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A proton number $Z$ or neutron number $N$ associated with a pronounced nuclear shell closure. The traditional sequence isThese count either species separately, rather than their sum $A=Z+N$. Calling a number magic summarizes a structural pattern; it does not promise that every isotope at that number is equally stable.

For example, in the chosen scheme the $20$-particle core plus the eight substates of $1f_{7/2}$ gives $28$. The next group contributes $4+6+2+10=22$, giving $50$. The following two groups contribute $8+6+4+2+12=32$ and $10+8+6+4+2+14=44$, giving $82$ and $126$. Counting verifies the capacities once the grouping has been chosen; it does not establish the physical energy gaps that justify that grouping.

Correlations that favor pairs of like nucleons coupled to zero total angular momentum in the simplest spherical picture. They help explain why an unpaired nucleon often dominates an odd nucleus's spin, and why neighboring nuclear separation energies show an odd-even staggering. Nuclear pairing belongs to the residual-interaction physics beyond merely filling independent energy levels. Mixing different occupations and deformation can also be necessary; the closed-core picture is most useful near suitable nuclear shell closures.

The energy needed to remove specified nucleons from a bound atomic nucleus in its ground state. It compares neighboring nuclear masses, and so can reveal a nuclear shell closure more clearly than a smoothly varying total nuclear binding energy.

A nuclear shell groups single-particle energy levels for protons or neutrons in a nuclear mean field. Filling all states below a sufficiently large gap gives a nuclear shell closure. These shells describe quantum-state occupation rather than thin spatial layers. Their discrete filling effects explain departures from the smooth semi-empirical mass formula.

Filling the accessible one-species nuclear shell model states below a substantial energy gap. A closed spherical $j$ level has all $m_j$ values occupied, so their sum is zero. In the independent-particle picture, a closed core therefore contributes zero total angular momentum; a single particle outside it, or a single missing particle inside it, supplies the leading spin assignment. Residual interactions can change this simple picture.

The three-dimensional isotropic harmonic oscillator alone has one-species shell capacities $(q+1)(q+2)$ for oscillator index $q=0,1,\dots$, including spin. Summing gives closures $2,8,20,40,70,112,\dots$, rather than the complete traditional magic number sequence. Strong nuclear spin-orbit splitting can separate a high-$j$ level from its former neighbors. Realistic radial potential energies and residual interactions also matter: a chosen oscillator plus an arbitrary single constant is not a demonstrated explanation of every real nuclear gap.

Traditional nuclear shell closures can weaken along isotope chains far from stability, as in neutron-rich magnesium.^[ref]

The relevant test is the actual energy gap and the correlated nuclear quantum states, rather than membership in an immutable numerical list.

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 operatorwhere $U_i$ 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 $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. 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 $n=1,2,\dots$ 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 $l=0,1,2,3,4,5,6$.

The qualitative explanation is nuclear pairing: ${}^{40}\mathrm{K}$ is odd-odd ($Z=19,N=21$), while both ${}^{40}\mathrm{Ar}$ and ${}^{40}\mathrm{Ca}$ are even-even. At $A=40$, moving from the odd-odd to either even-even branch lowers the pairing contribution to the atomic mass byTwo neighboring even-even nuclear isobars can therefore both be lower in mass than the intervening odd-odd member. The two possible directions areHowever, the particular coefficients printed in this paper do not quantitatively reproduce both allowed directions. Let $D=m_p+m_e-m_n\simeq -0.7823\mathrm{MeV}$. Using every supplied term, rather than pairing alone, givesThus this parametrization would forbid the calcium beta-minus branch. The physical atomic mass differences are instead approximately $+1.505\mathrm{MeV}$ to argon and $+1.311\mathrm{MeV}$ to calcium, so both physical directions are allowed.^[ref]

The valid interpretation is therefore pairing explains how both directions can be possible, but the supplied smooth mass formula is insufficient to establish both for this isotope. The semi-empirical mass formula has MeV-scale residual errors and omits nuclear shell closure corrections; ${}^{40}\mathrm{Ca}$ has the especially stable $Z=N=20$ shell closures. One must use measured masses or a model with those corrections for a quantitative proof, rather than changing the printed numbers silently.