# Is this a correct demonstration for why elements above untriseptium cannot exist?

With the confirmation that elements 113, 115, 117, and 118 are indeed fundamental elements that are now to be named on the periodic table, the next question is: what is the highest atomic number possible for an element? Feynman had a go at this years ago, and he derived (according to my limited understanding) that above element-137 (informally denoted 'Feynmanium') the electrons in the nearest orbit about the nucleus would be traveling at a velocity greater than the speed of light. (Note I am considering the Bohr model of the atom in this question.) I wished to demonstrate to my friends why this is and I came up with the following explanation.

According to Bohr's quantum condition, the angular momentum of an electron about the nucleus is given by $$L=m_evr_n=n\hslash \implies r_n=\frac{n\hslash}{m_ev},$$ where $m_e$ is the mass of the electron, $v$ its velocity, $n$ an integer, and $r_n$ the radius of the $n$th possible orbit. Because we are concerned with the nearest orbit to the nucleus, $n=1$; and thus $$r=\frac{\hslash}{m_ev}.$$

Now, according to Coulomb's law, if the electron orbits about the nucleus the centripetal motion can be described by $$\frac{Ze^2}{4\pi\epsilon_or}=m_ev^2,$$ where $Z$ denotes the number of protons in the nucleus (the atomic number), and $e$ the elementary charge. Solving for $Z$ and substituting in $r$ from above yields $$Z=\frac{4\pi\epsilon_o \hslash v}{e^2}.$$ But what is $v$? Well, the maximum velocity an electron could ever have is the speed of light, and we wish to find the atomic number associated with an orbiting electron traveling at this speed, so we set $v=c$ and obtain our final result of $$Z=\frac{4\pi\epsilon_o\hslash c}{e^2}\approx 137.521,$$ which implies that for $Z>137$, the electrons at a position of $n=1$ in the Bohr model would have a velocity $>c$; and thus the highest atomic number achievable on the periodic table is 137.

Again, I just want to make sure this is a correct method for deriving element-137 before presenting. Perhaps one could explain how relativity plays a role here. I know Feynman used the Dirac equation to get this result...so could anyone (subsequently of course) expatiate on this in a simplistic manner? Thanks!

• Possibly of interest (although not an answer to your question): rsc.org/chemistryworld/Issues/2010/November/…. – HDE 226868 Jan 8 '16 at 4:02
• This is certainly not correct. Bohr's model is basically just a lucky guess, even for non-relativistic atoms. Any non-relativistic treatment of the nucleus is essentially wrong by default. It's like trying to predict a black hole with Newton. – CuriousOne Jan 8 '16 at 4:04
• I realize that by today's understanding of this subject, my question is certainly not correct. But from a non-relativistic Bohr's model perspective, is it a correct method to effectively derive element-137? – fruitegg Jan 8 '16 at 4:06
• To be very honest... I don't think even QED is enough to make these kinds of predictions. I would like to see a full numerical QCD treatment of nuclei before I believe such "theorems". That, of course, is far outside our current capability. – CuriousOne Jan 8 '16 at 4:43
• As the atomic number is defined really by the number of nucleons in the nucleus, I agree with CuriousOne . What the electrons do is derived from the existence of the charge of the protons. You are just demonstrating a classical model's inconsistencies where quantum mechanics reigns. If a QCD solution would allow 140 protons then the electrons would follow. – anna v Jan 8 '16 at 5:03

No, electrons cannot impose any upper limit on the maximum $Z$ of atoms.

The whole research of heavy elements is the research of the nuclei, not the electrons that orbit them. Nuclear physics is about protons, neutrons (or quarks, gluons) and forces in between them and the typical speeds of the constituents are always rather close to the speed of light. Some nuclei classified by $(A,Z)$ are stable, some are short-lived, some are long-lived, some don't exist, and there are islands of stability etc.

For an arbitrarily charged nucleus, however, it's always possible to place an arbitrarily high number of electrons to the orbits.

Special relativity cannot prevent us from doing so and I am confident that people knew that it couldn't since the very discovery of special relativity in 1905. In practice, new quantum mechanics only existed from Heisenberg papers in 1925 but at that time, they already knew that electrons could have been added indefinitely. Since 1928, just 3 years later, they already had the Dirac equation which is enough to study how the motion of electrons in quantum mechanics is affected by speeds approaching the speed of light.

The main reason why the speed of light can't "forbid" some solutions is that relativity simply replaces the electrons that would be "increasingly superluminal" by electrons that are "increasingly close to the speed of light" but subluminal.

We should replace $v$ by $p$, the momentum. The uncertainty principle allows us to estimate the momentum $p$ for a given $Z$ and a given orbit i.e. principal quantum number $n$ etc. Above $Z=137$ or so, the calculated $p$ may indeed exceed $p_0 = m_e c$. However, that doesn't mean that the speed is predicted to be higher than $c$. This claim would be wrong because $p\neq mv$. Instead, in relativity, $$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$ For an arbitrarily high $p$, we may find a $v\lt c$ for which this equation is satisfied. So if the very heavy nuclei stayed around for a long enough time so that electrons have a chance to fill the orbits to create neutral atoms, they would do so and for a very large $Z$, the inner electrons would simply have speeds that are very close to the speed of light so that the momentum is very high. But the speed would never exceed $c$ and it wouldn't have to.

@LubosMotl Your answer that the electrons are not the limiting factor for the maximum possible atomic number (Z) is correct, but there are a couple of errors in your analysis that I would like to correct.

1) Nuclear physics is about protons, neutrons (or quarks, gluons) and forces in between them and the typical speeds of the constituents are always rather close to the speed of light. True for the low mass quarks, but not for neutrons and protons. These masses are heavy enough and the nuclear binding weak enough that the average velocities are still a relatively small percentage of c.

2) For an arbitrarily charged nucleus, however, it's always possible to place an arbitrarily high number of electrons to the orbits. Not true. You are forgetting that the electrons repel each other and that for a large enough number of electrons the positive repulsion will eventually overpower the attraction of the nucleus. It is true that for a neutral atom the number of bound states is infinite because of the infinite number of possible Rydberg states, but once the number of electrons exceeds Z that is no longer necessarily the case.

Nuclei are like atoms in that there are shell closures that enhance the stability (and binding energy) of certain isotopes. Because of a strong spin-orbit interaction in nuclei the shell closures occur at different numbers than is the case for atoms. These numbers were observed before the importance of spin-orbit effects was known and so they are called magic numbers. Of course there are magic numbers for both neutron and proton orbitals. The number 114 was predicted to be a magic number for protons long before its discovery in 1998. This element is called flerovium. It is missing from the list above because of its earlier discovery but all of the others (113, 115, 117, 118) are members of the island of stability associated with flerovium.

The number 126 is a strong magic number for neutrons (it is the neutron number in the exceptionally stable 208Pb isotope. There are good reasons to believe that it would also be a magic number for protons and so that is probably the next island of stability to be explored experimentally.