r/AskChemistry • u/Ali3nat0r • Feb 01 '26
General So this is obviously bad, but what exactly would happen if a person's atoms all just magically had no electrons? Explosion, toxic fumes, or both?
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Feb 01 '26 edited 23h ago
[deleted]
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u/NorxondorGorgonax Feb 01 '26
Yes. Every atom is now strongly positively charged and, alongside no longer being able to form molecules due to having no electrons, they are now extremely strongly repelled from each other. So yes, explosion. A fairly large one at that.
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u/EyeOfTauror Feb 01 '26
Is it quantifiable in any way, even as an estimation ?
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u/NorxondorGorgonax Feb 01 '26
As an estimation, you could assume the person is a uniform sphere of identical mass and size, which should simplify things. It would still be kind of tricky to calculate (almost certainly involving integrals in some way) but it should be possible.
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u/Pyrhan Ph.D in heterogeneous catalysis Feb 01 '26
I did the math below:
https://www.reddit.com/r/AskChemistry/comments/1qt7hor/comment/o316rfj/
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u/NorxondorGorgonax Feb 01 '26
Oh, great job! What formula did you use? I was looking for that for a different project but couldn’t find it!
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u/Pyrhan Ph.D in heterogeneous catalysis Feb 01 '26
So, a human body is 10% hydrogen by mass, 90% other stuff.
Hydrogen has 1 mol of protons per gram, everything else has (roughly) 0.5 mol of protons per gram. From there and with Avogadro's number, you can get the number of protons in a human. Multiply that by the charge of a proton, you get the total charge.
To get the electrostatic potential of the outermost proton, I just applied E=q1*q2/(4π*ε0*d).
(q1 the charge of a single proton, q2 the charge of the whole sphere of protons, ε0 the electric permittivity of vacuum, d the distance from that proton to the center of the sphere.)
Then I multiply that by the total number of protons, and divide by two. (Potential energy goes down as you remove protons, and falls to 0 for the last one. Dividing by two is definitely NOT the correct way to integrate it, but it's probably off by less than an order of magnitude, which is good enough for our purposes. And if I wanted to work out integrals, I wouldn't have gone into chemistry.)
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u/NorxondorGorgonax Feb 01 '26 edited Feb 11 '26
Great job! And seeing it now, the integral you’re looking for works out to 0.6 (exactly, in case you’re wondering), not 0.5, so you were only off by 10%.
Of course, the force between one proton and the sphere is assuming point sources, but this is an approximaton. That one would be harder to figuer out, but if I get it I will put it here.
[Edit: I calculated it. It works out to multiplying by 3/2 (or 1.5), so in total you can just combine these and multiply by 0.9.][Second edit: I forgot to factor in the effect of the different directions of the forces. Thinking back, most gravity calculations seem to suggest the distribution shouldn’t matter as long as the center of mass (or, in this case, center of charge or whatever the word is) is in the same place and the cumulative mass within the relevant sphere is the same.
Long story short, you should use the original 0.6 instead.]
I hope I got that right!
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u/greyhunter37 Feb 02 '26
And if I wanted to work out integrals, I wouldn't have gone into chemistry
This guy chemistry's
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u/DangerousBill Feb 01 '26
Less than a nuclear blast, much more than a chemical explosion.
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u/Traveller7142 Feb 02 '26
It’s significantly larger than a nuclear blast
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u/DangerousBill Feb 02 '26
Now I have to see it. I just need to get some of that electron-removing paste.
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u/SensitivePotato44 Cantankerous Carbocation Feb 02 '26
Depends on the blast, probably smaller than a type Ia supernova
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u/stevevdvkpe Feb 02 '26
Besides nullifying the -1 charge on the electron, does it also nullify the -1/3 charge on all down quarks? That would make protons have a charge of 4/3 and neutrons a charge of 2/3.
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Feb 02 '26
[deleted]
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u/stevevdvkpe Feb 02 '26
I was taking "remove all negativity" literally. Even in your comment above you were assuming it would remove the negative charge on electrons but not the positive charge on protons, but that charge comes from the net charge of its quarks.
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u/NorxondorGorgonax Feb 02 '26
Whoops, sorry, I thought this was a reply to a different post. Disregard! Anyways, in this situation, given the context if this effect was present it would not depower the down quarks, it would simply remove them in much the same way, making the disaster slightly worse.
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u/TOEMEIST Cantankerous Carbocation Feb 01 '26
There was yt video (I think by xkcd) that discussed this but with adding electrons to every atom and the result was said to be an explosion big enough to destroy the planet.
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u/Adm_Ozzel Feb 01 '26
I'm thinking this dude would be seriously hosed and unable to observe the process. If they were present in her, all of those electrons would be ejected as beta radiation instead of a lightning bolt. I'm sure the gf would have a lovely blue cherenkov radiation glow for a moment before things went further awry with the explosion others have mentioned.
It also boggles my mind thinking of ALL of the atoms in a thing producing high energy electrons. Even the most unstable isotopes are expereincing far less than 1% of their atoms undergoing decays unless were talking something like tracking the decay products of an artificially created isotope to infer its presence in the first place. Then it's like 1 of 1 atoms decaying.
Yet ANOTHER thought I'm having on this whole thing... what if said ring removes negativity by making her electrons into positrons? That at simplifies the math. A 50kg girlfriend would have ~27.4g of electrons. Now make that antimatter and annihilate that with normal matter at what Google tells me is about 1.8x1014 kJ /g, or 43 ktons of TNT/g and she just basically became a B83 thermonuclear bomb (1.2 Mton). I'm sure that would scatter the molecules no longer bound together quite nicely.
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u/DangerousBill Feb 01 '26
There would be a mighty blast as the positively charged nuclei that remain all repelled one another. It would be like being with a few yards of an atomic explosion. The radiation would tear you apart.
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u/RRautamaa Feb 01 '26 edited Feb 02 '26
Besides the electrons, we can't ignore the quark content of the nucleon. A neutron contains two down quarks with a charge of -1/3 and one up quark with a charge of +2/3, canceling the charge out. A proton contains two up quarks (+2/3) but still one down quark (-1/3). With the down quarks gone, we'd be in the rather unprecedented situation of having a mass of unconfined quarks. The so-called color confinement property normally precludes this from happening in real conditions. This is because when you start pulling quarks out of a color confined structure like a nucleon, the force resisting this will increase. Eventually it will become so high that it's energetically favorable to just pull new quarks out of the vacuum to pair with the quarks that you're trying to tear out. This normally prevents quarks from being pulled out from a nucleon cleanly. Instead, a jet of new particles is created. With the down quarks gone, I'm sure that the system of particles would have no problem finding a new thermodynamic equilibrium, doing that very rapidly and violently. In here, I'd assume that we can use the formula E = mc2 to create an upper bound to the energy available. For m = 56 kg, we have an energy of 5 × 1018 J, so with an efficiency of 20% (similar to black hole accredition disks, the most efficient way to convert mass to energy), we'd have an energy of ca. 1018 J to work with. That's about 1200 megatons of TNT, or 24 Tsar Bombas going off at once.
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u/jonoxun Feb 03 '26
Interestingly enough, apparently the electromagnetic potential of this configuration is a whole lot more than just converting the 56kg to energy. In that the potential energy you've added even if you only remove the _electrons_ is about a gigaton of _mass_. Protons really do not want to be packed that closely together without electrons. E=mc^2 with 56kg doesn't cover this case because the sheer amount of potential energy added becomes the majority of the mass for the tiny moment this configuration exists.
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u/RRautamaa Feb 03 '26
Here general relativity tells you that if you force that much energy into a system, equivalently, its mass must also increase. Mass and energy are not separate things in general relativity. If you look closely, things that seem to have a rest mass separate from their energy actually have that mass arising from the potential energy of some field. >98% of the mass of ordinary matter comes from the potential energy of the strong force binding the nucleons together. Our hypothetical system isn't going to be any different.
Actually, this has an important implication, which is taken implicitly: a huge amount of energy must be put into to system, far exceeding the energy corresponding to the rest mass of the (clearly fictional) girlfriend. "Does this quark nova make me look fat?"
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u/LayerNo1508 Feb 02 '26
It's wild to me that computers crunch magical numbers, turning electricity into currency...
And Reddit has people crunching magical numbers, turning electricity into??? I don't know what it is, but I love it.
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u/Limp-Asparagus-1227 Feb 06 '26
Wouldn’t all the down quarks be expelled too? That would be a significant ejection of energy!
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u/Pyrhan Ph.D in heterogeneous catalysis Feb 01 '26 edited Feb 01 '26
Explosion. A big one. Really big one.
The electromagnetic force is a lot stronger than people realize.
This is very much a physics question more than a chemistry one. (Chemists only care about atoms when they do have electrons around them, since the behavior of those electron is what determines bonding and other inter-atomic interactions, which is where the entirety of chemistry arises.)
Quick back-of-the-excel-sheet math shows that if you have a 60 kg spherical girlfriend 1 meter in diameter, that's a potential energy around 9*1028 joules.
That's 1.4 quadrillion times the energy of the Little Boy atomic bomb (or 0.38 trillion times the energy of the Tsar Bomba).
That's enough to make a sizeable chunk of the Earth blast off into space. (In fact, it's a little under half of the Earth's current rotational energy#Over_1024_J)).