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The deuteron is the bound proton–neutron system. Its ground state has $J^P=1^{+}$ and isospin $T=0$, with binding energy approximately $2.225\mathrm{MeV}$ and no excited bound state. The small binding compared with characteristic nuclear energy scales produces an extended tail. With reduced mass $\mu \simeq m_N/2$, the asymptotic radial wavefunction decays as $e^{-\kappa r}$, whereThe decay length is not a sharp deuteron radius or the force range. It illustrates weak binding outside the region of strong attraction.

The observed total spin is consistent with aligned nucleon spins, $S=1$, and a predominantly $L=0$ spatial state, ${}^3{S}_1$. A central attraction identical in both spin–isospin channels would not distinguish this channel from a spin-singlet proton–neutron channel. In fact the triplet channel binds while the corresponding singlet channel does not. A realistic nuclear force must therefore distinguish these spin–isospin channels, for example through a term $V_{\sigma }(r){\mathbf{\sigma }}_p\cdot {\mathbf{\sigma }}_n$, whose eigenvalues are $+1$ and $-3$ in triplet and singlet states. The absence of a bound dineutron and diproton reinforces the need to distinguish channels; proton–proton Coulomb repulsion is an additional effect, not a complete explanation of the singlet nuclear interaction.

The clearest evidence against a purely central force is the nonzero electric quadrupole moment. A pure isotropic S-wave charge distribution would give zero. The deuteron quadrupole evidence for a tensor force instead points to a nuclear tensor force,It depends on how the spins point relative to the line joining the particles. It preserves total $J$ but couples orbital angular momenta differing by two, producingBoth terms have $J^P=1^{+}$, and their interference permits a quadrupole moment. The 1939 theoretical analysis explicitly connected the measured quadrupole moment with such directional dependence.^[ref]

The nuclear magnetic moment provides a second check. A pure aligned-spin S wave would give approximately ${\mu }_p+{\mu }_n\simeq 0.880{\mu }_N$, close to but larger than the observed deuteron value approximately $0.857{\mu }_N$. The discrepancy is consistent with non-S-wave structure and interaction-current corrections; it is not by itself a model-independent measurement of a D-state probability. The deuteron D-state admixture depends on the adopted Hamiltonian and electromagnetic current. The magnetic moments and mass differences used for these scales are tabulated physical constants.^[ref]

A useful schematic interaction therefore contains central, spin-dependent and tensor terms, and possibly spin–orbit terms in other channels. Long-range one-pion exchange naturally supplies a tensor component, with its finite range set by the pion mass. The deuteron establishes an attractive triplet channel and a noncentral tensor interaction; a spin-independent central potential is insufficient.

The mirror nuclei have interchanged neutron and proton numbers. Neglect their strong-interaction binding difference and describe them with the same uniformly charged sharp sphere of radius $R$. The nuclear Coulomb energy isChanging one neutron into a proton decreases the nucleon rest-mass sum by $m_n-m_p=1.294\mathrm{MeV}/c^2$, but raises the Coulomb energy. The diagram gives $(M_F-M_O)c^2=2.25\mathrm{MeV}$ for nuclear masses, soWith the standard electromagnetic conversion $\alpha \hslash c\simeq 1.44\mathrm{MeV}\mathrm{fm}$, the uniform-sphere Coulomb displacement energy estimate isThis $R$ is the uniform sphere's outer radius. If “charge radius” means the rms nuclear charge radius, the same distribution givesThe numerical result is an estimate within the charge-density model, not a model-independent radius measurement. A discrete-proton correction replaces $Z^2$ by $Z(Z-1)$, giving $\Delta [Z(Z-1)]=16$ and $R\simeq 3.90\mathrm{fm}$. The main answer above follows the literal continuum charged-sphere model. Strong isospin-breaking, nonuniform density and changes in the valence-nucleon distribution also limit the extraction.

Beta decay converts a neutron to a proton or vice versa through the weak interaction, leaving the mass number fixed:Inside a nucleus the full initial and final nuclear masses determine whether either process is possible. Electron capture, $p+e^{-}\to n+{\nu }_e$, is the related bound-electron process. The emitted charged-lepton spectrum is continuous because energy is shared with the neutrino and nuclear recoil.

For allowed beta decay, the lepton wavefunctions' leading spatial terms are constant over the nucleus. A Fermi transition has $\Delta J=0$ and no parity change. A Gamow-Teller transition has $\Delta J=0,1$ and no parity change, excluding $0\to 0$. Thus a same-parity transition with nonzero initial and final spins and $\Delta J=0$ can contain both contributions. First-forbidden transitions change parity and permit $|\Delta J|=0,1,2$; the $|\Delta J|=2$ case is unique forbidden beta decay. Higher forbiddenness requires higher powers of the small lepton momentum times nuclear size. “Forbidden” denotes suppression of the leading allowed operator, not an impossible decay.^[ref]

Fermi's golden rule expresses a rate as a squared nuclear matrix element times leptonic phase space. Neglecting charged-lepton mass and Coulomb distortion, a constant allowed matrix element givesThis is the approximate Sargent rule, $\lambda \propto Q^5$ for comparable allowed strengths; it is not an exact law across arbitrary transitions. Finite electron mass, the Fermi function, forbidden shape factors and nuclear structure change the comparison. A microscopic first-forbidden unique operator introduces additional momentum factors.

Using bare nuclear masses $M_i,M_f$, the kinetic-energy release isThe daughter is different in the two cases: its charge is $Z+1$ for beta minus and $Z-1$ for beta plus. The relevant mass difference must exceed one electron mass. For neutral atomic masses, to leading order in small electronic binding corrections, the corresponding expressions are $Q_{{\beta }^{-}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}})c^2$ and $Q_{{\beta }^{+}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}}-2m_e)c^2$. Electron capture is energetically possible approximately when $M_i^{\mathrm{at}}>M_f^{\mathrm{at}}$, with available shell binding energy and daughter excitation included. Nuclear masses and atomic masses must not be mixed.

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]

A muonic atom binds a negative muon in the nuclear Coulomb potential. The muon is a spin one-half lepton with the electron's charge but about 207 times its mass. For a point nucleus of mass $M_N$ and charge $Ze$, the reduced mass and nonrelativistic hydrogenic scales, with fine-structure constant $\alpha$, areFor a heavy nucleus ${\mu }_r\simeq m_{\mu }$, giving a radius roughly $1/207$ of the corresponding electronic radius and transition energies roughly 207 times larger. For hydrogen, nuclear recoil matters: $m_{\mu }\simeq 106\mathrm{MeV}/c^2$ gives ${\mu }_r\simeq 95\mathrm{MeV}/c^2$ and $a_{\mu }\simeq 284\mathrm{fm}$ rather than the infinite-nuclear-mass value about $255\mathrm{fm}$.

A slowed negative muon is captured into an excited orbit and undergoes a radiative cascade, ejecting atomic electrons and emitting muonic X-rays. Inner orbits penetrate the nuclear charge distribution, particularly at large $Z$. Replacing a point charge by an extended charge makes the central attraction less singular and shifts states with substantial nuclear overlap. Measured transition energies consequently constrain the nuclear charge radius. Sensitivity is particularly strong for $S$ states, whose contact density scales as $(Z\alpha {\mu }_r{)}^3$.

An instructive experiment is Randolf Pohl's team's muonic-hydrogen spectroscopy. A stopped-muon target prepared a metastable $2S$ population; a tunable pulsed laser drove a $2S\to 2P$ transition, followed by a detectable $2P\to 1S$ X-ray. Scanning laser frequency located a resonance whose interpretation included the proton finite-size contribution to the Lamb shift. Delayed detection selected the long-lived population against prompt cascade backgrounds. Laser-off or detuned measurements provide a background control; energy and time selection alone would not establish a laser-induced resonance.^[ref]

Extracting a radius requires quantum electrodynamics, including vacuum polarization, recoil and nuclear-structure corrections. A free muon decays on a microsecond timescale; a bound negative muon can instead undergo muon capture, for example ${\mu }^{-}p\to n{\nu }_{\mu }$. These processes limit observation time. In heavy atoms, relativistic motion and nuclear overlap make the simple Bohr formulas only qualitative. Muonic spectroscopy converts a smaller orbital scale into enhanced sensitivity to nuclear charge structure, provided atomic and nuclear corrections are controlled.

The nuclear Fermi gas model treats protons and neutrons as two species of spin one-half fermions moving approximately independently in a confining nuclear mean field. At a temperature small compared with their Fermi temperature, the Pauli exclusion principle fills the lowest available momentum states. It does not allow all nucleons to sit at zero momentum.

For species $i=p,n$, two spin states occupy each momentum-space cell $(2\pi \hslash {)}^3/V$. Counting a filled sphere giveswhere $n_i=N_i/V$. The Fermi energy is measured from the mean-field well bottom; it need not equal the energy needed to remove a nucleon into vacuum. Occupied-state averaging givesEven at zero temperature there is kinetic energy and degeneracy pressure. As an explicitly illustrative input, total density $n=0.16{\mathrm{fm}}^{-3}$ and $N=Z$ give $p_{F}c\simeq 263\mathrm{MeV}$ and $E_F\simeq 37\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.

Separate proton and neutron seas explain a kinetic contribution to nuclear symmetry energy. At fixed $n$, put $\delta =(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. ThenIncreasing one species' population raises its Fermi surface and costs energy, favoring similar populations before electrostatic and interaction terms are included.

pauli blocking suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a nucleon into an empty state and reveal its initial Fermi motion, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent nuclear mean field.

A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete nuclear shells, nuclear spin-orbit splitting, nuclear pairing, short-distance correlations and deformation. It cannot assign all ground-state spins from a momentum sphere. Its useful prediction is the scale and occupancy of nucleon motion, with a surrounding mean field supplied separately.

The dominant interaction is strong. At hadron level, panel (iii) is an effective hadronic vertex for $\Delta p\pi$. The initial $uuu$ constituents rearrange into $uud$ and $u\bar{d}$ after creation of a $d\bar{d}$ pair:One original up quark enters the charged pion; the other two join the new down quark in the proton. Gluon-induced pair creation supplies the quark-flow diagram interpretation. This is nonperturbative hadronization: the effective hadronic vertex does not claim that one elementary perturbative quark graph calculates its width. Charge, baryon number and net quark flavor are conserved.

An S wave has $L=0$ and positive parity. The deuteron's $J=1$ therefore requires a spin-one-half triplet state, $S=1$. The two-nucleon isospin description makes this the antisymmetric $T=0$ channel. Its existence proves sufficient attraction to bind in the ${}^3{S}_1$-dominated channel.

A spin one-half pair could also form $S=0$. The absence of a bound ${}^1{S}_0$ deuteron indicates that the singlet interaction does not bind, not that it is everywhere repulsive. Together with triplet binding, this shows that a spin-independent central attraction is insufficient. No radial excited bound level also constrains the strength and range: the attraction supports only a shallow low-energy bound state.

For identical neutrons or protons, two-particle isospin has $T=1$ and is symmetric. At even $L$, spatial exchange is symmetric, so the Pauli exclusion principle requires an antisymmetric spin state, $S=0$. A low-energy $nn$ or $pp$ pair cannot occupy the deuteron's $S=1,T=0$ channel. With approximate charge independence of nuclear forces, their singlet strong interaction corresponds to the unbound neutron-proton singlet; proton-proton electrostatic repulsion further disfavors binding. There is no bound dineutron or diproton in these physical low-energy channels. The inference uses charge independence and singlet-channel behavior; the deuteron's quantum numbers alone do not constrain every imaginable potential.

In a pure $L=0,S=1$ stretched state both spins point along the total angular momentum, with no orbital contribution. The nuclear magnetic moment would beThe measured $0.857{\mu }_N$ differs. More decisively, a pure S wave spatial state is isotropic and has zero electric quadrupole moment. Its nonzero value requires a nonspherical component of the quantum charge distribution.

Positive parity allows even $L$; coupling $S=1$ to $J=1$ permits $L=0$ and $L=2$. The state contains a deuteron D-state admixture:with radial wavefunctions implicit. A nuclear tensor force couples the orbital components. The quadrupole moment involves their interference as well as a D-state contribution, making it sensitive to a small admixture. This is a quantum angular correlation, not two particles permanently fixed in space.

In the nonrelativistic one-body magnetic-moment approximation,The D-state coupling has $⟨S_z⟩=-1/2$ and $⟨L_z⟩=3/2$ at $J_z=1$; the proton supplies approximately half the relative orbital magnetic moment. ThusThe robust conclusion is a predominantly S wave state with a small tensor-induced D wave component. The 4% estimate is model dependent: exchange currents and other corrections also affect the magnetic moment, and a wavefunction component probability is not itself a uniquely measured observable.^[ref]

Use the inverse beta decay threshold, with the charged-lepton mass generalized to $m_{\ell }$. For a stationary proton and a massless incident antineutrino, the Mandelstam invariant is $s=m_p^2+2m_p{E}_{\nu }$. At threshold the final particles have no relative momentum in their centre-of-momentum frame, so $s_{\mathrm{thr}}=(m_n+m_{\ell }{)}^2$. ThereforeUsing the supplied, rounded masses givesThe second term is the recoil correction; setting the threshold equal just to the final rest-mass excess would miss it.

Reactor electron antineutrinos have energies of a few million electronvolts. They can produce a Positron through inverse beta decay above the first threshold, but cannot produce a muon. Thus reactor neutrino oscillations are measured primarily as electron-antineutrino disappearance, through the event rate and energy spectrum relative to the unoscillated prediction. An oscillated muon antineutrino cannot be identified by the proposed muon-production channel at reactor energies; disappearance alone does not distinguish all possible final flavours.

The separate proton and neutron shell capacities are $2j+1=2,4,2,6,\dots$. Filled shells and paired particles have zero net spin and magnetic dipole moment. The single-particle nuclear magnetic moment is $\mu =g_{j}j{\mu }_N$, with orbital factors $g_{\ell }^p=1$, $g_{\ell }^n=0$ and $s=1/2$.

For nitrogen, the eight neutrons form a closed shell. The seventh proton occupies $1p_{1/2}$, so $\ell =1$, $j=1/2$ and the Landé g-factor givesHence the nuclear shell model predictsIts single-particle spin and parity is $1/2^{-}$. The negative moment results from the proton spin being predominantly opposed to the total angular momentum in this $j=\ell -1/2$ level.

For oxygen, eight protons and the first eight neutrons are paired. The ninth neutron occupies $1d_{5/2}$, with $\ell =2$, $j=5/2=\ell +1/2$. Since its orbital factor is zero, $g_j=g_n/(2\ell +1)=-3.826/5$. ThereforeThese are ideal single-particle predictions in units of the nuclear magneton, not fits incorporating nuclear configuration mixing.

In both semileptonic decays, an active quark emits a virtual W boson, which produces an electron and an electron antineutrino:The unchanged $u,d$ spectator quarks combine with the produced $u$ into the final proton. The relevant Cabibbo-Kobayashi-Maskawa matrix elements are $V_{ud}\simeq \cos {\theta }_C$ and $V_{us}\simeq \sin {\theta }_C$.

Use the quoted masses in energy units, equivalently natural units $c=1$, and define the energy releases before subtracting the electron rest energy asThe prescribed density of states ansatz uses precisely these differences to the fifth power. A Cabibbo-angle extraction from semileptonic decay rates also needs a common reduced hadronic strength. In that approximation the partial decay rates, measured in ${\mathrm{s}}^{-1}$, areThe supplied mean lifetimes describe all decays, so multiply their inverses by the relevant branching fractions:Taking their ratio cancels $K$ and givesChoose the conventional first-quadrant Cabibbo angle:The Lambda baryon partial rate is much larger despite its smaller weak flavour coupling because the fifth-power phase space factor is enormous.

There is an assumption beyond the stated density of states ansatz: the neutron and Lambda reduced hadronic strengths have been set equal. If their ratio is $H=K_{\Lambda }/K_n$, the same data instead imply ${\tan }^2{\theta }_C=0.0588300/H$. Without specifying $H$, the angle is not uniquely determined. Unequal hadronic form factors, finite electron mass and recoil make the displayed common-strength result an exercise-level estimate, rather than a precision measurement.

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 original chart has $Z$ vertically and $N$ horizontally. Thus its region (a) is proton-rich relative to the beta-stable band, whereas (b) is neutron-rich.

Electron capture, $e^{-}+(Z,N)\to (Z-1,N+1)+{\nu }_e$, is the other proton-to-neutron weak process; it remains possible for some neutral atoms whose beta-plus decay lacks the $2m_e$ atomic-mass threshold.The emitted alpha particle contains two protons and two neutrons. Very heavy nuclei can also compete through nuclear fission; the chart alone does not determine each nuclide's branching fractions.

The quark transitions are $u\to d+W^{+}$ followed by $W^{+}\to e^{+}{\nu }_e$ in (a), and $d\to u+W^{-}$ followed by $W^{-}\to e^{-}{\bar{\nu }}_e$ in (b). The additional panel shows electron capture as $u+e^{-}\to d+{\nu }_e$ through W boson exchange. The remaining quarks in each nucleon are spectators.

Regions (c) and (d) lie beyond the proton and neutron binding limits, respectively. In (c), unbound protons can be emitted, $(Z,N)\to (Z-1,N)+p$; in (d), unbound neutrons can be emitted, $(Z,N)\to (Z,N-1)+n$. These are the proton drip line and neutron drip line limits, with one- or several-nucleon emission possible according to the nuclear separation energies. They are not merely more distant points on a beta-decay trajectory.

Use a high-energy proton beam on a target to produce secondary pions, then select and focus the negative ones. A magnetic horn and a charge-selecting dipole system remove positive pions, limiting wrong-sign ${\nu }_{\mu }$ contamination. Select a narrow range of magnetic rigidity; if additional species rejection is available, velocity or mass discrimination reduces negative kaon contamination at the same momentum.

The selected ${\pi }^{-}$ travel through an evacuated decay tunnel and undergo ${\pi }^{-}\to {\mu }^{-}{\bar{\nu }}_{\mu }$. Their forward Lorentz boost concentrates the antineutrinos within an angle of order $1/{\gamma }_{\pi }$. Choose a tunnel long enough for substantial pion decay but short compared with the boosted muon decay length. Since ${\tau }_{\mu }\simeq 2.2\mu \mathrm{s}$ is much longer than ${\tau }_{\pi }\simeq 26\mathrm{ns}$, this reduces secondary ${\mu }^{-}\to e^{-}{\bar{\nu }}_e{\nu }_{\mu }$ contamination.

Downstream magnets sweep the remaining charged particles into absorbers, and shielding stops residual hadrons and muons. The antineutrino beam continues straight through. Parent-particle collimation, charge selection, and a forward detector improve purity; neutral-kaon decays, remaining wrong-sign parents, and muon decays prevent perfect purity. Neutrinos themselves cannot be focused or charge-selected by these magnets. Neutrinos produced after stopped-muon decay or capture are much less forward and can be reduced by directional and energy selection.

There are three free-particle rest-energy terms and five nuclear corrections. In the paper's natural units, their meanings are:

The sign convention is significant: the quantity labeled $E_{\mathrm{bind}}$ in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive nuclear binding energy $B$:The coefficients must therefore retain their printed signs; using positive $B$ while still adding it to the masses would be inconsistent.

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.

Neon has $Z=10$ and sodium has $Z=11$, while $A=23$ is unchanged. The conversion is therefore beta-minus decay,A neutron is converted to a proton by the weak charged current. If an excited daughter is populated, subsequent gamma decay produces the observed photons without changing either $A$ or $Z$. The initial weak transition and the later electromagnetic nuclear de-excitation are distinct stages. A direct transition to the daughter ground state produces no nuclear de-excitation gamma ray.

The table gives the Xi baryon mean lifetime ${\tau }_{{\Xi }^0}=2.90\times 10^{-10}\mathrm{s}$. At $t_1=10^{-12}\mathrm{s}$, an initially resting particle has survival probabilityThus the most probable contents are one undecayed ${\Xi }^0$. The initial momentum is unspecified. For a moving parent, time dilation replaces the denominator by ${\gamma }_{\Xi }{\tau }_{{\Xi }^0}$, making the survival probability at least as large. The times below are characteristic proper mean lifetimes; laboratory times depend on the particle velocities.

The principal first step isIt typically occurs on the $10^{-10}\mathrm{s}$ scale. The neutral pion almost immediately converts to two photons, with ${\tau }_{{\pi }^0}=8.43\times 10^{-17}\mathrm{s}$ and $B_{\gamma \gamma }=0.98823$. The Lambda baryon has ${\tau }_{\Lambda }=2.632\times 10^{-10}\mathrm{s}$ and two dominant routes.

In the charged-pion route,The muon lifetime is a standard additional input, not present in the supplied five table pages.^[ref] The successive contents are approximately $\Lambda +2\gamma$, then $p+{\pi }^{-}+2\gamma$, then $p+{\mu }^{-}+{\bar{\nu }}_{\mu }+2\gamma$. After a few muon lifetimes the dominant stable endpoint isThe leading-path probability including the tabulated two-photon pion branch isHere the last factor is the tabulated dominant muonic charged-pion fraction; rare radiative corrections are being neglected in describing a fixed photon count. Before selecting pion decay channels, the probability of reaching the $p{\pi }^{-}{\pi }^0$ history is $0.99524\times 0.639=0.63596$.

In the neutral-pion route,This produces $n+4\gamma$ by roughly the nanosecond scale. It is a long-lived intermediate state, not the final endpoint: the free neutron has ${\tau }_n=879.4\mathrm{s}$ in the original table and eventually undergoes beta-minus decay,Thus after several thousand seconds the dominant stable endpoint isBefore selecting the two ${\pi }^0\to 2\gamma$ decays, the $n{\pi }^0{\pi }^0$ history has probability $0.99524\times 0.358=0.35630$. The familiar rough description is therefore a $64\%$ charged-pion route and a $36\%$ neutron route; including the specified leading pion branches gives approximately $62.8\%$ and $34.8\%$ for the two displayed endpoints. The remainder includes Dalitz decays and other rare channels, not a failure of probability conservation.

These are particle decay cascades, treating the separated final proton, electron, photons and neutrinos as stable asymptotic particles and neglecting subsequent collisions and rare extra radiation. An infinitely large box prevents escaping particles from being removed from the inventory. Neutrino labels specify their production flavours. Each lifetime describes a stochastic waiting time, so there is no finite sharp instant at which every unstable particle has certainly decayed.

Use the parton model, neglect antiquarks, binding and target mass corrections, quark masses, muon mass, QCD scaling violation and the W boson propagator correction. The struck quark is effectively free during the short interaction and carries parton momentum fraction $x$, giving $\hat{s}\simeq xs$. The charged current reaction is ${\bar{\nu }}_{\mu }{u}_i\to {\mu }^{+}{d}_j$. Its helicity amplitude has the same $(1-y)$ factor as in a. Define the effective parton distribution function by $q(x)={\sum }_{i,j}|V_{ij}{|}^2{u}_i(x)$, summing only kinematically accessible final quarks; for valence-only protons and accessible $d,s$ this is approximately $u(x)$. It is not the unweighted sum $u+d$.Here $y=p\cdot (k-k^{\prime })/(p\cdot k)$ is the inclusive deep inelastic scattering variable, and $x=Q^2/(2p\cdot (k-k^{\prime }))$ in the scaling limit. Integrating $0<y<1$ gives $1/3$, henceThe moment measures the momentum fraction in the active quarks. Approximate linearity is evidence for scattering from pointlike spin-one-half constituents with approximately scale-independent parton distribution functions, rather than coherent scattering from an extended proton. It is an intermediate-energy prediction: thresholds, experimental deep inelastic scattering cuts, QCD scaling violation and eventually the finite W boson propagator modify it.