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The low-energy localized-well basis gives the tight-binding modelA Bloch state with coefficients $e^{ipna}$ is an eigenstate because the two neighbouring coefficients add to $2\cos (pa)e^{ipna}$. ThusHere $p$ is a Bloch wavenumber; physical momentum is $\hslash p$, and writing $p$ for physical momentum would require $pa/\hslash$. Spectral evolution of these eigenstates is exactly the Fourier integral in the preceding solution. The negative hopping lowers the symmetric $p=0$ state, and coherent quantum tunnelling spreads a formerly degenerate family of localized levels into a band of width $4\Delta ϵ$. Using the printed positive Euclidean prefactor instead would give on-site energy $-\hslash \omega /2$, showing its contradiction directly.

For a slowly evolving spherical star, use a Lagrangian enclosed mass coordinate $m$. Let $u$ be specific thermal internal energy and $D/Dt$ follow a mass element. The stellar energy balance equation isThe outward luminosity is energy transported through the mass shell, ${ϵ}_{\mathrm{nuc}}$ is nuclear rest-mass energy released per unit mass and time, and ${ϵ}_{\nu }$ is escaping neutrino power per unit mass. Define the gravothermal stellar energy generation by ${ϵ}_{\mathrm{grav}}=-Du/Dt-PD(1/\rho )/Dt$. At fixed composition it is $-TDS/Dt$ by the first law of thermodynamics. During composition changes the full internal-energy derivative must use the appropriate equation of state, with nuclear rest-mass changes counted only once. If a tabulated nuclear rate already subtracts reaction-neutrino energy, that loss must not be subtracted a second time in ${ϵ}_{\nu }$.

In a nearly steady main sequence star, stellar nuclear fusion supplies most of the luminosity. Before stable hydrogen ignition, Kelvin-Helmholtz contraction releases gravitational energy. In a gas-supported hydrostatic star the stellar virial theorem gives $U\simeq -\Omega /2$, so about half the change in gravitational binding energy heats the star and half is available for radiation. The associated timescale is $G{M}^2/(RL)$. After nuclear burning ends, a white dwarf can shine by losing stored thermal energy; crystallization releases latent heat, and composition separation can add gravitational energy. These are heat and gravitational reservoirs rather than sustained hydrogen fusion.

Stellar nuclear fusion is possible because lighter nuclei can combine into products with larger nuclear binding energy. The energy per reaction is $Q=(m_{\mathrm{initial}}-m_{\mathrm{final}})c^2$, including the relevant particles consistently. Although thermal energies are below the Coulomb barrier, quantum tunnelling permits fusion. The thermonuclear reaction rate averages a cross-section over the thermal relative-speed distribution:For nonresonant charged-particle reactions the energy weighting contains $\exp [-E/(kT)-\sqrt{E_G/E}]$; the compromise between the thermal tail and penetration produces the Gamow peak. A resonance can greatly enhance the rate. Density, abundances and temperature therefore all affect the stellar energy-generation rate, and power laws such as $T^4$ or $T^{16}$ are local approximations.

For the proton–proton chain, first produce deuterium and helium-3:The initial reaction is slow because it converts a proton to a neutron through the weak interaction. It is the bottleneck that allows long hydrogen-burning lifetimes. The alternative pep reaction $p+e^{-}+p\to {}^2\mathrm{H}+{\nu }_e$ also feeds the same chain. There are three principal proton-proton chain branches. In pp I, two helium-3 nuclei terminate the chain:In pp II, helium-3 first captures an existing helium-4 nucleus, then electron capture and proton capture finish the branch:In pp III, the beryllium-7 instead captures a proton:Relative branch weights change with temperature and composition. All branches convert four protons into a net helium-4 nucleus, with two weak conversions and two neutrinos, allowing for the electron consumed in pp II. After including positron annihilation, the atomic-mass energy budget is approximately $26.7\mathrm{MeV}$ per net helium-4 nucleus, but the deposited heat is smaller by the branch-dependent escaping-neutrino energy. Near ordinary low-mass main-sequence conditions, ${ϵ}_{pp}\propto \rho X_H^2{T}^4$ is a useful local approximation.

The CNO cycle provides an alternative catalyzed hydrogen burning route in hotter cores. The CNO-I cycle, also called the CN cycle, consists ofThe carbon seed is regenerated: the net reaction again consumes four protons and produces one helium-4 nucleus, two positrons and two neutrinos. The slow ${}^{14}\mathrm{N}(p,\gamma ){}^{15}\mathrm{O}$ reaction governs the ordinary cycle and causes nitrogen-14 to accumulate. The CNO-II cycle branches through ${}^{15}\mathrm{N}(p,\gamma ){}^{16}\mathrm{O}$, ${}^{16}\mathrm{O}(p,\gamma ){}^{17}\mathrm{F}$, ${}^{17}\mathrm{F}\to {}^{17}\mathrm{O}+e^{+}+{\nu }_e$, and ${}^{17}\mathrm{O}(p,\alpha ){}^{14}\mathrm{N}$, after which the main cycle continues. Thus CNO nuclei are catalytic in the closed cycles, although their relative abundances change. A local rate is ${ϵ}_{\mathrm{CNO}}\propto \rho X_H{X}_{\mathrm{CNO}}{T}^{\eta }$ with a much steeper exponent than the PP chain; the preceding homology problem specifies $\eta =16$. The concentration of this heating near the centre favors a convective core in hotter, more massive hydrogen-burning stars.

Once hydrogen is exhausted in a core, contraction can raise its temperature enough for core helium burning. The Triple-alpha process overcomes the absence of stable mass-5 and mass-8 nuclei by maintaining a tiny transient beryllium-8 population:The Hoyle state, an excited carbon-12 resonance near the three-alpha threshold, makes the second capture efficient enough despite beryllium-8's very short lifetime. The net conversion is $3{}^4\mathrm{He}\to {}^{12}\mathrm{C}$, releasing about $7.27\mathrm{MeV}$. Because three alpha particles are required, the specific rate scales as ${\rho }^2{Y}^3$; a schematic resonant rate isIts local temperature exponent is $-3+44/T_8$, of order forty near $T_8=1$. A degenerate core cannot expand promptly to regulate this increase, giving a helium flash; in a nondegenerate core the stellar thermostat permits stable helium burning. The competing ${}^{12}\mathrm{C}(\alpha ,\gamma ){}^{16}\mathrm{O}$ reaction determines much of the eventual carbon-oxygen mixture.

If the core becomes hot enough, carbon burning follows, for example through ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{20}\mathrm{Ne}+\alpha$ and ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{23}\mathrm{Na}+p$. Neon burning begins with photodisintegration, ${}^{20}\mathrm{Ne}+\gamma \to {}^{16}\mathrm{O}+\alpha$, followed by alpha capture such as ${}^{20}\mathrm{Ne}+\alpha \to {}^{24}\mathrm{Mg}+\gamma$. The first step consumes heat, but the combined rearrangement can release net energy. Oxygen burning includes ${}^{16}\mathrm{O}+{}^{16}\mathrm{O}\to {}^{28}\mathrm{Si}+\alpha$ and channels producing phosphorus and other nearby nuclei. At still higher temperatures, silicon burning proceeds through photodisintegration and particle captures in a reaction network approaching quasi-statistical equilibrium, producing iron-group nuclei. It is not simply direct silicon-plus-silicon fusion; the products depend on the electron fraction and weak-interaction timescale.

The increasing nuclear binding energy per nucleon supplies progressively less energy as products approach the iron group. Further fusion past that region cannot provide sustained net heat to support a core. Massive stars can therefore end with an unstable iron-group core; lower-mass stars do not attain all these burning stages and instead leave remnants such as carbon-oxygen white dwarfs. Heavy-element neutron captures can synthesize nuclei beyond the iron group without being the principal hydrostatic power source.

Finally, stellar neutrino energy loss competes with every heat source. Reaction neutrinos accompany the proton–proton and CNO chains. Hot or dense material also emits neutrino pairs through $e^{-}+e^{+}\to \nu +\bar{\nu }$, plasmon decay, $e^{-}+\gamma \to e^{-}+\nu +\bar{\nu }$, and electron-ion bremsstrahlung. These neutrinos usually escape, unlike the photons whose transport is diffusive; neutrino trapping requires much more extreme collapse conditions. Late burning has smaller fuel-energy reservoirs and strong neutrino cooling, so its duration is much shorter than core hydrogen burning. The surface luminosity is the escaping photon power, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$, not the sum of photons and unobserved neutrino power. Radiative diffusion in a star, convection and electron conduction redistribute energy, while the energy-balance equation distinguishes true sources, losses and storage. Accretion or mass loss adds boundary energy and mechanical work when present, rather than changing the nuclear reaction budget.

In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

For a plane wave $A{e}^{i(kx-Et/\hslash )}$, the probability current isThe incident amplitude is one and the transmitted amplitude is $T$, soThis is the transmission probability for quantum tunnelling. In the stationary state, $\partial \rho /\partial t=0$, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

For a stationary scattering state, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. The left side contains the incident and reflected waves, while the solid has no incident wave arriving from positive infinity:Multiplying by $e^{-iEt/\hslash }$ gives the time-dependent wavefunctions. The kinetic energy rises inside the lower-potential solid, so $k_2>k_1$ and the transmitted wavelength is shorter. A downward step permits a propagating transmitted wave for every $E>0$, but still generally reflects some incident flux. There is no forbidden region merely because $E<V_0$; the relevant transmitted kinetic energy is $E+V_0$.

Classically every particle crosses a downward step and accelerates, so this partial reflection has no classical counterpart. If the step is instead upward to $U>0$, $E>U$ gives oscillatory transmission with possible reflection, while $E<U$ gives a decaying evanescent wave. A semi-infinite forbidden region carries no transmitted stationary flux; a finite barrier can permit quantum tunnelling to a propagating region beyond it.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.

The common energetic principle behind nuclear fission and nuclear fusion is the nuclear binding energy. If the products are more tightly bound in total than the reactants, their smaller rest mass releasesThe two binding-energy expressions assume that the same numbers of protons and neutrons are counted on both sides; emitted free nucleons have zero nuclear binding. The energy appears mainly as product kinetic energy, with radiation and subsequent decays contributing according to the reaction.

The binding energy per nucleon rises rapidly through light nuclei, has a broad maximum in the iron/nickel region, and decreases more slowly for heavier nuclei. Combining suitable light nuclei can therefore increase their total binding, whereas splitting a sufficiently heavy nucleus into moderately heavy fragments can do the same. This is an energy comparison, not a rule that every fusion or fission channel is exothermic: the actual nuclear masses fix the sign of $Q$.

In nuclear fission, the liquid-drop model illustrates the competition between attractive short-range nuclear cohesion and long-range proton repulsive Coulomb interaction. Deformation increases surface area and costs surface energy, but separating the charge can reduce nuclear Coulomb energy. Their competition creates a fission barrier. An excited nucleus may pass over the barrier; spontaneous fission can occur through quantum tunnelling. The resulting fragments are often excited and neutron-rich, so they may emit prompt neutrons and photons and undergo later beta decay. The typical total energy released by a heavy-nucleus fission is of order a few hundred MeV, rather than an electronic-scale energy. Its size follows from an approximately MeV-per-nucleon binding gain summed over a heavy nucleus. A neutron from one fission can induce another; a nuclear chain reaction is the repeated physical process, distinct from the energy-release condition for one event.

In nuclear fusion, the initial charged nuclei repel one another. Exothermicity does not remove this entrance Coulomb barrier. A reaction requires sufficient overlap of their wavefunctions, achieved by energetic encounters and quantum tunnelling. For nonresonant low-energy charged-particle fusion, a useful separation isThe astrophysical S factor isolates the nuclear part from much of the rapidly varying Coulomb penetration probability. In a thermal population, the competition between the decreasing high-energy population and increasing penetration produces the Gamow peak. This explains how fusion can proceed below the classical barrier without requiring the mean thermal energy to equal its height.

Both processes conserve total energy, charge, baryon number and angular momentum. Their very different entrance barriers and mass ranges explain why a favorable $Q$ value by itself does not specify a reaction rate. Short-range attraction gives approximately saturated nuclear binding, while surface, Coulomb, symmetry and pairing terms account for the broad trend and its isotope-dependent departures.

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.

quantum tunnelling is transmission through a finite region in which the classical kinetic energy $E-V$ is negative. The wavefunction need not vanish there: the Schrödinger equation permits evanescent solutions. The transmitted particle retains its conserved energy; it does not borrow energy to surmount the barrier.

For a rectangular barrier of height $V_b$ on $0<x<d$, with $0<E<V_b$ and zero exterior potential energy, define $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_b-E)}/\hslash$. A left-incident scattering solution has the three piecesContinuity of $\psi ,{\psi }^{\prime }$ at both boundaries imposesEliminating $b,c,r$ givesEqual incident and exit velocities justify $T=|t{|}^2$; different exterior potential energies require the velocity ratio. Probability-current conservation gives $R+T=1$. Although either real exponential alone has zero current, their complex-coefficient combination can carry the current crossing the barrier.

For $\kappa d\gg 1$, $T\simeq 16E(V_b-E)V_b^{-2}{e}^{-2\kappa d}$. A slowly varying barrier generalizes this to the WKB approximation $T\sim \exp [-2{\int }_{x_1}^{x_2}\sqrt{2m(V-E)}dx/\hslash ]$, with matching prefactors and turning-point limitations. This exponential sensitivity explains why very small geometric or energy changes can produce large changes in a tunnelling rate.

In alpha decay, a quasibound alpha particle penetrates the nuclear Coulomb barrier; penetration probability, attempt frequency and nuclear preformation determine the rate. In a scanning tunnelling microscope, a bias between a conducting tip and sample gives current approximately proportional to $e^{-2\kappa d}$. Piezoelectric positioning and constant-current feedback probe surfaces, while vibration isolation and electronic-state dependence limit a purely geometric interpretation. Atomic-scale steps and surface reconstruction were resolved in the original surface-imaging study.^[ref] In a Josephson junction, coherent tunnelling of Cooper pairs permits $I=I_c\sin \phi$, a phase-dependent supercurrent through a weak link. Thus the same barrier physics explains natural decay and underpins sensitive surface and superconducting devices.

In a classically forbidden region, the Schrödinger equation has evanescent solutions rather than forcing the wavefunction to vanish. Matching at both sides of a finite barrier permits nonzero transmission. For a broad barrier the WKB approximation giveswhere $C$ is a matching prefactor. Transmission is exponentially sensitive to barrier width and height. A single decaying exponential function carries no probability current; the matched scattering solution includes the combination needed for transmitted current. energy is conserved throughout.

In alpha decay, a quasibound alpha particle penetrates the nuclear Coulomb barrier. Multiplying the penetration probability by an assault frequency estimates the decay rate; the strong dependence on alpha energy explains the wide range of lifetimes. In a scanning tunnelling microscope, a voltage drives electron quantum tunnelling between a conducting tip and sample. Approximately $I\propto V{e}^{-2\kappa d}$, so constant-current feedback resolves changes in gap width, with electronic density of states also affecting the image. Piezoelectric motion, vibration isolation and stable voltage are necessary; a bright feature is not automatically a taller atom. Atomic-scale imaging was demonstrated with a vacuum-tunnelling microscope in the original surface study.^[ref] In a Josephson junction, Cooper pair quantum tunnelling through a weak link gives $I=I_c\sin \phi$, allowing supercurrent without an applied voltage and phase-sensitive interference. These are distinct applications of barrier penetration and coupling with quantum coherence, not particles borrowing energy to climb over a barrier.

The mass dependence of fixed-barrier tunnelling follows by holding energy and the potential-energy barrier fixed: quantum tunnelling has the leading WKB approximationThe alpha-particle mass is about $7.3\times 10^3$ electron masses, so its decay exponent is about 85 times larger for the same $V-E$ profile. Exponential suppression therefore makes transmission much smaller. This comparison holds the potential energy and incident energy fixed: the same electrostatic potential is not the same potential-energy barrier for charges of different magnitude and sign.

Quantum tunnelling is supported by transport and decay through barriers that classical particles of the same energy cannot traverse. Three complementary examples are:

Together these test barrier penetration in nuclear decay, vacuum-gap transport and semiconductor transport. Their detailed currents depend on state supply and matching as well as the barrier transmission probability.

Quantum tunnelling is transmission through a finite classically forbidden region, where $E<V(x)$. The stationary Schrödinger equation permits an evanescent wavefunction inside such a barrier rather than forcing it to vanish at its entrance. Matching to propagating states beyond a barrier of finite width gives a nonzero transmitted probability current.

For a rectangular barrier of height $V_0$ and width $a$, with equal exterior potentials and $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The exact rectangular-barrier tunnelling probability isFor an opaque barrier its dominant dependence is $e^{-2\kappa a}$. For a slowly varying barrier, the WKB approximation generalizes this to $T\sim \exp [-(2/\hslash ){\int }_{x_1}^{x_2}\sqrt{2m(V(x)-E)}dx]$, apart from matching prefactors. Barrier width, particle mass and energy therefore have a strong influence. Energy is conserved in a static barrier: tunnelling does not require the particle to borrow energy or acquire a classical path with negative kinetic energy.

In alpha decay, a nuclear alpha-particle configuration is confined by the short-range nuclear interaction but separated from the exterior by a Coulomb barrier. Its energy can be positive relative to infinity while still lying below the barrier maximum. Escape by tunnelling produces alpha particles at that conserved energy, with an exponentially small decay probability per nuclear encounter. The pronounced dependence of half-life on alpha energy is a characteristic consequence of the barrier exponent, unlike a simple classical escape threshold.

In scanning tunneling microscopy, a conducting tip is brought within a nanometer-scale vacuum gap of a conducting surface and biased weakly. Electron tunnelling gives a current that depends approximately exponentially on separation, $I\propto V{e}^{-2\kappa s}$ in a simple low-bias model. Feedback adjusts the tip height to keep current fixed while scanning, producing a surface map sensitive to both geometry and electronic states. Finite current through the non-contact gap, and its steep dependence on separation, are the direct tunnelling signatures; current is not a direct measurement of topographic height alone.

A Josephson junction provides a collective superconducting example. Two superconductors separated by a thin barrier can carry a coherent pair current $I=I_c\sin \phi$ even at zero voltage. At constant voltage the AC Josephson effect gives $d\phi /dt=2eV/\hslash$ and oscillation frequency $2eV/h$. The coupled condensate quantum phases and pair transfer distinguish this from independent normal-electron transmission. These examples involve different microscopic barriers, but all require quantum coupling across a region that suppresses classical passage. Comparing weak-temperature tunnelling behavior with thermally activated escape helps distinguish the mechanisms experimentally.

Inside the rectangular potential barrier, the electron's wavefunction is evanescent, with decay constantFor width $a=1\mathrm{nm}$, the squared exponential suppression in the quantum tunnelling probability isThis is the usual leading-exponential estimate. The rectangular-barrier tunnelling probability, retaining the interface-matching prefactor for equal potentials on the two sides, isHere that prefactor is $16(8)(2)/10^2=2.56$, so the more accurate result is $\boxed{T\simeq 1.30\times 10^{-6}}$. Both estimates show that transmission is of order one part in a million. A classical particle with this kinetic energy cannot cross the barrier at all.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.