Hi, I’ve been wondering this for quite a while and cannot seem to dig up the answer… here goes:
If anti-hydrogen were to come into contact with regular H2O, would the annihilation take place, leaving pure oxygen?
What about anti-noble gas, like anti-helium? Can it safely encounter regular matter with no reaction because it’s stable?
I really want to know these things-- please share what you know!
–Ray
No to both. Matter/antimatter mutual annihilation occurs at the subatomic level, with anti-electrons, anti-protons, and anti-neutrons all interacting with electrons, protons, and neutrons; the chemical properties of the atoms those particles are bound up in would be secondary at most.
Anti-matter atoms will destructively react with matter without regard to stability. The anti-electrons (positrons) in the shell of the anti-atom will electrostatically attract the electrons in the atom, and (regardless of chemical stability) will combine, annihilate, and release energy. Ditto the nuclei - the anti-protons will attract the protons and collision/annihilation will occur.
The reactions will not be neat and tidy, either. A molecule of anti-H[sub]2[/sub] will not react with a molecule of H[sub]2[/sub]O to leave O - there will be a reaction, energy and sub-atomic particles will fly off in all directions, and in the end, there will be enough mass for an O, but there will almost certainly not be an O just sitting there.
and ninja-ed by Derleth
Si
So anti-H can react with other atoms besides H?
All that matters is that anti-H is made up of an anti-electron and an anti-proton. The anti-electron can react (annihilate) with one electron, and the anti-proton can react (annihilate) with one proton. If you, say, dropped on anti-H into a cup of water, the anti-electron would very quickly annihilate with one of the electrons orbiting an H20 molecule. This would leave an ionized H20 molecule with a net positive charge, and an anti-proton, with a negative charge. The anti-proton might annihilate with any of the protons in the H20 molecule, leaving either H+O or H2 + a (stable) isotope of nitrogen.
Short answer: yes. It will basically annihilate with any normal-matter atom (but obviously stuff might be left over / pushed away depending on the specific atoms and the way they collide).
Also note that anti-atoms and particles can still be deflected by EM fields so it is possible to make an antimatter container. But simply a container of inert, normal matter is insufficient because positrons and electrons attract.
I was thinking this, but I think the negatively charged anti-proton will be repelled by the electrons in the filled non-bonding orbitals of the O atom. It will have to interact with a H[sup]+[/sup] (or H[sub]3[/sub]O[sup]+[/sup]) because it can’t get close to a nucleus otherwise (at low reaction energy, anyhow).
Si
I agree that in this case it is much more likely for the anti-proton to annihilate with an H’s proton rather than an O’s proton, but in principle the anti-proton from the anti-H doesn’t care whether it annihilates with a proton from an H or an O or any other atom. If, for example, we were talking about an anti-O dropped into a cup of H20 rather than an anti-H, the anti-protons and anti-neutrons in the anti-O would probably annihilate with those in the H and O somewhat more democratically.
EDIT: although in that case it would be a bigger mess, since a single anti-proton annihilation would scatter the other anti-protons and anti-neutrons before they could all together annihilate with those in a single H20 molecule
My only question is, how are you getting 20 Hydrogen atoms into a stable molecule with nothing else!?!
H20 is the chemical symbol for water: Look Around you - Water
Woosh.
His joke is that certain people in this forum have been writing it H20 not H2O. (That is, H-twenty instead of H-2-Oh).
Did you click the link at all?
But it is impossible to describe!
It’s lower-level than that, even, since protons and neutrons are made up of quarks: An anti-proton will also react with a neutron, producing either two neutral pions and a positive pion, or two positive pions and a negative pion. The pions, in turn, will then decay, into a pair of photons for the neutral one, or to a muon and a neutrino for the charged ones. The neutrinos will be effectively lost, and the muons, in turn, will eventually decay into more neutrinos and either an electron or a positron. And the positrons will probably find matching electrons before too long, and annihilate with them for more photons.
Even proton-antiproton isn’t always simple: In that case, you can end up with three neutral pions (and thus, eventually, nothing but photons), but you can also (and more likely) end up with one each positive, negative, and neutral, with results as described above.
People always talk about electrons and positrons as the textbook example of matter-antimatter annihilation, and thus folks end up thinking of that as typical, but electron-positron is only the textbook example because it’s so simple. Most matter-antimatter reactions are really, really messy.
I’ve seen it before, it’s some British spoof science show. I don’t see how that’s relevant to what I posted though, other than you got whooshed by a joke so you responded by posting another joke?
Whatever.
So you didn’t click the link. The narrator says “Water – chemical symbol H-twenty…”.
First thing I thought of as well, though it helps that I have their fake periodic table in plain sight.
I understand that - once the electron/positron pair annihilate, if the antiproton hits a nucleus it will be messy in all sorts of ways, but surely (for a cold antimatter interaction) wouldn’t electrostatic forces be dominant - i.e the antiproton would not get near an O nucleus due to the surrounding nonbonding electrons. I guess I am asking whether any other nuclear forces will come into play to trump the electrostatic repulsion.
Yup, right there between the elements red (Rd) and iodine (I).
Yes, electrostatic forces would be dominant.
To my understanding of the term here, I think that’s not a ninja but a beat-me-to-it.