# Why these two hypothetical nuclei have no differ by the asymmetry energy?

Consider two hypothetical nuclei both with the mass number $$A=6$$ but one with $$Z=N=3$$ and the other with $$Z=2, N=4$$. Let us assume that each nucleon moves in an average nuclear potential as in the shell model, which is a simple harmonic potential equally spaced levels having energies $$E_0, E_0+\delta, E_0+2\delta,...$$, etc.

In the case of the 1st nucleus, two protons (spin-up and spin-down) and two neutrons (spin-up and spin-down) will go into the ground state, and the remaining one proton and one neutron will go into the first excited state. Therefore, its energy will be $$4E_0+2(E_0+\delta)$$.

In the case of the 2nd nucleus, two protons (spin-up and spin-down) and two neutrons (spin-up and spin-down) will go into the ground state, and the remaining two neutrons will go into the first excited state. Therefore, its energy will again be $$4E_0+2(E_0+\delta)$$.

But don't we expect that the second nucleus to have higher asymmetry energy due to asymmetry in the neutrons and proton numbers? Forget about pairing energy and all that. Why isn't the asymmetry energy making any contribution here or is my caricature of the shell model too naive to capture this? Isn't $$N\neq Z$$ supposed to make an additional contribution?

• I am not a specialist but I would say that there should be at least Ep and En, rather than E zero. Or you can't forget pairing and interaction. At the end higher Z requires even more higher N. Oct 31, 2021 at 9:24

First of all, these nuclei are not hypothetical, they are known as $$^6$$Li and $$^6$$He, with masses $$m(^6{\rm Li})=6.015u, \qquad m(^6{\rm He})=6.018u.$$ $$^6$$He is unstable. The main decay channel is beta decay to $$^6$$Li.
As you discuss, in a naive shell model these two nuclei have the same mass. Part of the mass difference is that the neutron is heavier than the proton, but this only explains part of the difference (about 0.0013$$u$$). The rest must be due to residual correlations. Chief among them is probably the $$pn$$ interaction in the deuteron (spin one, isospin zero) channel. Indeed, $$^6$$Li has spin one.