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Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

It is enough to choose an admissible parameter, without solving the minimizing equation. Take $a=V_0^2>0$ in the Gaussian variational bound for an attractive Gaussian well. The Rayleigh-Ritz variational principle giveswhere an explicit remainder isThe denominator tends to $2$ as $V_0\to 0$, proving $\epsilon =O(V_0^4)$ in Big O notation. More precisely the binomial series givesand thereforeThis is an upper bound on the exact ground state energy, rather than an assertion that the leading coefficient of its exact weak-potential expansion is $-1/2$.

The diagram refers to ground state, $L=0$, color singlet baryons. Their spatial wavefunction is symmetric and their color wavefunction is antisymmetric. The Pauli exclusion principle therefore requires a symmetric combined spin-and-flavor wavefunction. One cannot independently attach every three-spin coupling to each flavor assignment.

For three spin-$1/2$ particles, there are formally two ways to couple to total spin $1/2$. In the ground state quark model these are tied to the corresponding mixed-symmetry flavor functions. Their required symmetric combination yields one physical spin-$1/2$ flavor octet, not two independent species at each peripheral weight. In spin-flavor symmetry notation,under flavor $SU(3)$ and spin $SU(2)$. Thus there is one peripheral species in the baryon octet, each still possessing its two ordinary spin projections. The two abstract flavor-octet copies in $3\otimes 3\otimes 3$ do not double the physical ground state octet.

At the dotted positions, the flavor content would be $uuu$, $ddd$ or $sss$. These flavor wavefunctions are completely symmetric. The spin wavefunction must then also be completely symmetric, which means total spin $3/2$, not $1/2$. The corresponding baryon decuplet states are ${\Delta }^{++}$, ${\Delta }^{-}$ and ${\Omega }^{-}$. This exclusion is for the stated ground state spin-$1/2$ multiplet; it is not an exclusion of all excited baryons at those weights.

Both central baryons have valence quarks $uds$, baryon number $1$, strangeness $-1$ and isospin projection zero. Their total isospins differ:The $ud$ flavor part is symmetric for the Sigma baryon and antisymmetric for the Lambda baryon. In the usual ground state coupling, the associated $ud$ pair has spin one for ${\Sigma }^0$ and spin zero for ${\Lambda }^0$, with the strange quark giving total spin $1/2$ in both cases. Distinct total isospin makes these orthogonal states despite their coincident position in the weight diagram.

A ground state spin-zero meson is a color singlet $q\bar{q}$ with $L=0$ and constituent spin zero, hence a pseudoscalar meson. The nine flavor combinations form $3\otimes \bar{3}=8\oplus 1$. Here $B=0$, so the vertical coordinate $B+S$ equals strangeness. The six peripheral weights in the right panel of the figure in part (a) are

At the center, $u\bar{u},d\bar{d},s\bar{s}$ all have $I_z=S=0$. A useful orthonormal flavor basis isThe neutral pion has $I=1$; ${\eta }_8$ and ${\eta }_1$ have $I=0$. The Eta octet state belongs to the octet, while ${\eta }_1$ is the flavor singlet. The physical eta and eta prime mesons are predominantly mixtures of these two isoscalar basis states,Unequal quark masses and the chiral anomaly prevent exact flavor symmetry and a degenerate nonet. The three central physical species are consequently ${\pi }^0,\eta ,{\eta }^{\prime }$, rather than three separately observable pure diagonal flavor pairs. Small isospin-breaking admixtures are neglected here.

Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.

Set $u=r\psi$, choose $\alpha >0$ for a negative bound-state energy, and write $\kappa =2mK/{\hslash }^2$. Multiplying the radial equation by $r$ givesFor $u=e^{-\alpha r}f$, its second derivative is $e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. Substitution cancels the two ${\alpha }^2$ terms and givesFor the physical hydrogen problem, the angular eigenvalue has $\ell$ a nonnegative integer. Regularity at the origin selects the Frobenius exponent $u\sim r^{\ell +1}$, rather than the singular branch. PutThe equation for the series factor $v=f/r^{\ell +1}$ is $r{v}^{\prime \prime }+(2\ell +2-2\alpha r)v^{\prime }+(\kappa -2\alpha (\ell +1))v=0$, and coefficient comparison yieldsThe regular confluent-hypergeometric solution generically grows like $e^{2\alpha r}$ in this series factor; after the extracted $e^{-\alpha r}$ it is not normalizable. A bound state requires termination at some degree $q\ge 0$, namely $2\alpha (q+\ell +1)=\kappa$. Thus $n=q+\ell +1$ is an integer greater than $\ell$, andRegularity, normalizability and the integer angular index are the physical conditions selecting this spectrum; the differential equation without those conditions would admit many other solutions.

For $n=1$, necessarily $\ell =0$ and $q=0$. With $a_0={\hslash }^2/(mK)$,A normalized spatial ground state is ${\Psi }_1=(\pi a_0^3{)}^{-1/2}{e}^{-r/a_0}$. In spherical coordinates the radial probability has weight $r^{2}dr$. Constants and the angular integral cancel in the tail ratio. The supplied integral with $\lambda =2/a_0$ gives

The Gaussian integral gives ${\int }_{{\mathbb{R}}^3}{e}^{-\gamma |y{|}^2}{d}^{3}y=(\pi /\gamma {)}^{3/2}$. Thus the normalized ground state is, up to an overall phase,Its probability density is even in each shifted coordinate, so all three expectations $⟨y_i⟩$ vanish by oddness. ThereforeThe mean position is the minimum of the displaced potential, rather than the original coordinate origin.

The Gaussian integral gives ${\int }_{{\mathbb{R}}^3}{e}^{-\gamma |y{|}^2}{d}^{3}y=(\pi /\gamma {)}^{3/2}$. Thus the normalized ground state is, up to an overall phase,Its probability density is even in each shifted coordinate, so all three expectations $⟨y_i⟩$ vanish by oddness. ThereforeThe mean position is the minimum of the displaced potential, rather than the original coordinate origin.

A nuclear excited state has internal energy above its ground state, with levels labelled by spin and parity $J^{\pi }$ and, where useful, approximate isospin. In the shell model, nucleons occupy orbitals split by spin-orbit coupling; nuclear particle-hole excitation and broken pairs produce excited configurations. An orbital has parity $(-1{)}^l$, and an even-even paired ground state is commonly $0^{+}$. Residual interactions mix configurations, so shell spacings are not a complete spectrum.

Collective motion gives distinct patterns. A deformed nucleus can have a rotational band with $E_J-E_0={\hslash }^{2}J(J+1)/(2\mathcal{I})$, so a simple even-even $0^{+},2^{+},4^{+}$ band has $E_4/E_2=10/3$. Near spherical shape, a quadrupole vibration gives a one-phonon $2^{+}$ excitation; two such phonons can couple to $0^{+},2^{+},4^{+}$ near twice that energy in the harmonic limit. Anharmonicity and coupling to single-particle motion modify these ideal patterns.

De-excitation occurs by gamma emission, internal conversion, or particle emission if energetically open. photon multipolarity $\lambda \ge 1$ must satisfy the angular-momentum triangle rule. Electric multipoles change parity by $(-1{)}^{\lambda }$ and magnetic multipoles by $(-1{)}^{\lambda +1}$; a single photon cannot connect $0\to 0$. Large multipolarity or a small transition matrix element can make a long-lived nuclear isomer. Gamma energies, coincidence cascades, angular correlations and lifetimes constrain level schemes; inelastic scattering and transfer reactions supply complementary spin/configuration information. One observed line alone need not identify a unique level assignment.

At energies far below the $W$ mass, replace its propagator by a constant. The modern V-A interaction form of Fermi's contact theory has, in $\hslash =c=1$ units,Nucleons are treated nonrelativistically within nuclear states. To leading order in their velocity, the vector charge and axial spatial current give the two nuclear operatorswhere ${\tau }^{+}$ here means neutron-to-proton conversion. (i) The scalar Fermi operator permits $\Delta J=0$, including $0\to 0$, without parity change. The rank-one Gamow–Teller operator permits $\Delta J=0,1$ without parity change, excluding $0\to 0$. Both terms can contribute when the same nuclear states permit them; their strengths contain nuclear matrix elements, not just phase space.

(ii) Expand the lepton waves across the nucleus in powers of their momenta times nuclear radius. When $pR/\hslash \ll 1$, the constant spatial term gives allowed transitions. If its matrix element vanishes by spin/parity selection, higher terms supply lepton orbital angular momentum: “forbidden” means suppressed, not impossible. First-forbidden terms change parity and can have $\Delta J=0,1,2$; the rank-two $\Delta J=2$ case is unique first-forbidden. Higher orders carry additional small spatial factors and different energy-dependent shape factors. The contact approximation alone does not imply the allowed approximation.

For a single branch, neglect recoil and neutrino mass and first assume an energy-independent nuclear shape factor. Fermi's golden rule integrates the neutrino phase space with $E_{\nu }=Q-T$, while the electron momentum volume is proportional to $p^{2}dp$. Including Coulomb distortion,This momentum-bin Kurie ordinate is linear in kinetic energy and vanishes at its endpoint. If instead using counts per unit kinetic energy, $dN/dT\propto F(p)p{E}_e(Q-T{)}^2$ in $c=1$ units; converting with $dT/dp=p/E_e$ recovers the supplied momentum-bin definition. No extra factor of $E_e$ should be left inside that definition of $N^{\prime }$.

The graph's horizontal coordinate is $T/(m_e{c}^2)$, not $T$ in MeV. Its intercept is approximately $3.9$–$4.0$, soFor a ground-to-ground branch between bare nuclei, the emitted electron's rest energy must also be supplied:For neutral atomic masses the mass difference instead equals approximately $Q\simeq 2.0\mathrm{MeV}$, neglecting small electron-binding corrections. The graph supports this approximate precision, not a precise evaluated isotope mass. Interpreting a branch to an excited daughter would require adding its excitation energy; the ground-state comparison assumes the ground-state branch intended here.

The shell-model prediction for yttrium-91 beta decay follows from shell filling: the proton orbitals above 28 fill $2p_{3/2}$ with four protons and $1f_{5/2}$ with six, reaching 38. Yttrium has its 39th proton in $2p_{1/2}$, giving $J_i^{\pi }=1/2^{-}$. Its 52 neutrons include the closed 50-neutron core and a pair in $2d_{5/2}$, coupled to zero in the simple paired ground state. Zirconium has 40 paired protons and 51 neutrons, leaving an unpaired $2d_{5/2}$ neutron, so $J_f^{\pi }=5/2^{+}$. ThereforeThe single-particle change is neutron $d_{5/2}$ to proton $p_{1/2}$; the parity change and rank two prevent an allowed Fermi or Gamow–Teller contribution.

The straight plotted line is a constant-shape idealization. For the shell-model transition just inferred, a leading unique-forbidden shape factor contains $C(T)\propto p_e^2+p_{\nu }^2$ in $c=1$ units, with Coulomb/finite-size refinements. Consequently the uncorrected quantity specified in the paper is generally proportional to $(Q-T)\sqrt{C(T)}$, not exactly linear. A consistent shape-corrected Kurie plot also divides its ordinate by $\sqrt{C(T)}$. The source's exact straight-line construction and its forbidden-transition inference cannot both be exact with a Coulomb-only $F$; the approximate graphical endpoint can still be read as requested. Allowed and unique-forbidden spectral shapes are experimentally distinguishable.^[ref]

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]

Use the relative coordinate and reduced massTake $V=-V_0$ inside, $V=0$ outside, and $E=-B$. For $V_0>B>0$, define inverse lengthsThe radial equation becomes $u^{\prime \prime }+k^{2}u=0$ inside and $u^{\prime \prime }-{\kappa }^{2}u=0$ outside. Finiteness of the full wavefunction at the origin imposes $u(0)=0$; normalizability excludes the growing exterior exponential. HenceContinuity of $u$ was built in. Continuity of its derivative at the finite step givesFor the nodeless ground state, $\pi /2<kb<\pi$ selects the first branch. Higher roots would correspond to different wells with more bound states. Normalization is

For the requested weak-binding estimate, $B/V_0\to 0$ makes $\kappa /k\to 0$. On the first branch $kb\to \pi /2$, and $k^2\simeq 2\mu c^2{V}_0/(\hslash c{)}^2$. The leading threshold estimate isThis is the zero-binding threshold of a spherical square well, so nonzero positive binding needs a depth above it. Corrections start at order $\sqrt{B}$, not just at order $B$: setting $kb=\pi /2+ϵ$ gives $ϵ\simeq \kappa b/(\pi /2)$ and

The actual numbers permit an accuracy check: $\kappa =0.231608{\mathrm{fm}}^{-1}$ and $\kappa b=0.486377$ are not extremely small. Retaining the exterior decay, solveThis gives $x=kb=1.830503$, and the full square-well result isNeglecting $B$ only in the interior kinetic relation while retaining the measured exterior decay instead estimates $V_0\simeq 31.50\mathrm{MeV}$. These approximations should be identified: 23.2 MeV is the leading $B\to 0$ threshold, whereas matching the supplied nonzero binding requires a boundary phase shift and a deeper well.

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.

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 supplied schematic has four daughter energy levels, including the ground state, and all six downward pair transitions. Thus the observed gamma ray energies must be the six positive pairwise differences ofThe largest difference fixes $E_3=2.982\mathrm{MeV}$. Since $E_1,E_2$ are themselves transitions to the ground state, each must occur among the measured energies. Testing those pairs gives exactly two level schemes:They are reflections of one another about half the total span: replacing every level $E$ by $E_3-E$ preserves the set of differences. The assignment of every observed line is

These original diagrams plot the inferred excitation energies to scale. They do not reproduce the source schematic. The reconstruction treats the supplied three-decimal gamma ray energies as the transition gaps, neglecting their much smaller nuclear recoil corrections.

Using the observed counts,ThusTogether with the $67\%$ direct ground state branch, these sum to unity. No measurement uncertainties were supplied for these ratios, so the extra digits only document the arithmetic.

Let $K$ be the common normalization converting the quoted $N_i$ to photon numbers per parent decay. Apply nuclear level population balance to scheme A, starting with the top level:For example, the $3\to 2$ transition contributes $K{N}_2$ to the population of level 2; it must be subtracted from the total leaving level 2 to obtain its direct beta feeding. Similarly, a photon in the strong $1\to 0$ line need not imply a direct beta transition to level 1.