Why is 82208Pb\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$${}^{208}_{82}\ extrm{Pb}$$\\end{document} 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 d quarks in light stable nuclei is much the same as that of u quarks, while for heavier nuclei beginning with 2040Ca\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\mathrm{{}^{40}_{20}Ca}$$\\end{document}, the number of d quarks is greater than the number of u 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 ZmaxH≈82\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$Z_{\\max }^{\ extrm{H}}\\approx 82$$\\end{document}.