Oddly specific number of elements

  • #1
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Main Question or Discussion Point

There are 118 elements known to man, and some scientists like Feynman think that element 137 might be the end of the Periodic Table.

Isn't that oddly specific? To me, it feels like it is completely random and of no significance. What is going on here? Is there a constant that relates to this number in any way? Or is there something that I'm missing completely?
 

Answers and Replies

  • #2
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There are 118 elements known to man, and some scientists like Feynman think that element 137 might be the end of the Periodic Table.

Isn't that oddly specific? To me, it feels like it is completely random and of no significance. What is going on here? Is there a constant that relates to this number in any way? Or is there something that I'm missing completely?
The "bigger" the atom, the "bigger" its nucleus is, the "bigger" the nucleus, the more unstable the balance between the binding forces and the scattering forces becomes. This is what causes an atom to emit radiation(particles) and decay into other atoms.

Feynman was probably theorizing that atomic nuclei would be too unstable to form stable atomic structures beyond 137, at least, too unstable to maintain integrity for very long, seconds or minutes maybe, before it decayed or lost integrity.
 
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  • #3
Borek
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Think about an analogy. Take a steel rod of a 1 cm diameter and put it upright. When the rod is short - say, 1 m - there is no problem, it keeps up. When it becomes longer and longer at some point it will be too heavy and it will bend and break. We can calculate - from known physic properties of the solid - maximum length of such an erected rod.

And it is a very similar situation with nuclei. In the case of atoms it is not length that has its maximum limit, but number of protons. At some point the nucleus becomes too large to stay intact. We can estimate this limit from known properties of the nucleons, and that's the number Feynman referred to.
 
  • #4
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Think about an analogy. Take a steel rod of a 1 cm diameter and put it upright. When the rod is short - say, 1 m - there is no problem, it keeps up. When it becomes longer and longer at some point it will be too heavy and it will bend and break. We can calculate - from known physic properties of the solid - maximum length of such an erected rod.

And it is a very similar situation with nuclei. In the case of atoms it is not length that has its maximum limit, but number of protons. At some point the nucleus becomes too large to stay intact. We can estimate this limit from known properties of the nucleons, and that's the number Feynman referred to.
Ohh, so there could be more elements, but they would certainly not be stable. Thanks!
 
  • #5
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Actually, that´s tied to fine structure constant.
But the reasoning does not hold water.
 
  • #6
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After ~115, IMHO, it is currently 'Beyond Theory'. The long-anticipated 'Island of Stability' has yet to appear, IIRC, due failures of the 'magic number' shell model due increasingly complex relativistic corrections and arcane nuclear lobe formation...
But, more and more, looks like that '137' really is the end-stop.
Pending 'New Physics', of course, of course...
 
  • #7
TeethWhitener
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The current understanding (according to QED calculations) is that, around Z=173 or so, the 1s orbital will have sufficient energy (1.02 MeV) to self-populate via electron-positron pair production. So you can’t have a bare Z=173 nucleus, according to current theory. However, synthesis of such a nucleus in the first place is so far out of the realm of possibility that it’s a moot point.
 
  • #8
HAYAO
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The current understanding (according to QED calculations) is that, around Z=173 or so, the 1s orbital will have sufficient energy (1.02 MeV) to self-populate via electron-positron pair production. So you can’t have a bare Z=173 nucleus, according to current theory. However, synthesis of such a nucleus in the first place is so far out of the realm of possibility that it’s a moot point.
I'm a complete layman here in particle collisions, but let's say that the production of Z=173 atom is technologically possible. Wouldn't an electron-positron pair production require a photon (at least larger than 1.022 MeV)? If there is no photon, then wouldn't a bare 173 nucleus still be possible? What would be the source of the photon with sufficient energy?
 
  • #9
TeethWhitener
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I'm a complete layman here in particle collisions, but let's say that the production of Z=173 atom is technologically possible. Wouldn't an electron-positron pair production require a photon (at least larger than 1.022 MeV)? If there is no photon, then wouldn't a bare 173 nucleus still be possible? What would be the source of the photon with sufficient energy?
Good question. You need a photon to conserve momentum, but the electric field of the nucleus supplies enough energy by itself to produce the electron-positron pair. I don’t know if this means you still need a 1MeV photon or if, because of the 1s orbital energy, any photon will do. Maybe someone who knows more than me can sort it out.
 
  • #10
HAYAO
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Good question. You need a photon to conserve momentum, but the electric field of the nucleus supplies enough energy by itself to produce the electron-positron pair. I don’t know if this means you still need a 1MeV photon or if, because of the 1s orbital energy, any photon will do. Maybe someone who knows more than me can sort it out.
Thank you for the reply.
If my understanding is correct, a photon needs certain amount of energy for the electron-positron pair to produce in the presence of an atom. The requirement for the atom, on the other hand, is less severe because the pair-production is a probability if with a sufficient photon energy. In another words, heavier atom is preferred but is not completely necessary.

So my layman understanding was that photon with sufficient energy is a necessity. Maybe a photon is inevitably produced in a collision experiment...I don't know.

(I feel that this thread belongs in the physics section)
 
  • #11
TeethWhitener
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From Fricke and Greiner, Theor. Chim. Acta, 1971:
...a physical limit is given by the binding energy of ##-mc^2## because at this value the level drops into the continuum of filled states of negative energy. This occurs approximately at ##Z =Z_{critical} \approx 175##. It is not yet clear whether this really happens or - what is more likely - that quantum electrodynamical effects e.g. vacuum polarization and fluctuation, become so large that the single particle Dirac equation breaks down completely. One possibility is, that the levels do not drop into the negative continuum but they will asymptotically reach the lower continuum for very large Z which would mean that in nature a maximal field strength would exist [24]. Another possibility is that spontaneous electron-positron production occurs at the critical Z-value. Theoretical considerations for the quantum electrodynamics of strong fields are given by Reinhardt [5] et al., but practical calculations are still missing.
So I guess no one is really sure (as of 1971, probably some better QED calculations have been done since then).
 
  • #12
TeethWhitener
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There's also this, from Popov, 2001:
https://link.springer.com/content/pdf/10.1134/1.1358463.pdf
Thirty years ago, there arose interest in the predictions of quantum electrodynamics (QED) in ultrastrong Coulomb fields—in particular, in the effect of spontaneous positron production from a vacuum (see, for example, [1–30] and the review articles [7, 31–39]). A feature peculiar to this process is that it has no bearing on the frequency of an electric field and can occur in the case of an arbitrarily slow (adiabatic) growth of the nuclear charge in the region ##Z > Z_{cr}##, a point where it differs from any other positron-production mechanism known so far.
This is in line with what I anticipated in post 9.
Edit: this is V.S. Popov, not the famous V.N. Popov of Faddeev-Popov ghost fame.
 
  • #13
HAYAO
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  • #14
TeethWhitener
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(I wonder how the momentum is preserved)
Sorry, in post 9 I meant angular momentum, not linear momentum.
 
  • #15
HAYAO
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Sorry, in post 9 I meant angular momentum, not linear momentum.
Oh yeah, you are right. Sorry about that.
 

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