= Nuclear shell model
{wiki}
Why should adding one <neutron> be easy for one <atomic nucleus> and much less favorable for its neighbor? In the <nuclear shell model>, <nucleons> occupy discrete <energy levels> in a <nuclear mean field>. Filling all available states below a sufficiently large gap produces a <nuclear shell closure>. This connects <nuclear separation energies> with the counting of <quantum states>, and connects the <spin> of an odd <atomic nucleus> with its last unpaired <nucleon>.
The starting point is an approximate <Hamiltonian operator>
$$
H=\sum_{i=1}^{A}\left[\frac{\mathbf p_i^2}{2m_i}+U_i(r_i)-a_i(r_i)\frac{\mathbf L_i\cdot\mathbf S_i}{\hbar^2}\right]+V_{\mathrm{res}},
$$
where $U_i$ is a confining <potential energy>, $a_i$ specifies <nuclear spin-orbit splitting>, and $V_{\mathrm{res}}$ contains interactions left out of the <nuclear mean field>. This is an effective organization of the nuclear <many-body problem>, rather than a derivation of the <nuclear force>. In a spherical approximation, <addition of angular momentum> combines the <orbital angular momentum> $l$ with <spin> $s=1/2$ to give $j=l\pm1/2$; for $l=0$, only $j=1/2$ exists.
A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\ldots,j$. The <Pauli exclusion principle> permits one <neutron> in each such state and, separately, one <proton>. Thus a $j=7/2$ level holds eight <neutrons>, or eight <protons>; it does not hold sixteen identical <neutrons>. Here $n=1,2,\ldots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal <quantum number>. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.
Back to article page
