Alternative particles in Nucleus?

[Algr]

Tony Stark: I've tried every element.
Nick Furry: You haven't tried all of them.

Are there any other particles besides protons and neutrons (and their antis) that could be stable in the nucleus of an atom?

 

[mfb]

Strangelets might be possible, but there is no evidence for their existence.

 

[newjerseyrunner]

There are other types of "atoms" but not like that. For example: an electron and positron can go into a long stable orbit with each other.

 

[rkolter]

f your goal is to create a new element that Tony Stark hadn't tried by adding different particles to existing elements to create stable new elements - that's not how elements work. Elements are defined by the number of protons they contain - Period. if you are not adding or removing protons from the nucleus, you are not creating a new element.

Tony Stark and his vast resources and near mythical technological prowess probably did try every element we can create and maintain for long enough to experiment with. Nick Fury was trolling Tony.

 

[snorkack]

How much is known about chemical properties of quarks?

One basic property of a quark: you cannot neutralize it, period. (Unless you have more quarks). A quark is always part of an ion.

Yet existence of quarks will not stop electrons from filling orbitals.

Boron has its individual properties from having a nucleus of charge +5 - they are not just an interpolation between beryllium and carbon.
Would nuclei of charge +1/3, +2(2/3) et cetera have their own chemical properties?

Also - elements are defined by charge of their nuclei. Positive integers from 1 to almost 100 are fairly well known.
Number of "protons" as such is irrelevant. If you could stabilize a sigma hyperon or some beautiful hyperon in a nucleus, it would be simply a strange, charmed, beautiful or truthful isotope of some known element.
What could be new is if you

1. stabilize a nuclear charge appreciably over +100
2. stabilize a fractional charge
3. create nuclei that possesses non-negligible strong interaction across internuclear distances (which coloured nuclei might have...)

 

[mfb]

A quark is always part of an ion.

What does that mean?
Do mean hadron? In that case, a quark gluon plasma is an exception.
Isolated quarks do not exist.

Yet existence of quarks will not stop electrons from filling orbitals.

Neither will the existence of bananas stop them. Because the two things are completely unrelated.

Would nuclei of charge +1/3, +2(2/3) et cetera have their own chemical properties?

They would probably be somewhere between existing elements. A +8/3 nucleus, if such a thing could exist, would have two tightly bound electrons like helium, and could attract a weakly bound third electron. It would probably behave a bit like lithium, but with weaker bonds because its charge as ion is lower. Make the charge 11/3 and you probably get an atom that can bind with 2 or 3 electrons remaining: like lithium or beryllium.

 

[snorkack]

What does that mean?
Do mean hadron? In that case, a quark gluon plasma is an exception.
Isolated quarks do not exist.

Cannot easily be produced and have not been found.
A quark cannot be destroyed for conservation of charge reasons alone (and is colour conserved?).

Neither will the existence of bananas stop them. Because the two things are completely unrelated.
They would probably be somewhere between existing elements. A +8/3 nucleus, if such a thing could exist, would have two tightly bound electrons like helium, and could attract a weakly bound third electron. It would probably behave a bit like lithium, but with weaker bonds because its charge as ion is lower. Make the charge 11/3 and you probably get an atom that can bind with 2 or 3 electrons remaining: like lithium or beryllium.

Yes - but with its own individual properties.
Plus the impossibility to neutralize. Lithium is easy to ionize, but it can attract a weakly bound third electron and become a neutral lithium atom. A nucleus with charge of 8/3 cannot: add a third electron and it is not a neutral atom, it is an anion.

 

[mfb]

Cannot easily be produced and have not been found.
A quark cannot be destroyed for conservation of charge reasons alone (and is colour conserved?).

Quarks can easily be produced, they have been found, and they can be destroyed. I have no idea what you are talking about.

They can be destroyed in direct annihilation (e. g. up + antiup -> gluon) or via the weak interaction (e. g. up + anti-down -> W+), and they can be produced via the reverse interactions.

Color is conserved, but there are anti-colors.

 

[snorkack]

Quarks can easily be produced, they have been found, and they can be destroyed. I have no idea what you are talking about.

They can be destroyed in direct annihilation (e. g. up + antiup -> gluon) or via the weak interaction (e. g. up + anti-down -> W+),

... all of which require availability of an antiquark.
Even an antiquark cannot be neutralized. World is full of quarks; but these quarks are in nucleons. Annihilate an antiquark with one of the quarks in a nucleon, and you have got rid of the antiquark... but you still have the two quarks of the three that the nucleon had. You still have fractional electric charge, and colour.

and they can be produced via the reverse interactions.

No, they cannot. Because of colour confinement.

Color is conserved, but there are anti-colors.

And world is full of nucleons... which are white. You cannot neutralize the colour of a quark with white nucleons, either.
Would a coloured nucleon be stable? An ordinary white proton has integer charge (+1), and contains a red, a blue and a green quark.
Now suppose that you had a volume of space containing three quarks to an integer charge (+1), but they are not white, because all three are red.
Would three red quarks remain bound to each other, or would strong interaction cause them to repel to different ends of world as fractionally charged lone quarks?

 

[mfb]

There are no free quarks around. All quarks come in quark/antiquark pairs (mesons), groups of three quarks or antiquarks (baryons), more exotic combinations (tetra-, pentaquarks) or as part of a quark gluon plasma. The scenario you want to consider does not exist.

... all of which require availability of an antiquark.

So what? You made a general statement. That general statement is wrong.

No, they cannot. Because of colour confinement.

Color confinement just tells us the produced quarks cannot be free. No one ever claimed that.
##e^+ e^- \to p \bar{p}## is a perfectly valid reaction and creates quarks (3 quarks, 3 antiquarks). ##e^+ e^- \to \pi^+ \pi^-## is another process, producing 2 quarks and 2 antiquarks.

Would a coloured nucleon be stable?

Colored nucleons cannot exist, the question is meaningless.

Now suppose that you had a volume of space containing three quarks to an integer charge (+1), but they are not white, because all three are red.

They cannot be bound. They can be part of a quark gluon plasma, but then they are not a combined particle.

 

[snorkack]

There are no free quarks around. All quarks come in quark/antiquark pairs (mesons), groups of three quarks or antiquarks (baryons), more exotic combinations (tetra-, pentaquarks) or as part of a quark gluon plasma.

We have not found a free quark. We also have not found a mechanism to destroy a free quark, should one exist.

 

[mfb]

We have not found a free quark. We also have not found a mechanism to destroy a free quark, should one exist.

That is correct, but that is completely different from what you claimed earlier.

 

[rkolter]

How much is known about chemical properties of quarks?
Also - elements are defined by charge of their nuclei. Positive integers from 1 to almost 100 are fairly well known.
Number of "protons" as such is irrelevant.

Elements are, by definition, substances whose atoms all contain the same number of protons in their nucleus. Helium is an element and any atom in the universe you find with exactly two protons in its nucleus is a helium atom. It may be an isotope. It may be ionized. But it's still the element helium.

If you could stabilize a sigma hyperon or some beautiful hyperon in a nucleus, it would be simply a strange, charmed, beautiful or truthful isotope of some known element.

It would not be an Isotope, by the definition of an Isotope. But you are correct - If you started with the element Helium and changed out one Proton in the atom's nucleus for one of your newly stable particles, the resulting atom WOULD be a variant of a known element - specifically it would be a variant of Hydrogen, since there was just one proton in the nucleus. It might have different characteristics. But it would still be Hydrogen.

 

[Algr]

Elements are, by definition, substances whose atoms all contain the same number of protons in their nucleus. Helium is an element and any atom in the universe you find with exactly two protons in its nucleus is a helium atom. It may be an isotope. It may be ionized. But it's still the element helium.

I think you are getting into semantics here, rather then chemistry. It sounds like if you had a substance that somehow had a nucleus of +3 charge despite having only two protons, it would behave in all respect like lithium. So (assuming that is correct) why call it helium? Helium does not behave half way between hydrogen and lithium, so why would 2.3nium behave half way between helium and lithium?

Edit: Duoettertium! Now it looks like I know what I'm talking about.

 

[Algr]

Hmmm... Duoettertium would be electrically neutral if you had three atoms sharing seven electrons. Something like how Oxygen usually appears as O₂? Although the reason isn't the same.

 

[mfb]

It would not be an Isotope, by the definition of an Isotope. But you are correct - If you started with the element Helium and changed out one Proton in the atom's nucleus for one of your newly stable particles, the resulting atom WOULD be a variant of a known element - specifically it would be a variant of Hydrogen, since there was just one proton in the nucleus. It might have different characteristics. But it would still be Hydrogen.

It looks like hypernuclei are named based on the nucleus charge, not based on the number of protons. A proton plus an additional positively charged baryon would be a helium hypernucleus. Here are some examples.

 

[snorkack]

Hmmm... Duoettertium would be electrically neutral if you had three atoms sharing seven electrons. Something like how Oxygen usually appears as O₂? Although the reason isn't the same.

Yes, very different.
And you have potential for problems due to strong interaction.
If you have three quarks whose charges combine to +1, and their colours combine to white, then even if they are at initial positions at internuclear distances, due to colour confinement they would promptly produce quark-antiquark pairs so as to form one nucleon and two mesons.
But what happens if you have 3 quarks whose colours do not combine to white, e. g. they are all red? Do they attract, repel or neither?

 

[mfb]

But what happens if you have 3 quarks whose colours do not combine to white, e. g. they are all red? Do they attract, repel or neither?

You cannot have those in isolation either.

Can we please come back to things that can at least theoretically exist? "What do the laws of physics predict in situations that are impossible according to those laws" is a meaningless question.

 

[snorkack]

You cannot have those in isolation either.

No. You cannot create them.
You cannot create baryons either, except in pairs with antibaryons (which easily annihilate again). Yet baryon number is conserved. And happens to not be zero.

A free quark cannot be destroyed for charge conservation reasons alone. And if colour is conserved, as you admit, then neither can a free gluon.

A colour combination of, e. g. red-antiblue could be possessed by a lone free gluon, or a coloured "meson" of a red quark and antiblue antiquark, or two free quarks, the said red quark and antiblue antiquark far from each other.
Which of these is the most stable state of free colour?

 

[mfb]

No. You cannot create them.

Isolated quarks cannot be created and they don't exist today (and never did in the past). Which means you cannot have them.

Which of these is the most stable state of free colour?

There is no such thing.

 

[rkolter]

It looks like hypernuclei are named based on the nucleus charge, not based on the number of protons. A proton plus an additional positively charged baryon would be a helium hypernucleus. Here are some examples.

Thanks for that. I didn't know there was a naming convention for it. +1 like.

To the OP's original question though and all the talk since - HyperNucleic Helium is still a type of helium - a known element. Tony Stark probably did "try every element". Nick Fury was either trolling Tony or meant to say something less damaging to his chemistry credentials like "Not every isotope or hypernuclei of every element."

Case closed?

 

[Algr]

Case closed?

Nah, I still don't like the idea of "a type of helium" that has a +3 nucleus and behaves in most/all respects like lithium.

 

[snorkack]

Isolated quarks cannot be created and they don't exist today (and never did in the past). Which means you cannot have them.

They cannot be created, any more than they can be destroyed. And baryons likewise can be neither created nor destroyed.
Baryons are very rare compared to photons (about 1 baryon per 1 000 000 000 photons), but they are numerous enough that we have observed baryons (and consist of them).
We have not yet reliably observed an isolated quark, showing that they must be rare enough compared to baryons. Does not mean they do not exist in low numbers.

 

[PeterDonis]

We have not yet reliably observed an isolated quark, showing that they must be rare enough compared to baryons.

This is not a valid analogy. Baryons are rare compared to photons because almost all of the baryons in the early universe were annihilated by antibaryons, producing photons. The baryons we see today are the small leftover excess of baryons over antibaryons.

Isolated quarks are not "rare"; they are nonexistent, because of the way the strong interaction works. Claiming that they could exist in very low numbers is contradictory to current mainstream physics and is off limits here.

 

[Algr]

Oops, I didn't quite connect rkolter's reply to what MFB said. Sorry.

I had a nasty thought that if twoandathirdnium existed, it might share a property with plutonium in which it was extremely poisonous because Earth biology never evolved any tolerance to it's presence. Oh well. Thank you everyone. :)