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Calcium-40 is an even-even nucleus with the closed-shell values $Z=N=20$. Its extra stability illustrates nuclear shell corrections beyond a smooth semi-empirical mass formula.

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.

At fixed mass number $A$, the smooth semi-empirical mass formula makes the atomic mass approximately quadratic and convex in atomic number $Z$. Neighboring nuclear isobars undergo energetically allowed beta decay toward a discrete local minimum, subject to the electron-mass threshold for beta-plus decay. For odd $A$ there is one smooth branch; for even $A$, nuclear pairing splits the even-even and odd-odd branches. A local minimum against a single change of $Z$ need not be stable against double beta decay.

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.

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.

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.

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.

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.

The semi-empirical mass formula approximates nuclear binding energy using volume, surface, electrostatic, neutron-proton asymmetry, and nuclear pairing contributions. At fixed mass number, its quadratic dependence on atomic number gives a nuclear mass parabola. It describes broad trends and does not resolve individual nuclear shell effects.

Nuclear symmetry energy is the energy cost associated with proton-neutron imbalance in nuclear matter or finite nuclei. A finite-nucleus semi-empirical mass formula includes the positive asymmetry term $a_A(N-Z{)}^2/A$. The Pauli exclusion principle and strong-interaction dynamics both contribute.