I have heard the argument that elements with atomic numbers above 137 are not possible but I am unsure if it's true or why.
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7$\begingroup$ Is this the argument based on an electron having angular velocity $Zc\alpha$? If so, that'll just be a non-relativistic approximation for $Z\ll1/\alpha$. $\endgroup$– J.G.Commented Jul 28 at 21:31
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3$\begingroup$ Not quite a duplicate of Is There a highest possible atomic number ? $\endgroup$– StephenG - Help UkraineCommented Jul 28 at 22:09
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1$\begingroup$ See also this citation in Wikipedia's page on the extended periodic table. $\endgroup$– StephenG - Help UkraineCommented Jul 28 at 22:12
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3$\begingroup$ related physics.stackexchange.com/q/228183 $\endgroup$– MithoronCommented Jul 29 at 12:22
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1$\begingroup$ "I hav heard" requires you to say where you heard this. $\endgroup$– James KCommented Jul 31 at 3:29
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3 Answers
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Empirically, nuclei composed from magic numbers of nucleons ( either proton number or neutron number, separately) which follows Otto Haxel's relation $$a(n) = \frac{n(n^2 + 5)}{3} \tag 1$$ Starting from $n=0,1,2,\ldots$ should be stable as nucleons are arranged into complete shells within atomic nucleus, even if they fall-out of general atom island of stability. And so elements with atomic numbers like $184, 258, \ldots$ should be possible.
By the way, this magic sequence assumes spherical symmetry of nuclei, which by current research is shown to be false, and as such real predictions of super-heavy stable elements can be a little bit different than is pictured by Otto Haxel's formula. But still, the basic premise of magic numbers extension of island of stability should hold.
answered Jul 28 at 22:20
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1$\begingroup$ You say in the first sentence "... numbers of nucleons" but then say "numbers like 184, 258, ... should be possible". But aren't these already known if we are talking about nucleons and not the atomic number? E,g, 258 is Mendelevium 258 - half life 52 days. Or am I confused about something? $\endgroup$– rghomeCommented Jul 30 at 11:31
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3$\begingroup$ @rghome Magic number corresponds to either proton number or neutron number, separately, but not to the sum of two. If number of protons AND number of neutrons in the atom follows magic rule,- then such atom is considered "doubly magic". As for example atom ${{}^{12}}_6C$ (standard Carbon isotope) has 6 protons and 6 neutrons, so according to Haxel rule,- Carbon is "doubly magic". $\endgroup$ Commented Jul 30 at 12:43
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3$\begingroup$ Standard 258-Mendelevium isotope has $101$ protons and $157$ neutrons, neither of which are Haxel magic numbers and so ${}^{258}Md$ is not considered to be very stable atom, and indeed it has half life of 52 days. $\endgroup$ Commented Jul 30 at 13:03
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1$\begingroup$ What should we do with the argument from @my2cts answer for such atomic numbers as 184 and 258? I doubt that finite-size effects allow exceeding the $1/\alpha$ limit that much. $\endgroup$ Commented Jul 30 at 14:34
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1$\begingroup$ @E.Anikin Well, either Haxel empirical sequence or ground state catastrophe mentioned by Itzykson and Zuber for a finite size nucleus is wrong. I do not have required grounds to decide which argument is stronger, however as Zuber argument feels like an attempt for theoretical analysis instead of plain Haxel empirics,- that may be stronger argument than just some correlations between atomic numbers. Or maybe these two arguments are both incomplete. Anyway, my duty was to present this empirical argument to the public. $\endgroup$ Commented Jul 30 at 14:44
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For a point nucleus with $Z \gt 1/\alpha$ the Dirac equation has no ground state solution for a single electron, that is a +(Z-1) ion. Itzykson and Zuber describe this catastrophe on page 75. For a finite size of the nucleus this occurs at a higher value.
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2$\begingroup$ As I understand it, this lack of electron ground states does not prevent such an element from existing--it would simply always be in an excited state (or a fully-ionized bare nucleus), correct? It probably wouldn't be stable for other reasons, but the lack of a ground state for bound electrons doesn't make the nucleus unstable. $\endgroup$– HearthCommented Jul 30 at 19:32
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1$\begingroup$ There may be pair formation, leading to electron capture and positron emission. So the questions of nuclear stability and ground state catastrophe may be related. $\endgroup$– my2ctsCommented Jul 30 at 19:57
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2$\begingroup$ My answer gives 'the argument that elements with atomic numbers above 137 are not possible'. This argument is valid for the point charge approximation to the nucleus. For a realistic size nucleus the limiting value of Z in the context of this argument is higher. I believe that the limit obtained in this way is hard. From nuclear physics there may be arguments that set the limiting Z to lower values. $\endgroup$– my2ctsCommented Jul 31 at 15:59
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There is also a mechanism predicted by QED that "imples"/motivates that atomic nuclei with $Z>137$ are absent in the periodic table, which is called Schwinger pair production.
You can do some back-of-the-envelope calculation as for example shown in section 33.4.3 of Ref. 1 that for $Z=137$ the electric field "near"$^1$ the nucleus becomes strong enough to produce Schwinger pairs.
$^1$: Validity of the Euler-Heidenberg Lagrangian requires $r \sim 1/m_e $.
Ref. 1: "Quantum Field Theory and the Standard Model", M. D. Schwartz
answered Jul 31 at 9:40
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1$\begingroup$ But would electron pair production near the nucleus keep such a nucleus from existing? $\endgroup$ Commented Aug 1 at 1:24