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Material capable of sustaining a nuclear chain reaction with slow neutrons in a suitable configuration. Fissility involves capture and fission probabilities, neutron production and losses, rather than positive fission Q value alone. Uranium-235 and plutonium-239 are common examples; fissile is more specific than fissionable by sufficiently energetic neutrons.

The increase in a nucleus's collective deformation potential energy between a local minimum and a saddle on a path toward separated fragments. A positive final-state Q value does not remove this intermediate barrier. A nucleus can cross it after sufficient excitation energy, or undergo quantum tunnelling. The relevant barrier for neutron-induced fission belongs to the compound nucleus formed after capture, not automatically to the target before capture.

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.

For two spherical fragments with the same proton fraction as the parent, let $(A_1,Z_1)=y(A,Z)$ and $(A_2,Z_2)=(1-y)(A,Z)$. In the semi-empirical mass formula without nuclear pairing, nucleon rest energies, the volume term and the nuclear symmetry energy cancel in the parent-minus-fragment mass difference. The remaining Q value is the displayed competition between increased surface energy and reduced nuclear Coulomb energy. Its symmetric value is positive whenThis is an energy-release condition, not the disappearance of a fission barrier or a prediction of a short mean lifetime.

In two-body alpha decay of a parent at rest, momentum conservation makes daughter and alpha momenta equal and opposite. The nonrelativistic kinetic energies therefore satisfy $T_d/T_{\alpha }=m_{\alpha }/M_d$, givingAn excited daughter reduces the available Q value. To leading recoil order, an alpha line separation $\Delta T_{\alpha }$ corresponds to excitation $E_x=(1+m_{\alpha }/M_d)\Delta T_{\alpha }$. Thus ratios of excitation energies inferred from line separations do not need the common recoil factor. These formulas assume kinetic energies small compared with rest masses; an exact relativistic two-body decay calculation can replace them when necessary.

A sequence in which particles released by one nuclear reaction induce further reactions. In a neutron-driven fission chain, neutron capture, leakage and non-fission absorption compete with neutron production. A sustained chain requires an effective reproduction factor of one; whether material is fissile material cannot be determined from a single positive fission Q value.

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.

Smooth-model fission energy. Use natural units $c=1$, so the quoted nuclear masses are expressed as rest energies. Conservation of mass number and atomic number fixes the second fragment to $(A_2,Z_2)=((1-y)A,(1-y)Z)$. First omit nuclear pairing to compare a continuous range of $y$. Pairing can subsequently add ${\delta }_1+{\delta }_2-{\delta }_{\mathrm{parent}}$ for integer fragments; it has no smooth parity assignment for arbitrary real $y$.

In the parent-minus-daughters mass difference, the nucleon rest energy and volume terms cancel. The nuclear symmetry energy also cancels because both fragments have the same $Z/A$:The remaining liquid-drop fission energy isSplitting increases the surface energy cost, but reduces nuclear Coulomb energy. Both contributions are symmetric under $y\leftrightarrow 1-y$, so the symmetric split is stationary. Symmetry alone would not establish a maximum, since the two contributions compete.

For a full check, put $S=a_S{A}^{2/3}$, $x=a_C{Z}^2/(a_{S}A)$ and $t=y^{1/3}+(1-y{)}^{1/3}$. On $0\le y\le 1/2$, $t$ increases from $1$ to $2^{2/3}$. Expressing the two fractional-power sums through $t$ givesThe derivative can change sign only from negative to positive, so any interior stationary point in $t$ is a minimum. The largest value is therefore either the no-split endpoint $Q=0$ or the symmetric split. Thus symmetric fission maximizes the positive liquid-drop energy release: whenever this smooth model permits an exothermic split, its largest release occurs at $y=1/2$.

There is a qualification to the paper's unqualified wording. For arbitrary low $Z^2/A$, the symmetric split need not maximize $Q$; indeed it is a local minimum when $Z^2/A<(2/5)a_S/a_C=10$. The symmetric-maximum conclusion is valid in the heavy, fission-favorable regime relevant here, rather than a universal property of the formula.

At $y=1/2$,Requiring $Q_{\mathrm{sym}}>0$ gives

Barrier and fragment yields. Positive final-state Q value does not remove the fission barrier along a deformation path. Surface energy initially rises as the nucleus deforms; spontaneous fission then requires collective quantum tunnelling unless enough excitation is present. Its rate is exponentially sensitive to barrier height, width and collective inertia. Thus nuclei close to the exothermicity threshold can have immeasurably long spontaneous-fission mean lifetimes, while sufficiently heavier nuclei have smaller effective fission barriers or larger quantum tunnelling probabilities.

The fission-fragment mass distribution of low-excitation actinide fission is usually double peaked: complementary fragments have mass fractions roughly $0.4$ and $0.6$ rather than both exactly $1/2$. Nuclear shell model effects and the fission dynamics help favor asymmetric fragments. Maximizing the final smooth-model energy release is not a calculation of the fragment yield. The following yield curve is schematic, with no measured counts implied:

Capture excitation energies. Capturing a negligible-energy neutron forms a compound nucleus with the rest energy of the target plus the free neutron. Relative to the compound ground state, the neutron-capture excitation energy is thereforewhere $B$ is positive nuclear binding energy and $S_n$ the compound nucleus's neutron separation energy. The free neutron rest energy cancels in the binding-energy difference. With the supplied semi-empirical mass formula,At $Z=92$, uranium-235 has odd $A$, while uranium-236 is even-even. The pairing contribution is $+33.5(236{)}^{-3/4}=+0.55637\mathrm{MeV}$. By contrast uranium-238 is even-even and uranium-239 has odd $A$, giving $-33.5(238{)}^{-3/4}=-0.55286\mathrm{MeV}$. The individual contributions areThus the model predictsOf the $1.48305\mathrm{MeV}$ predicted difference, $1.10922\mathrm{MeV}$ comes from the change in nuclear pairing. It is the primary reason the even-even capture product is more strongly excited; the smooth terms account for the remaining $0.37383\mathrm{MeV}$.

Thermal versus fast-neutron fission. The last comparison uses the supplied observed capture excitations, $6.5$ and $4.8\mathrm{MeV}$, rather than replacing them by the preceding smooth-model estimates. Interpreting the quoted $6.2$ and $6.6\mathrm{MeV}$ activation energies as the barriers in the respective capture channels givesA simple neutron-induced fission threshold estimate for the second channel requires about $6.6-4.8=1.8\mathrm{MeV}$ additional centre-of-mass kinetic energy. Sub-barrier tunnelling is possible, but it does not give the large thermal-neutron fission probability of the first channel.

The barrier labels in the original PDF need care: it assigns them to ${}^{235}\mathrm{U}$ and ${}^{238}\mathrm{U}$, whereas the states formed after capture are ${}^{236}{\mathrm{U}}^{\ast }$ and ${}^{239}{\mathrm{U}}^{\ast }$. The comparison above uses the intended capture-channel interpretation. If the quoted numbers are literally barriers of the uncaptured target nuclei, the compound-nucleus barriers are missing and this numerical comparison does not follow from those data alone.

In a thermal reactor, neutron moderation slows fission neutrons so the large low-energy fission cross section of uranium-235 can support a nuclear chain reaction. Fuel composition, moderator choice and geometry must balance production against capture and leakage; capture resonances of uranium-238 during slowing are important losses from that neutron population. The reaction rate per target nucleus depends on the spectrum through $\int \varphi (E){\sigma }_f(E)dE$, where $\varphi (E)$ is differential neutron flux per unit energy. A fast spectrum can access the uranium-238 fission channel and changes this balance. Also, uranium-238 is a fertile material: capture followed by two beta-minus decays can produce plutonium-239, a fissile material, so capture can contribute to breeding even when it causes no immediate fission.

Dating principle. Atmospheric cosmic rays produce secondary neutrons, which can generate carbon-14 through ${}^{14}\mathrm{N}(n,p){}^{14}\mathrm{C}$. Carbon then enters atmospheric carbon dioxide and living organisms. While an organism exchanges carbon with its environment, its carbon-14 to carbon-12 ratio approximately tracks that reservoir. After exchange stops, surviving carbon-14 undergoes beta-minus decay, whereas the stable carbon remains. On the closed-sample, constant-initial-ratio model, radioactive decay givesThus radiocarbon dating infers elapsed time from the loss of an isotope ratio or of radioactive activity per fixed carbon mass. It dates cessation of carbon exchange, which need not equal the date an object was manufactured or buried. Actual calendar dating also requires radiocarbon calibration and control of contamination and reservoir differences.

Mean lifetime. For exponential decay, the half-life obeys $e^{-\lambda T_{1/2}}=1/2$, while the mean lifetime is $\tau =1/\lambda$. HenceUsing $1\mathrm{yr}=365.25\times 86400\mathrm{s}$ gives $\lambda =3.83325\times 10^{-12}{\mathrm{s}}^{-1}$ and $\tau =2.60875\times 10^{11}\mathrm{s}$.

The mass minimum and beta stability. Treat the supplied masses as nuclear rest energies in $\mathrm{MeV}$, and omit nuclear pairing as instructed. At fixed mass number $A$, the volume and surface terms of the semi-empirical mass formula are constant. Expanding the remaining terms gives a convex nuclear mass parabola,The continuous minimum in atomic number is thereforeAt $A=14$, the quadratic coefficient is $7.01302\mathrm{MeV}$ and $Z_{\ast }=6.79450$. The nearest integer is $Z=7$; the nuclear isobar of minimum mass is nitrogen-14. Maximizing nuclear binding energy alone would omit the neutron-proton rest-mass difference; its continuous optimum is slightly different, but also selects integer $Z=7$ here.

The values most useful for checking neighboring decay thresholds areConsequently carbon-14 can lower its nuclear mass by changing a neutron into a proton:For bare nuclear masses, the Q value must include the emitted electron mass. Taking $m_e{c}^2=0.511\mathrm{MeV}$,The opposite proton-to-neutron conversion would lead to $Z=5$, whose nuclear mass is $18.1567\mathrm{MeV}$ higher than that of carbon-14. Neither beta-plus decay nor electron capture can bridge that gap by supplying an electron-mass correction. For the nitrogen-14 daughter, both neighboring nuclear masses are higher: $Z=6$ by $4.13070\mathrm{MeV}$ and $Z=8$ by $9.89535\mathrm{MeV}$. Even electron capture to $Z=6$ has $Q=0.511-4.13070<0$. Thus the daughter has no further energetically allowed single beta decay in this model; physical nitrogen-14 is stable.

These threshold numbers belong to the supplied smooth semi-empirical mass formula, not to an accurate prediction of the carbon-14 electron spectrum or its long half-life. Semi-empirical mass errors can change beta-decay thresholds. Also, the instruction to omit nuclear pairing is a model simplification: even $A=14$ does not make physical pairing vanish, and even-even carbon and odd-odd nitrogen need not have the same pairing correction. No omitted pairing coefficient is guessed here.

Activity of modern carbon. Neglect the small contribution of other stable carbon isotopes when counting atoms in a gram. The Avogadro constant givesThe expected radioactive activity of that one gram isor approximately $15.0$ decays per minute per gram of carbon. A becquerel is one decay per second; the normalization is per gram of carbon, rather than per gram of the entire organic sample.

Artefact age. Its remaining activity fraction is $0.13/0.25008=0.51983$. Inverting the exponential decay law,This is the radiocarbon age on the supplied constant-initial-activity, $5730$-year half-life model. It is not a calibrated calendar date; the sample's historical initial ratio and any later carbon exchange are not supplied.

All three processes are weak charged current transformations between neighboring nuclear isobars. Neglecting neutrino masses, and writing $M_a(A,Z)$ for a neutral atomic mass, their Q values are:

These formulae use the paper's natural units. With energies in MeV, beta-plus decay requires an atomic mass-energy difference exceeding $2m_e{c}^2=1.022\mathrm{MeV}$. The extra electron-mass subtraction arises from using neutral atomic rather than bare nuclear masses. Electron capture can occur when the atomic mass-energy difference is positive but below this threshold. For a particular capture shell or excited daughter, the available energy must also cover the electron-hole and nuclear excitation energies, and the small nuclear recoil. Small atomic binding corrections are suppressed in the displayed threshold formulae.

The question's fixed-$A$ family consists of nuclear isobars, rather than isotopes: isotopes have fixed $Z$. The semi-empirical mass formula gives a convex nuclear mass parabola as a function of $Z$ at fixed $A$. For odd $A$, nuclear pairing does not split the family into two branches, so there is typically one beta-stable isobar, at the discrete minimum. Neighboring members can lower their mass by changing $Z$ toward it.

For even $A$, nuclear pairing lowers even-even nuclei and raises odd-odd nuclei, producing two offset nuclear mass parabolas. There are often two beta-stable even-even isobars, separated by an odd-odd member to which neither can energetically decay. The number is not an exact universal rule: some even-$A$ chains have one or more than two such minima against single beta decay, and a few light odd-odd nuclei are stable. Stability against single beta decay also does not guarantee stability against double beta decay.

Both formal processes use one weak charged current transition, with a virtual W boson producing the lepton pair. Two representative lowest-order quark-line diagrams are shown below. They use the $d\bar{d}$ component of ${\pi }^0$; attachment to the antiquark line in the $u\bar{u}$ component gives the other contribution to the same hadronic current.

In the first diagram, $d\to u+W^{-}$ and $W^{-}\to e^{-}{\bar{\nu }}_e$, with $\bar{d}$ a spectator quark. In the second, $u\to d+W^{+}$ and $W^{+}\to e^{+}{\nu }_e$, again with a spectator $\bar{d}$. The diagram for the first process describes the coupling but cannot make an energetically forbidden on-shell decay occur.

From the original tables,Therefore the first process has Q valueand is forbidden for an isolated pion. The second hasThe table therefore lists only the second, pion beta decay, withIts tiny branching fraction combines a weak interaction with an exceptionally small available energy. An allowed semileptonic transition has a decay width scaling approximately as $G_F^2|V_{ud}{|}^2{\Delta }^5$, where $\Delta =m_{{\pi }^{+}}-m_{{\pi }^0}$, with electron-threshold and recoil corrections. The fifth power makes the few-MeV gap especially restrictive. Unlike direct ${\pi }^{+}\to e^{+}{\nu }_e$, this semileptonic channel is not forced to have an amplitude proportional to $m_e$; its primary extra suppression is the small phase space. Its approximately Cabibbo-favoured $u\leftrightarrow d$ current does not supply a further large flavour suppression. The observed branching fraction corresponds to a partial decay rate of about $(1.036\times 10^{-8})/(2.6033\times 10^{-8}\mathrm{s})=0.398{\mathrm{s}}^{-1}$, competing with the much faster dominant muonic decay.

The Q value of a reaction or particle decay is the initial total rest energy minus the final total rest energy. Positive $Q$ releases kinetic energy; a negative $Q$ requires energy supplied by incident motion or another source. Atomic and bare nuclear mass conventions must be distinguished when counting Electrons in beta decay.

Beta decay Q values depend on small differences between much larger atomic masses. A smooth semi-empirical mass formula can therefore approximate absolute masses reasonably while predicting the wrong sign for a MeV-scale decay threshold. Nuclear pairing explains offset even-even and odd-odd mass branches, but does not guarantee that every specific even-even neighbor lies below the intervening odd-odd nucleus. Measured masses or additional nuclear shell corrections are needed for quantitative threshold tests near a model's residual error scale.