Why is ${\rm Pb}^{208}$ the heaviest stable nuclide?
Abstract
In an effort to understand nuclei in terms of quarks we develop an effective theory to low-energy quantum chromodynamics in which a single quark contained in a nucleus is driven by a mean field due to other constituents of the nucleus. We analyze the reason why the number of quarks in light stable nuclei is much the same as that of quarks, while for heavier nuclei beginning with , the number of quarks is greater than the number of quarks. To account for the finiteness of the periodic table, we invoke a version of gauge/gravity duality between the dynamical affair in stable nuclei and that in extremal black holes. With the assumption that the end of stability for heavy nuclei is dual to the occurrence of a naked singularity, we find that the maximal number of protons in stable nuclei is .
Keywords
Cite
@article{arxiv.2309.13082,
title = {Why is ${\rm Pb}^{208}$ the heaviest stable nuclide?},
author = {B. P. Kosyakov and E. Yu. Popov and M. A. Vronsky},
journal= {arXiv preprint arXiv:2309.13082},
year = {2024}
}
Comments
19 pages, 4 figures; 2nd ed. minor grammar corrections; 3rd ed. minor stylistic corrections; 4th ed. the form which is accepted in EPJC