mass of muonium compared to regular hydrogen leads to a dynamical regime with saddle points in P.E vs K.E (as pointed out earlier).
Isn't saying muonium is "a hydrogen isotope" sort of like saying a car is "a horse isotope"?
I thought proton equivalency was the fundamental definition of an isotope, though perhaps that's just a lie you feed chemistry undergrads.
To describe it as "a hydrogen isotope" is just about the optimal mix of uninformative and misleading!
For the purposes of explanation, the antimuon-electron pair is chemically similar enough to a proton-electron pair to call it an isotope. It has a +1 massive center, with a probability distribution for a less-massive -1 charge cloud.
Deuterium and tritium, two true isotopes of hydrogen, increase the mass of the center by adding neutrons. An antimuon is, I think, about 10% the mass of a proton. So muonium would act a lot like extra-light hydrogen for the instant before it decays.
It would have been clearer to call it an "exotic atom, chemically similar to hydrogen", though. That wouldn't conflict with the common definition of isotope as an atom with the same number of protons as the reference atom, but with a different number of neutrons.
I'm actually not aware of any exotic atoms that are not chemically similar to hydrogen or antihydrogen, as getting an exotic helium-like atom would require forming a +2 exotic center and making two -1 charges interact with it before it decays. That is likely beyond our current capabilities.
Ha!
So, does that mean the idiotic idea behind that magic material from The Core called Unobtanium might actually be real?