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The Geiger-Nuttall law expresses the extreme sensitivity of an alpha decay half-life to its Q value. In a restricted range of heavy nuclei, a useful local form is ${\log }_{10}[T_{1/2}/T_{\ast }]\simeq a{Z}_d/\sqrt{Q_{\alpha }}+b$, where $Z_d$ is the daughter atomic number, and the constants depend on the chosen units and nuclear family. It is named for Hans Geiger and John Mitchell Nuttall.

The Coulomb barrier outside the daughter has $V(r)=2Z_d\alpha /r$ in natural units. Its outer turning point is $r_t=2Z_d\alpha /Q_{\alpha }$. The WKB approximation gives a penetration probabilitywhere $\mu$ is the daughter-alpha reduced mass. The leading large-barrier exponent is $2\pi (2Z_d\alpha )\sqrt{\mu /(2Q_{\alpha })}$, producing the approximate $Z_d/\sqrt{Q_{\alpha }}$ dependence. Finite nuclear size modifies the exponent and the slowly varying intercept. The nuclear decay constant also contains an assault frequency and an alpha-cluster formation probability. Extra orbital angular momentum raises the barrier further, and odd-nucleon structure can introduce substantial hindrance. Thus extrapolation from even-even decays gives an order-of-magnitude reference rather than a precision prediction for an odd-odd nucleus.

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.