The refitted formula ( , , , , ) gives these binding energies for fluorine ( )

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear constituents, size and binding · Nuclear constituents, size and binding

The Nucleus Resolved, Ch How big a nucleus is, and what the mass surface knows

Problem

The refitted formula (aV=15.76a_V = 15.76, aS=17.91a_S = 17.91, aC=0.72a_C = 0.72, aA=23.25a_A = 23.25, aP=12.51 MeVa_P = 12.51\,\mathrm{MeV}) gives these binding energies for fluorine (Z=9Z = 9), in MeV: N=13N = 13: 168.02168.02; N=14N = 14: 174.13174.13; N=15N = 15: 173.82173.82; N=16N = 16: 177.57177.57; N=17N = 17: 175.39175.39; N=18N = 18: 177.30177.30. Fluorine-31 (N=22N = 22) has been observed as a bound nucleus. ((a)) Form the formula's SnS_n for N=14N = 14 to 1818 and S2nS_{2n} for N=15N = 15 to 1818. Where does S2nS_{2n} first turn negative? ((b)) A nuclide is bound only if it can emit neither one neutron nor two. Which is the heaviest fluorine isotope the formula binds against both? ((c)) By how many neutrons does each criterion fall short of the measured chain? The same formula puts tin's S2n=0S_{2n} = 0 near N=112N = 112, some 2727 neutrons past the last measured tin mass. What should be said when that number is quoted?

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