I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors.

  • $\begingroup$ I have read the last element possible is 137 as this element will have electrons speed greater than speed of light which is not possible. $\endgroup$ – Kamlesh Patel Jan 28 '14 at 14:06
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    $\begingroup$ @KamleshPatel I have converted your post to a comment… it's not really an answer to the question, as it's only one line and you don't link to any source or explain in detail. See Nicolau's fine answer for an example of a fuller discussion of that same point. $\endgroup$ – F'x Jan 28 '14 at 20:14
  • $\begingroup$ @KamleshPatel By that reasoning, wouldn't the last element possible then be 136? $\endgroup$ – Melanie Jan 22 '18 at 3:44

Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.

As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.

EDIT: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today, so no one can hope to provide a rigorous answer from that angle.

There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.

Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.

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    $\begingroup$ Using the naïve Bohr model of the atom, we run into trouble around $Z = 2$... $\endgroup$ – leftaroundabout Jan 27 '14 at 20:28
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    $\begingroup$ @leftaroundabout Only with respect to the accuracy of energy levels, not the stability of the atom itself! $\endgroup$ – Nicolau Saker Neto Jan 27 '14 at 21:35
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    $\begingroup$ With respect to any property these atoms have. The Bohr model simply doesn't work out for anything but 2-body systems, so it can't really apply to atoms other than hydrogen (though it can well apply to $\ce{He}^+$ etc.). $\endgroup$ – leftaroundabout Jan 27 '14 at 22:02
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    $\begingroup$ @leftaroundabout Fair enough. I guess Bohr's model is just often mentioned for historical reasons, to show that models can set limits (even if wrong) and because $v^{1s}_e=Z\alpha c$ is a very simple result. Of course, the Dirac equation itself is also an approximation (a much better one, no doubt). We don't even need a new theory to overturn its conclusions; at some point even more subtle QED effects will become appreciable, and how they will alter the final picture is still unknown, as far as I understand. $\endgroup$ – Nicolau Saker Neto Jan 28 '14 at 22:10

An "element" must be defined as the set of all atomic nuclei having a specified number of protons. Definitions based on electrons (or other leptons) can't be used because how many electrons are associated with an element changes with the atom's environment.

Defining an "atomic nucleus" as a set of protons and neutrons, in a common nuclear potential well, whose mean life is large with respect to the time it took the set to form. (A nuclear interaction takes place over a span of time on the order of $1\times10^{-23}$ sec.)

If you add neutrons to a nucleus, each is more weakly bound than the last. Eventually, the last neutron added is unbound, so it comes right back out. Usually, this happens within a time comparable to $1\times10^{-23}$ sec. For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons. The set of nuclides $(Z,Nd)$ is a curve on a Z,N plane known as the neutron dripline. The neutron dripline defines the maximum size a nucleus with a given number of protons may have.

If a nucleus with Z protons has too few neutrons, one of two things will happen. It may eject a proton or it may fission. Large nuclei will almost invariably fission, though, so that's the important criterion. The simplest workable model of an atomic nucleus is the "liquid drop model". Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play. Electric repulsion varies as $(Z^2 / r_{eff})$ where reff is distance between equivalent point charges. What pulls the nucleus together is what amounts to surface tension - unbalanced nuclear cohesion - and the total "surface energy" stored varies as $(r^2)$, where r is nuclear radius. The ratio between Coulomb and surface energies is defined by $(Z^2 / r_{eff})*(1/ r^2) = K$. Set $r_{eff} = r$. Nuclear volume is proportional to the total number of particles, $A = Z+N$, in a collection. That means r varies as $A^{1/3}$, so $(Z^2 / r^3) = K = (Z^2)/A$. K is called a "fissility parameter". A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission. For specified value of K, $N(Z) = (1/K)*(Z^2) - Z$ defines a curve of constant fission barrier height on the $(Z,N)$ plane. One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not. In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.

At least one nuclear model includes nuclei with up to $330$ neutrons and $175$ protons(1). An equation for the neutron dripline as a function of Z can be extracted from their dripline. A second equation for $N/Z$ as $f(Z)$ can be used to construct an alternate dripline curve. KUTY's neutron dripline does not show any dramatic changes below $N=330$. Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be $1/4$ order of magnitude ($1.77$) times larger.

Liquid-drop theory predicts immediate fission for $K>50$; however, the liquid drop model does not account for the additional binding produced by nuclear structure. The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections. Taking that value as the geometric mean between $K=50$ and the value of K to be used gives $K=102$. (This was the highest of three techniques tried).

For large Z, the fission curve rises faster than the dripline curve. The point at which they meet is the largest possible nucleus. Anything larger will immediately decay by neutron emission or fission. Nominally, the largest nucleus $Z=592$, $N=2846$ - but that is way too much precision for this sort of calculation. It's reasonable to say that the largest possible nucleus has $Z <600$ and $N < 3000$.

It's entirely possible that my remarks are completely wrong. I hope so, because that would mean somebody who knows more about it than I do has come up with a better answer.

  1. "Decay Modes and a Limit of Existence of Nuclei"; Hiroyuki Koura; http://tan11.jinr.ru/pdf/10_Sep/S_2/05_Koura.pdf
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    $\begingroup$ I like your last sentence ;) This is nice analysis, but you go too far with "Definitions based on electrons (or other leptons) can't be used" While you can't define element by number of electrons, they are actually needed for these nuclei to become part of element. Nucleus with half-time too short to bind electrons around it isn't enough to get element. This is a matter of definition which might be changed, though. $\endgroup$ – Mithoron Jan 21 '18 at 1:01

Naively, the nuclear electric field at Z ~ 137 or greater, reciprocal of the Fine Structure constant, would "spark the vacuum." Vacuum would be torn into electron-positron pairs. The electrons go in to convert protons to neutrons plus neutrinos. As stated above, non-classical treatment suggests we'll never get near a cold nucleus that sparks the vacuum. RHIC and LHC rip the vacuum hot by colliding deeply relativistic gold or lead nuclei.

The big problem in making new heavy elements is getting enough neutrons in there while colliding the heaviest nucleus with the lightest to do the job. Fusion charge repulsion is the product of the two charges. It must be minimized. Ca-48 is stable for a footnote, and it's run its course for fusion with available transuranics. The isotopes produced are way too neutron deficient to stick around. One might do something clever with a big custom-configured H-bomb - lots of compression and neutron density there - but sample retrieval is problematic.

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The real thing is that we do NOT know for sure. Accepting Quantum Mechanics, we have some parts of the answer, but we can not know the rest until we test it. Curiously, the higher Z is, the more relativistic the atom is. So, relativistic quantum mechanics do matter in the Periodic Table.

Option 1. Something avoids to allow elements with Z>118. That is quite unlikely. Pekka Pykko did simulations until 173...

Option 2. The maximum Z is somewhere between 122-173. It seems unlikely too! The only caveat is that Z=137 is special because the Dirac equation implies that the electron 1s has a speed greater than light for Z>137 due to the value of the fine structure constant. However, this is done neglecting the finite extension of the nuclei. So, here we realize that the ultimate fate of the atom depends on the stability of the nuclei...

Option 3. Elements are stable until 173, nuclear physics acts to avoid higher elements some point between 137 and 173...This is more likely, but nobody knows...

Option 4. Elements are allowed in principle, even at supercritical Z or above until the nuclei can not support more shells. We don't know really what happens here due to limited simulations. Maybe quantum computers can help us here (I hope so) to simulate elements and atoms in these quantum realm.

Option 5. There is no limit on Z. I consider this thing unlikely unless something is discovered beyond the current quark-lepton periodic table of fundamental particles.

I have written about the Bohr model at my blog, and written about the last element too. Here: http://www.thespectrumofriemannium.com/2013/06/30/log113-bohrs-legacy-i/, http://www.thespectrumofriemannium.com/2013/07/10/log114-bohrs-legacy-ii/, http://www.thespectrumofriemannium.com/2013/07/10/log115-bohrs-legacy-iii/, http://www.thespectrumofriemannium.com/2013/12/31/log150-bohr-and-doctor-who-amc%c2%b3/, http://www.thespectrumofriemannium.com/2014/05/26/log151-bohrlogy-i/ , http://www.thespectrumofriemannium.com/2014/05/26/log152-bohrlogy-ii/, http://www.thespectrumofriemannium.com/2015/07/04/log171-from-bohrlogy-to-dualities/, and http://www.thespectrumofriemannium.com/2017/07/11/log183-bohrlogy-some-pocket-formulae/

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It might be possible that they can keep on going to find new elements by using cycloterons and fusing atoms together till they reach a blackhole. So I believe that may be the limit for atomic number can be calculated by the singularity of Kerr Newman metric. But by assuming periodic table pattern if one other element could be found there comes at least 32 others to complete one other row of periodic table to reach may be 150 atomic number.

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