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The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

Ben Mottelson helped connect independent-particle nuclear motion with collective excitations. In a simple positive-parity, $K=0$ ground state nuclear rotational band of an even-even nucleus, the excitation energies approximately follow $E_J={\hslash }^{2}J(J+1)/(2\mathcal{I})$ for $J=0,2,4,\dots$, where $\mathcal{I}$ is a moment of inertia. Thus $E_4/E_2=10/3$ in this idealization. Comparing such patterns and transition strengths with observation distinguishes organized collective motion from an arbitrary list of separate particle excitations.^[ref]

A physicist.^[ref]

The axial electric quadrupole moment of an intrinsic nuclear charge distribution, with the origin at its center and the $z$ axis along its symmetry axis:This convention includes electric charge in $Q_0$; a tabulated moment divided by the elementary charge $e$ instead has units of area. For the uniform spheroid, mapping a uniform unit ball to $(ax,ay,cz)$ gives $⟨x^2⟩=a^2/5$ and $⟨z^2⟩=c^2/5$, soThe formula explains the signs in the figure. It measures an anisotropy of the nuclear charge distribution, rather than counting individual nucleons on a surface. An illustrative choice $\eta =\sqrt{8}$ gives $a=R/\sqrt{2}$, $c=2R$ and $Q_0=7Ze{R}^2/5$, while preserving $a^{2}c=R^3$ exactly.

James Rainwater's lecture contrasts measured laboratory electric quadrupole moments with intrinsic moments inferred through Coulomb excitation, including even-even nuclei. Extracting intrinsic shape therefore uses a nuclear model, rather than equating every measured moment to $Q_0$.^[ref]

Departure of an intrinsic nuclear shape from spherical symmetry. Here intrinsic means referred to axes attached to a model nuclear charge distribution. A nuclear quantum state with definite spin need not have the same laboratory charge density as that intrinsic distribution. In particular, the ground state of a deformed even-even nucleus can have $I=0$, a spherical laboratory expectation value, and a nonzero intrinsic nuclear quadrupole moment.

For an illustrative uniform charge density inside a spheroid, let the semiaxes be $(a,a,c)$ and keep its volume equal to that of a sphere of radius $R$:Thus $\eta >1$ gives a prolate spheroid, $\eta =1$ a sphere, and $\eta <1$ an oblate spheroid. These are three-dimensional equal-volume shapes: equal areas of their drawn cross-sections would be a different constraint. Real nuclear charge distributions have diffuse surfaces and need not be uniform, axially symmetric or confined to one shape.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

A positive-parity $K=0$ nuclear rotational band in an even-even nucleus has $J=0,2,4,\dots$ and excitation energiesThe four nonzero levels give $E_J/[J(J+1)]=15.17,14.85,14.48,14.01\mathrm{keV}$. These approximately constant coefficients, together with the even-spin positive-parity sequence, support a rotational interpretation. The roughly eight-percent decline means a constant-inertia rigid rotor is only an approximation: the effective moment of inertia increases along the sequence.

Using the lowest excited level givesUsing the highest level instead gives $2.48\times 10^{-54}\mathrm{kg}{\mathrm{m}}^2$, showing the scale of the departure from a constant inertia. This is a comparison of nuclear rotational inertia relative to a rigid sphere, not evidence for an exactly spherical rotating intrinsic state.

For a conventional size estimate, explicitly take a uniform sharp radius $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ and mass $M\simeq A{m}_u$. These are standard scale approximations, not additional measured properties of this isotope supplied by the paper. They give $R\simeq 6.70\mathrm{fm}$ andThus the fitted rotational inertia is substantially smaller than the rigid-body scale. Nuclear pairing provides a physical reason collective nuclear inertia need not equal the classical rigid-body value. The comparison's numerical radius dependence is explicit; it is not an independently determined radius.

In the nuclear shell model, a level with total angular momentum $j$ has $2j+1$ substates for each nucleon species. The first three levels accommodate $2+4+2=8$ protons or eight neutrons. nuclear pairing makes a filled shell or a pair of like nucleons contribute $J=0$, and an orbital has spatial parity $(-1{)}^{\ell }$.

For oxygen-17, eight protons and eight neutrons form a closed core; the remaining neutron occupies $1d_{5/2}$, so $J^P=5/2^{+}$. For nitrogen-15, eight neutrons close their shell, while the seventh proton occupies $1p_{1/2}$. Equivalently it is a proton hole relative to the closed eight-proton shell, giving $J^P=1/2^{-}$. beryllium-10 is an even-even nucleus; pairing its four protons and six neutrons gives $J^P=0^{+}$.

boron-10 needs an additional qualification. The fifth proton and fifth neutron each occupy $1p_{3/2}$. addition of angular momentum permitsThe experimentally correct ground-state assignment is $3^{+}$, corresponding to aligned unpaired $j=3/2$ contributions in this simple description. However, one-particle level ordering alone does not select it over the other couplings: their splitting requires the nuclear residual interaction. Assuming a preference for the aligned proton-neutron coupling gives the intended prediction; without that assumption, the strictly deducible result is the allowed positive-parity multiplet. The evaluated ground state is $3^{+},T=0$, with a $1^{+},T=0$ level above it.^[ref]