# Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible?

We currently know that there are atoms with atomic number up to 118 are possible.

Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible? Or are there theoretical arguments or maybe heuristic arguments why this can't happen?

I suspect the answer to be yes, because technetium and promethium have no stable forms while the elements surrounding it in the periodic table do have stable forms. A similiar thing might happen here. The more general question: Does there exist $k > l > 118$ such that atoms with $k$ protons are possible, but that atoms with $l$ protons aren't possible?

It is not a duplicate of The last element's atomic number. I wonder whether there could be 'gaps' in the periodic table, not what the last element's atomic number is.

• Please define 'possible' - do you mean a short lifetime, or completely unbound? – Jon Custer Oct 11 '15 at 18:23
• @JonCuster What I meant: An element is possible if there is at least one isotope with a half live greater than the Planck time. I'm not sure if it really works, and whether there are isotopes with half live smaller than the Planck time, but I think it will do for this question. – wythagoras Oct 11 '15 at 18:32
• Your 'definition' doesn't make much sense, becuse for sth to decay in time shorter than Planck time, it would need to be about Planck radius small, and even single proton in comparison huge. – Mithoron Oct 15 '15 at 17:35
• This is of course more of a nuclear physics question rather than a chemistry question. I'd expect just about any atomic number is possible, but that all elements above lead z=82 are radioactive. So for z>82 then it is a question of how long is the half-life. – MaxW Dec 15 '15 at 0:58
• Look at "Island of stability" that might help too – David Wyn Williams May 23 '18 at 13:46

• Imagine that electrons follow Bohr's orbits (which they do not, but let's forget about that for a moment). Then we have to balance the centripetal force $mv^2\over r$ against the electrostatic force $kZe^2\over r$. As Z exceeds ~137, the necessary $v$ exceeds the speed of light, so we might be tempted to think this is the end. But no, this is derived in the assumption that $m$ does not depend on $v$, which is not quite so (as per special relativity). – Ivan Neretin Dec 15 '15 at 15:17