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February 28

Semi-empirical mass equation

I used the semi-empirical mass formula to estimate the neutron drip line for nuclides A=1 to A=100 by using Myers and Swiatecki's version of the formula, substituting it into the formula for neutron separation energy, setting S_n to zero and solving the resulting quadratic:
$B_{tot}(Z,A)=c_{1}A\left[1-k\left({\frac {A-2Z}{A}}\right)^{2}\right]-c_{2}A^{2/3}\left[1-k\left({\frac {A-2Z}{A}}\right)^{2}\right]-{\frac {c_{3}Z^{2}}{A^{1/3}}}+{\frac {c_{4}Z^{2}}{A+{\frac {11}{A^{1/2}}}}}$
$S_{n}=B_{tot}(Z,A)-B_{tot}(Z,A-1)=0$
$Z^{2}\left({\frac {4c_{2}A^{2/3}k}{A^{2}}}-{\frac {4c_{2}(A-1)^{2/3}k}{(A-1)^{2}}}+{\frac {c_{4}}{A+{\frac {11}{A^{1/2}}}}}-{\frac {c_{4}}{(A-1)+{\frac {11}{(A-1)^{1/2}}}}}-{\frac {4c_{1}k}{A}}+{\frac {4c_{1}k}{A-1}}-{\frac {c_{3}}{A^{1/3}}}+{\frac {c_{3}}{(A-1)^{1/3}}}\right)-Z\left({\frac {4c_{2}A^{2/3}k}{A}}-{\frac {4c_{2}(A-1)^{2/3}k}{A-1}}\right)+c_{1}-c_{1}k-c_{2}A^{2/3}+c_{2}(A-1)^{2/3}+c_{2}A^{2/3}k-c_{2}(A-1)^{2/3}k=0$
Given c₁=15.677MeV, c₂=18.56MeV c₃=0.717MeV, c₄=1.211MeV and k= 1.79, the quadratic coefficients can be numerically determined for any given A, and positive solutions for Z can be found. Using this method it is predicted that ⁴
₁H
would be a possible nucleus, which contradicts the Wikipedia article that states the neutron drip line comes after ³
₁H
(that article then appears to be contradicted again by the article isotopes of hydrogen, which lists very, very short lived nuclei up to ⁷
₁H
). This is not unexpected, and it didn't surprise me that the formula did not yield correct results for hydrogen. What did surprise me, is that the formula correctly placed the neutron drip line after ⁸
₂He
, ¹¹
₃Li
, ¹⁴
₄Be
and ¹⁷
₅B
. Especially in the cases of ¹¹
₃Li
and ¹⁴
₄Be
, these are halo nuclei, and as such, the model on which the semi-empirical mass formula is based doesn't take into account their structure at all. The formula assumes a spherical nucleus with fully coordinated nucleons in its center and less strongly binding nucleons on the surface. It has no way to account for neutrons so loosely bound that they basically orbit around the nucleus at a distance of several femtometers. How is it that the formula is so good at predicting the location of the drip line when it doesn't seem to make any attempt to account for these edge cases? 202.155.85.18 (talk) 02:08, 28 February 2019 (UTC)[reply]

Are you looking for references on this topic, or simply seeking general discussion of the issue?

Our article cites several reference books that provide a lot more detail than our encyclopedia.

Nimur (talk) 18:01, 28 February 2019 (UTC)[reply]

Shoot, those isotopes of hydrogen are teaching me basic metric system stuff. Yoctoseconds, who knew??? Reading isotopes of hydrogen it appears H-4 decays by neutron emission, which is literally a neutron dripping, so there is no contradiction in our articles. The neutron drip line apparently differs for one vs two neutron cases, but I didn't look into that. According to isotopes of beryllium, Be-13 decays by neutron emission but not Be-14, so go figure. Li-9 can emit a neutron according to isotopes of lithium, Li-10 will, Li-11 won't, so go figure. As a rule, a formula will be approximately right ... the question is just how approximately. ;) Wnt (talk) 02:03, 3 March 2019 (UTC)[reply]