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large radioactive atoms


lemur
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If the periodic table can be extended to increasingly larger atomic numbers, I was wondering if such heavy atoms could survive longer under certain conditions. If they are subject to very little radiation/energy, could the resulting calm/cold allow them to remain intact longer? Are there absolute conditions that govern the possibility of such large atoms remaining intact anywhere or could there be huge atoms floating around that will only split if triggered in some way?

 

edit: could such atoms form very strong bonds because of their large electron shells, and if so could being more strongly bonded together chemically stabilize them against decay?

Edited by lemur
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what am i doing trying to talk about chemistry? I just know that there is this concept hypothesized called the island of stability, but I believe it is most likely a dream, but who knows.

Edited by Realitycheck
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Atoms and molecules are subject to radiation and mechanical shocks that have energies of the order of kT where k is Botzman's constant and T is the temperature.

Near room temperature that's something like the equivalent of 0.025eV

If you heat things up to a few thousand degrees you raise the average energy to about 0.25eV.

Nuclear transitions take place with energies of the order of 10,000 to 10,000,000 eV

 

From the nuclear point of view, any atom outside of the core of a star is already somewhere calm and cold.

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I think larger atoms would be less stable. The evidence for that.... atoms of lower molecular weight are more numerous, the atoms of very high molecular weight are rare, and larger atoms means more subatomic particles to order. However, I don't know too much about nuclear physics, but using a Newtonian view, I'd say entropy doesn't not favor the formation or stabilization of extremely large atoms. In that way, entropy is the bully at the beach, he'd rather have your sandcastle reduced to mound of unrecognizable sand.

Edited by MrSandman
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I think larger atoms would be less stable. The evidence for that.... atoms of lower molecular weight are more numerous, the atoms of very high molecular weight are rare, and larger atoms means more subatomic particles to order. However, I don't know too much about nuclear physics, but using a Newtonian view, I'd say entropy doesn't not favor the formation or stabilization of extremely large atoms. In that way, entropy is the bully at the beach, he'd rather have your sandcastle reduced to mound of unrecognizable sand.

That's a good metaphor for entropy. In terms of organic chemistry though, think how complex the "sandcastles" are that are molecules formed by organic processes. You could say that Earth's biosphere is particularly suited to nurture such molecular growth, so maybe there is some other physical situation that nurtures the growth and sustainment of giant atoms?

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Tests have shown that the only effect on decay is for electron capture at exceedingly high pressure, owing to the deformation of the atoms. The spontaneous decays have not been shown to be affected, and there is no mechanism by which one would think they are affected. Spontaneous decay, by its very nature, does not rely on an interaction with other particles.

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That's a good metaphor for entropy. In terms of organic chemistry though, think how complex the "sandcastles" are that are molecules formed by organic processes. You could say that Earth's biosphere is particularly suited to nurture such molecular growth, so maybe there is some other physical situation that nurtures the growth and sustainment of giant atoms?

 

A state of high energy would be required to make these molecules, but as for this hypothetical physical situation... in order for it to exist it would need to defy our understanding of physics and chemistry. Not saying it couldn't, I'm all for our theories being proven wrong, but saying the likelihood is highly improbable. Of course a rebuttal would be if our universe is infinite then there is infinite possibilities, and yes those atoms exist.

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There are measurable effects of the environment on the half life of some elements. IIRC it's the k capture decay of some Be isotope, probably 7Be.

The half life depends on the chemical environment because that affects the electron density at the nucleus.

It's small, obscure effect.

For a heavier atom (and almost all atoms are heavier than Be) it would be even smaller because the chemistry only affects the outer shell electrons and the nucleus can capture the inner shell ones.

 

The instability of heavy atoms isn't an entropy effect- it's brought about by the electrostatic repulsion of all the positive charges on the protons.

The repulsive force is overruled in the presence of lots of neutrons because they stick to the protons using the strong nuclear force, however the strong force is only effective at short distances so the nuclei cannot be physically big.

There is a cop-out way round this.

You can thing of a neutron start with a bunch of extra protons as the nucleus of a truly giant atom, held together by gravity.

 

It is difficult to do chemistry with neutron stars because you need rather bigger test tubes than the normal suppliers stock.

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There are measurable effects of the environment on the half life of some elements. IIRC it's the k capture decay of some Be isotope, probably 7Be.

The half life depends on the chemical environment because that affects the electron density at the nucleus.

It's small, obscure effect.

For a heavier atom (and almost all atoms are heavier than Be) it would be even smaller because the chemistry only affects the outer shell electrons and the nucleus can capture the inner shell ones.

 

The instability of heavy atoms isn't an entropy effect- it's brought about by the electrostatic repulsion of all the positive charges on the protons.

The repulsive force is overruled in the presence of lots of neutrons because they stick to the protons using the strong nuclear force, however the strong force is only effective at short distances so the nuclei cannot be physically big.

There is a cop-out way round this.

You can thing of a neutron start with a bunch of extra protons as the nucleus of a truly giant atom, held together by gravity.

 

It is difficult to do chemistry with neutron stars because you need rather bigger test tubes than the normal suppliers stock.

So as nuclei grow, the electrostatic repulsion of the protons gains significance relative to the ability of the nuclear force to hold them together? If this is the case, I would think that the electrons would be significant in that the protons are attracted to them and so tidal forces would result from the pull of the electrons on the protons. Why wouldn't this be the case?

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Compared to the repulsion between the protons, the electrons are very far away (Also they are light and moving. On average any given electron doesn't pull the nucleus in any particular direction.)

I know this, but logically I would think that the electrons' pull form an equal and opposite reaction to the protons' pull on them. I know that sounds very classical and Bohr-ish, but equal-and-opposite force is a physical law I have a hard time ignoring. So either through momentum or some other force-impetus, I would think the electrons have to be pulling on the protons, since the protons seem to be pulling on them. If that's the case, then why wouldn't force-changes occurring with the electrons transfer to the protons in the nucleus?

 

 

 

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I know this, but logically I would think that the electrons' pull form an equal and opposite reaction to the protons' pull on them. I know that sounds very classical and Bohr-ish, but equal-and-opposite force is a physical law I have a hard time ignoring. So either through momentum or some other force-impetus, I would think the electrons have to be pulling on the protons, since the protons seem to be pulling on them. If that's the case, then why wouldn't force-changes occurring with the electrons transfer to the protons in the nucleus?

 

I don't know much about the strong nuclear force. But I know that it is WAY stronger than the Coulombic force over very short distances, distances on the inter-nuclear scale. Remember that Coulombic forces also drop off as [math] \frac {1}{r^{2}} [/math]. So the Coulombic proton-proton repulsion contributes a lot more to opposing the strong nuclear force than the Coulombic proton-electron attraction does. That, and as John Cuthber said, electrons are much lighter than nucleons.

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I don't know much about the strong nuclear force. But I know that it is WAY stronger than the Coulombic force over very short distances, distances on the inter-nuclear scale. Remember that Coulombic forces also drop off as [math] \frac {1}{r^{2}} [/math]. So the Coulombic proton-proton repulsion contributes a lot more to opposing the strong nuclear force than the Coulombic proton-electron attraction does. That, and as John Cuthber said, electrons are much lighter than nucleons.

 

Let's do a back-of-the-envelope calculation

 

Protons and neutrons attract and form a bound system despite the repulsion of the protons, so even at that scale we know that the nuclear force is at least as strong as the electrostatic force. But the nucleus is of order 10^-15 m across, while the electrons tend to be a million times further away, on average. Which means, in very rough terms, the electrostatic force from the electrons is at least around 10^11 - 10^12 times smaller. In terms of energy, the binding of a proton is measured in MeV while an electron is measured in eV; energy is a common proxy for interaction strength. On those terms, the electrostatic interaction is a million times smaller.

 

Then, you can add in the previosuly-mentioned fact that there is no net charge distribution, so whatever force is present from a single electron acting on a proton goes to zero in an atom.

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Then, you can add in the previosuly-mentioned fact that there is no net charge distribution, so whatever force is present from a single electron acting on a proton goes to zero in an atom.

 

Upon thinking about it for a second. It's worth mentioning that even in the case where the distribution of electrons is not spherically centro-symmetric, [math]( \ell>0 )[/math]. It still has centro-symmetry about some plane or axis. So the net force acting on a proton going to zero, what you said, would in fact hold in all cases for a single atom. Obvious but interesting.

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Upon thinking about it for a second. It's worth mentioning that even in the case where the distribution of electrons is not spherically centro-symmetric, [math]( \ell>0 )[/math]. It still has centro-symmetry about some plane or axis. So the net force acting on a proton going to zero, what you said, would in fact hold in all cases for a single atom. Obvious but interesting.

 

Another way of putting this is that atoms do not have permanent electric dipole moments. They are experimentally excluded at a very small level (something like 10^-28 e-cm)

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I don't know much about the strong nuclear force. But I know that it is WAY stronger than the Coulombic force over very short distances, distances on the inter-nuclear scale. Remember that Coulombic forces also drop off as [math] \frac {1}{r^{2}} [/math]. So the Coulombic proton-proton repulsion contributes a lot more to opposing the strong nuclear force than the Coulombic proton-electron attraction does. That, and as John Cuthber said, electrons are much lighter than nucleons.

 

If you want the less classical version then the electrons are delocalised all round the nucleus and pull it in every direction so the net effect is nil.

How can the force connecting the electrons and nucleus be less than inelastic if energy is to be transferred from one atom to another in collisions? Do the electrons somehow contain all or most of the momentum of the atom and only drag the nucleus along as a solid afterthought of vector-changes? If so, does that mean the electrons' force on the nucleus pulls the nucleus from all points simultaneously? Is it at all analogous to compare it to the gaseous part of Jupiter somehow violently shifting direction and pulling the solid core along as a whole without exerting any shift in the cohesion-forces of that core?

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