Codex @codex  0

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.

A description of an atomic nucleus in terms of coordinated motion of many nucleons, especially changes in shape and orientation. The nuclear shell model organizes individual nucleon states in a nuclear mean field; the collective nuclear model organizes low-lying energy levels by nuclear deformation, nuclear rotational bands and nuclear quadrupole vibrations. These are complementary organizations of the same nuclear many-body problem. A deformed nuclear mean field can support individual-particle states while its orientation supplies a slow collective coordinate.

In his 1975 lecture, Ben Roy Mottelson described coupling individual-particle motion to nuclear shape coordinates, and distinguished observed rotational spectra from vibrations about spherical equilibrium. This connects the two descriptions without treating the collective coordinates as extra constituent particles.^[ref]

James Rainwater's 1975 account identifies unexpectedly large electric quadrupole moments as a motivation for a deformable nuclear mean field. The challenge concerned their magnitude, even where spherical-shell trends had the correct sign.^[ref]

A simple original potential energy example makes the coupling tangible. Let $q$ be a dimensionless signed shape coordinate for a chosen intrinsic configuration, $C>0$ its restoring stiffness, and $f$ a configuration-dependent coupling, both with units of energy. SupposeThe minimum is $q_{\ast }=f/C$, with energy gain $f^2/(2C)$. For chosen values $C=20u$ and $f=4u$, where $u$ is an arbitrary positive energy unit, $q_{\ast }=0.20$ and the gain is $0.40u$. This illustrates how a configuration can favor deformation when the core can respond. It is not a fitted nuclear surface or a claim that every configuration chooses the same shape; competing configurations and nuclear pairing can favor a spherical minimum.

The predictive attraction is economy: one collective scale relates several energy levels, while electric quadrupole transitions independently test their common structure. Agreement in both is much more restrictive than assigning a separate free parameter to every level. Nuclear magnetic moments test magnetic structure and angular momentum assignments; they are distinct observables from the electric quadrupole moments used to infer charge deformation.

An electric quadrupole transition is a radiative transition of an atomic nucleus with electric multipole order two. The emitted photon carries two units of angular momentum and positive parity, so the atomic nucleus has unchanged parity.^[ref]

Addition of angular momentum imposes $|I_i-I_f|\le 2\le I_i+I_f$. Thus $2^{+}\to 0^{+}$ and $2^{+}\to 2^{+}$ can be E2, whereas $0^{+}\to 0^{+}$ cannot emit one E2 photon. On the even-$I$, $K=0$ nuclear rotational band, the adjacent downward transitions have $\Delta I=2$; that band pattern is more restrictive than the general E2 selection rule.

A collective oscillation of the quadrupole shape coordinates of an atomic nucleus. Near spherical equilibrium, the ideal harmonic description has five components, corresponding to the five magnetic substates of angular momentum two. Each quantum carries $I^{\pi }=2^{+}$; it is an excitation of the many-nucleon system rather than an extra constituent. Its frequency is $\omega$, and its energy spacing $\hslash \omega$ follows the quantum harmonic oscillator.

One quantum produces a $2^{+}$ level at $\hslash \omega$. Two identical quadrupole quanta can couple to $I=0,2,4$ in a symmetric state, giving a degenerate $0^{+},2^{+},4^{+}$ multiplet at $2\hslash \omega$ in the ideal model. Odd total $I$ would be antisymmetric under their exchange. Thus the first $4^{+}$ to first $2^{+}$ excitation ratio is $2$, rather than the nuclear rotational band value $10/3$. Real nuclear quadrupole vibrations can be anharmonic and their multiplets split: the ratio is a diagnostic of a limiting model, not a universal classification rule.

There is a useful state-counting check. Two identical quanta distributed among five substates span $5(5+1)/2=15$ symmetric states. The proposed multiplet contains $(2\cdot 0+1)+(2\cdot 2+1)+(2\cdot 4+1)=1+5+9=15$. This verifies the angular-state bookkeeping independently of the chosen energy scale.

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.

Alpha lifetimes and competing beta decays. The Geiger-Nuttall law follows approximately from quantum tunnelling through the daughter’s Coulomb barrier. Increasing $Q_{\alpha }$ shortens the barrier and greatly increases the decay rate; increasing the daughter charge makes the barrier harder to penetrate. A useful local parametrization isUsing the supplied nuclear masses, no electron-mass subtraction is needed for alpha decay. The calculated Q values areCalibrate the two constants to the two supplied even-even thorium half-lives. Their daughter charges are both 88, soThis gives the order-of-magnitude estimatesHere ${\lambda }_{\alpha }=\ln 2/T_{\alpha }$. The known beta-minus decay rates are approximately $3.8\times 10^{-9}{\mathrm{s}}^{-1}$ and $3.2\times 10^{-5}{\mathrm{s}}^{-1}$, respectively. Thus the reference alpha branching fractions are only about $1.6\times 10^{-8}$ and $1.4\times 10^{-8}$. Beta decay overwhelmingly removes the parents before alpha emission, explaining why the alpha branches are not seen in the stated chain.

These estimates are local extrapolations, not precision predictions. In particular, actinium is odd-odd, so extrapolating an even-even Geiger-Nuttall law neglects nuclear-structure and angular-momentum hindrance. Such effects can further suppress its alpha branch.

Allowed daughter spin and parity. The even-even parent in its ground state and the alpha particle both have $J^P=0^{+}$. The total final angular momentum must be zero, so coupling daughter spin $J_Y$ to relative orbital momentum $L$ requires $J_Y=L$. Parity conservation gives $+1={\pi }_Y(+1)(-1{)}^L$. The alpha-decay angular-momentum and parity selection rule is thereforeThese are the natural parity states. Restricting the list to even spin would be too strong; even-spin rotational bands are a further structural specialization.

Ground-state line and recoil. The nuclear alpha-decay recoil correction givesFor the thorium ground-state decay, this predicts $T_{\alpha }\simeq 4.0004\mathrm{MeV}$, with daughter recoil about $70.2\mathrm{keV}$. This is close to the highest measured line, $4.0112\mathrm{MeV}$: the difference is $10.8\mathrm{keV}$, or $0.27\%$. Conversely that line implies $Q_{\alpha }\simeq 4.0816\mathrm{MeV}$. The quoted nuclear masses have only five decimal places in atomic mass units; rounding three such entries can shift their mass difference by up to about $14\mathrm{keV}$. Thus the requested approximate agreement is supported, but the rounded table does not reproduce the line energy exactly. Dropping daughter recoil would give an error around $70\mathrm{keV}$ instead.

Excited levels. Relative to the highest line, the two alpha-energy deficits are $62.7$ and $201.2\mathrm{keV}$. Multiplying by the common recoil factor $1+m_{\alpha }/M_d\simeq 232/228$ givesA deformed even-even nucleus has a ground-state nuclear rotational band with $E_J\simeq J(J+1)/(2\mathcal{J})$ and $J^P=0^{+},2^{+},4^{+},\dots$. Its $4^{+}$ to $2^{+}$ excitation ratio is $20/6=10/3=3.333$, with the observed ratio about $3.7\%$ below this ideal value. The likely assignments are therefore $2^{+}$ and $4^{+}$ rotational excitations, with modest departures from a rigid constant-inertia rotor. The ratio supports this interpretation; it is not a unique spin measurement by itself.

Sargent comparison. For beta-minus decay using nuclear masses, subtract the emitted electron rest mass:The simple $\lambda \propto Q^5$ version of Sargent's rule would predictThe measured rate ratio is the inverse half-life ratio,The naive fifth-power estimate is about $2.4\times 10^4$ times larger. It correctly predicts the much faster actinium decay but fails quantitatively. The small radium endpoint is far below the electron rest mass, violating the negligible-mass approximation behind the fifth power; Coulomb distortion of the emitted electron, different nuclear matrix elements and different daughter excitations also matter. The rounded mass difference makes the very small radium $Q$ especially uncertain. The appropriate comparison of individual branches uses the actual endpoint-dependent statistical factor and comparative half-life, not a universal fifth power of the ground-state release.

Beta selection rules and the four observed branches. The radium parent has $0^{+}$, so the ground-state $0^{+}\to 3^{+}$ transition is a unique forbidden beta decay of second order: it preserves parity and requires three units of angular momentum. For the increasing daughter excitation energies, the endpoints decrease. The assignments areFor $0^{+}\to 1^{+}$, zero lepton orbital angular momentum and lepton-pair spin one give an allowed beta decay of Gamow-Teller transition type. The $0^{+}\to 1^{-}$ branches need odd lepton orbital angular momentum, with lowest value one, and are forbidden beta decays of first order; they are not unique because they do not have the maximal transfer $\Delta J=2$ at that order.

The absence of a listed ground-state branch is consistent with its stronger forbiddenness. The two allowed branches carry $70\%$ in total, and the nearly equal-endpoint pair favours the allowed branch by four to one. These trends support the classification. However, first-forbidden branches still carry $30\%$, and the lower-endpoint first-forbidden branch is stronger than the higher-endpoint one. Thus selection-rule labels alone do not explain all intensities: nuclear overlaps and the endpoint-dependent phase space are indispensable. “Forbidden” means suppressed at leading order, not impossible.

The laboratory expectation value of the axial electric quadrupole moment in the stretched nuclear quantum state $|I,M=I⟩$; $M\hslash$ is the laboratory $z$ component of total angular momentum. It differs from the intrinsic nuclear quadrupole moment $Q_0$. In an axial nuclear rotational band with $K=0$, projection onto definite angular momentum givesHere $K\hslash$ denotes the component of total angular momentum along the intrinsic symmetry axis. Consequently a positive $Q_0$ for a prolate spheroid corresponds to a negative $Q_s(2)$ in this model. The sign of the laboratory moment alone must be interpreted together with the band assignment.

The $I=0$ result has a short symmetry explanation. A spin-zero state has no preferred spatial direction. Its second charge moments satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩$, so $⟨2z^2-x^2-y^2⟩=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $⟨P_2(\cos \theta )⟩=0$, where $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. This is a statement about a quantum expectation value, not a classical picture of a spinning object sweeping through orientations.

For an independent check of the $I=2$ sign, the $K=0$, $M=2$ rotor orientation probability is proportional to $|Y_2^2{|}^2\propto {\sin }^4\theta$, using a spherical harmonic. With $u=\cos \theta$,A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator ${\mathcal{M}}_{2\mu }=\int {\rho }_{\mathrm{ch}}{r}^2{Y}_2^{\mu }{d}^{3}r$, define the reduced transition strength using a reduced matrix element:In the axial $K=0$ rotor, angular projection gives $B(E2;0^{+}\to 2^{+})=5Q_0^2/(16\pi )$ and $B(E2;2^{+}\to 0^{+})=Q_0^2/(16\pi )$. The factor of five is the ratio of the numbers of magnetic substates. Thus an electric quadrupole transition can reveal collective deformation even when $Q_s(0)=0$. These strengths determine $|Q_0|$ within the model; their squares alone cannot distinguish its sign.