Nuclear Forces and Nuclear Structure

Derivation. Derives Phenomenon 107.2. Suppose only that each nucleus has, in any short interval \(\dd t\), the same probability \(\lambda\,\dd t\) of decaying, independently of its age and of what the other nuclei do. This is a statement about a collection of identical, memoryless, independent systems, and it is the whole physical input; everything else is arithmetic. The number of decays expected in \(\dd t\) is then \(\lambda N\,\dd t\), which is Equation (107.6), and the solution follows by separation of variables. The half-life is the time for \(N\) to halve,

and the mean life is \(\tau=1/\lambda=T_{1/2}/\ln 2\).

Two properties of this law are worth stating exactly, because both are routinely misread. First, memorylessness is not an extra assumption but a consequence: the probability that a nucleus surviving to time \(t\) survives a further \(s\) is \(\ee^{-\lambda s}\), independent of \(t\), so a nucleus does not age. The exponential is the unique distribution with that property. Second, \(\lambda\) is a statement about an ensemble and says nothing whatever about when a given nucleus will decay. The number \(\mathcal{N}\) counted in a fixed interval is a Poisson variable of mean \(\mu=\lambda N\,\Delta t\) and variance \(\mu\), so the fractional uncertainty on a measured activity is \(\mathcal{N}^{-1/2}\); this is the statistics of Probability and Statistics, and it is why a counting experiment quotes \(\sqrt{\mathcal{N}}\) as its error bar.

The activity \(\mathcal{A}=\lambda N\) is measured in becquerels, \(\mathrm{Bq}=/\mathrm{s}\), one disintegration per second. The older unit, the curie, was defined as the activity of one gram of \(^{226}\mathrm{Ra}\), and the definition can be checked: with \(T_{1/2}=1600\,\mathrm{yr}=5.05\times 10^{10}\,\mathrm{s}\) [Kondev:2021] and \(N=N_{A}/226\) atoms per gram,

against the defined \(3.7\times 10^{10}\,\mathrm{Bq}\); the small discrepancy is the improvement in the measured half-life since 1910.

Derives Proposition 107.3. The daughter is created at the rate at which the parent decays and destroyed at its own rate, \(\dd N_{2}/\dd t=\lambda_{1}N_{1}-\lambda_{2}N_{2}\), a linear first-order equation whose integrating factor is \(\ee^{\lambda_{2}t}\). With \(N_{1}=N_{1}(0)\ee^{-\lambda_{1}t}\),

which integrates to Equation (107.9). When \(\lambda_{1}\ll\lambda_{2}\) the second exponential dies first and \(N_{2}\to\lambda_{1}N_{1}/\lambda_{2}\), so the two activities are equal. This is why a sealed uranium ore assays as though every member of the chain were a separate element of fixed abundance, and why the Curies could isolate radium at all: its abundance in pitchblende is set by Equation (107.9) at \(\lambda_{1}/\lambda_{2}\approx3.6\times 10^{-7}\) of the uranium by number, which is about one gram of radium in three tonnes of uranium.

Derivation. Derives Phenomenon 107.5. The interaction is the repulsive Coulomb potential \(V(r)=Z_{1}Z_{2}e^{2}/4\pi\varepsilon_{0}r\). It is central, so the orbit is a hyperbola and the deflection of a particle incident with impact parameter \(b\) and kinetic energy \(E\) is fixed by

[Goldstein:2002] [Landau:1976], the hyperbolic-orbit relation of the classical repulsive Kepler problem. Write \(\kappa=Z_{1}Z_{2}e^{2}/8\pi\varepsilon_{0}E\), so that \(b=\kappa\cot(\theta/2)\). The deflection decreases monotonically with \(b\), so the particles arriving in the annulus between \(b\) and \(b+\dd b\) are exactly those leaving between \(\theta\) and \(\theta-\dd\theta\), and the cross-section is the ratio of the two areas,

From \(b=\kappa\cot(\theta/2)\) one has \(\abs{\dd b/\dd\theta}=\tfrac{1}{2}\kappa/\sin^{2}(\theta/2)\), and with \(\sin\theta=2\sin(\theta/2)\cos(\theta/2)\),

which is Equation (107.10). Two features carry the physics. The distribution diverges as \(\theta\to0\) because the Coulomb force has infinite range, so no total cross-section exists and only a differential rate is meaningful. And it falls only as \(\sin^{-4}(\theta/2)\): backscattering is rare but not exponentially rare, which is precisely what distinguishes a single point charge from a charge spread over the whole atom, for which small deflections compound at random and large ones are suppressed exponentially.

Derives Corollary 107.6. Energy conservation in the head-on collision gives \(E=Z_{1}Z_{2}e^{2}/4\pi\varepsilon_{0}d\), which is Equation (107.13); and Equation (107.10) was derived assuming a pure Coulomb field down to the turning point, so its persistence at \(\theta=\pi\) certifies that nothing else was met. For the \(7.7\,\mathrm{MeV}\) \(\alpha\) particles from \(^{214}\mathrm{Po}\) incident on gold, \(Z_{2}=79\), and Equation (107.2) gives

about \(2\times 10^{-4}\) of the atomic radius. The volume ratio is \(10^{-11}\): essentially all of the atom is empty, and that is the content of the experiment. Geiger and Marsden went on to verify every separate dependence of Equation (107.10) — on angle, on foil thickness, on the kinetic energy and on \(Z_{2}^{2}\) — which is what converted a suggestive anomaly into a measurement [Geiger:1913].

Derivation. Derives Phenomenon 107.8. Take a neutral particle of mass \(m\) and speed \(v\) striking a nucleus of mass \(M\) at rest. In a head-on elastic collision, non-relativistic kinematics gives the target the maximum speed

so the ratio of the maximum recoil speeds of hydrogen and nitrogen is

in which the unknown \(v\) has cancelled. Measured, the left-hand side is \(3.3\times 10^{7}\,\mathrm{m}/\mathrm{s}/4.7\times 10^{6}\,\mathrm{m}/\mathrm{s} =7.0\); with \(M_{\mathrm{H}}=m_{p}\) and \(M_{\mathrm{N}}\approx14m_{p}\),

a mass within the experimental error of the proton's. The modern value is \(m_{n}c^{2}=939.56542\,\mathrm{MeV}\) [Navas:2024], larger than the proton's by Equation (107.5).

The alternative had to be excluded quantitatively. A photon of energy \(E_{\gamma}\) scattering elastically from a mass \(M\) at rest imparts at most

the Compton kinematics of The Photon: Photoelectric and Compton Effects. The observed proton speed corresponds to \(T=5.7\,\mathrm{MeV}\), which by Equation (107.16) needs \(E_{\gamma}\approx55\,\mathrm{MeV}\); the observed nitrogen recoil corresponds to \(T=1.6\,\mathrm{MeV}\) and needs \(E_{\gamma}\approx10^{2}\,\mathrm{MeV}\). One radiation cannot have two energies, and neither energy is available: the reaction \(^{9}\mathrm{Be}+\alpha\) releases about \(10\,\mathrm{MeV}\) in total. The photon hypothesis fails by an order of magnitude and by an internal inconsistency, and the neutral-particle hypothesis fails by neither.

Derives Proposition 107.9. Three independent arguments close the question, and each is quantitative.

Localization. Confining an electron to a region of size \(\Delta x\approx5\,\mathrm{fm}\) requires, by the uncertainty relation, a momentum spread \(\Delta p\gtrsim\hbar/2\Delta x\), that is

so the electron would be ultrarelativistic with an energy of order \(20\,\mathrm{MeV}\). Nothing in the nucleus binds at that scale — the total binding energy per nucleon is \(8\,\mathrm{MeV}\) — and the electrons actually observed emerging in \(\beta\) decay carry about \(1\,\mathrm{MeV}\). The same estimate for a nucleon, heavier by \(m_{p}/m_{e}\approx1836\), gives a non-relativistic kinetic energy \((\Delta p)^{2}/2m_{p}\approx0.2\,\mathrm{MeV}\), comfortably below the binding: a nucleon fits in a nucleus and an electron does not.

Spin and statistics. The proton–electron model assigns \(^{14}\mathrm{N}\) fourteen protons and seven electrons, twenty-one spin-\(\tfrac12\) particles, hence half-integer total spin and Fermi–Dirac statistics for the molecule \(^{14}\mathrm{N}_{2}\). Band spectra show integer spin \(1\) and Bose–Einstein statistics. The proton–neutron model assigns seven protons and seven neutrons, fourteen fermions, integer spin: correct. The argument is decisive because the alternation of intensities in a band spectrum measures the statistics directly and admits of no adjustment; the general theory of identical particles belongs to Identical Particles.

Magnetic moment. A nuclear magnetic moment is measured in units of the nuclear magneton \(\mu_{N}=e\hbar/2m_{p}=5.0507837\times 10^{-27}\,\mathrm{J}/\mathrm{T}\) [Mohr:2025], some \(1836\) times smaller than the Bohr magneton. An electron inside the nucleus would contribute of order a Bohr magneton. No nuclear moment of that size exists [Fermi:1930].

Derivation. Derives Theorem 107.14. Let \(\phi\) obey the Klein–Gordon equation of The Klein–Gordon Equation with a source,

in which \(1/\lambda\) has dimension \(/\mathrm{m}\), as it must: it is the only combination of \(m\), \(c\) and \(\hbar\) with that dimension, and its appearance is the whole of Yukawa's idea. For a static source the time derivative drops and

with a strength \(g\) whose dimension is fixed below. Away from the origin the solution is spherically symmetric, and the substitution \(\phi=u(r)/r\) turns \(\nabla^{2}\phi\) into \(u''/r\), so that \(u''=u/\lambda^{2}\) and \(u\propto\ee^{\pm r/\lambda}\). Only the decaying sign is admissible. The coefficient is fixed by integrating Equation (107.22) over a small ball: the \(1/\lambda^{2}\) term contributes nothing as the radius shrinks, and the divergence theorem gives \(\oint\nabla\phi\cdot\dd\vect{S}=-g\), which for \(\phi=C\ee^{-r/\lambda}/r\) is \(-4\pi C\). Hence \(C=g/4\pi\) and

As \(\lambda\to\infty\) this is the Coulomb field, which is the check that the algebra is right.

A second source at distance \(r\) has energy \(-g'\phi(r)\) in this field, and for two identical nucleons \(g'=g\). Writing the product \(g^{2}\) so that the result carries the dimension of energy fixes the normalization: with \(g\) dimensionless the natural coefficient is \(g^{2}\hbar c/4\pi\), since \(\hbar c\) has dimension \(\mathrm{J}\,\mathrm{m}\) by Equation (107.1) and dividing by \(r\) gives an energy. That is Equation (107.19). The convention is deliberately the electromagnetic one: comparing with \(V_{\mathrm{Coulomb}}=e^{2}/4\pi\varepsilon_{0}r=\alpha\hbar c/r\) shows that \(g^{2}/4\pi\) is the exact analogue of the fine-structure constant, dimensionless and directly comparable with \(\alpha=7.3\times 10^{-3}\).

For Equation (107.20), do the angular integral,

the last step by elementary integration (or by closing a contour, as in Complex Analysis). Multiplying by \(-g^{2}\hbar c/4\pi\) and using \(\hbar/\lambda=mc\) gives Equation (107.20). The denominator \(\vect{q}^{2}+m^{2}c^{2}\) is a squared momentum, in \(\mathrm{kg}^{2}\,\mathrm{m}^{2}/\mathrm{s}^{2}\), exactly as the mass-shell condition of Notation 100.1 requires; it is the propagator denominator \(p^{2}-m^{2}c^{2}\) continued to a purely spacelike momentum transfer, which is what a static force is.

Derives Corollary 107.15. Immediate from \(\lambda=\hbar/mc\) and Equation (107.1). The inference runs both ways, and Yukawa ran it in this direction: the range was known from the size of light nuclei — the deuteron is bound and the diproton is not, which places the range within a factor of two of a femtometre — and the mass was the prediction. It is one of the few predictions in physics of a new particle's mass, made in 1935 and confirmed in 1947 [Lattes:1947] with the observed \(m_{\pi^{\pm}}c^{2}=139.57039\,\mathrm{MeV}\) [Navas:2024]. The equivalent range is

Derivation. Derives Phenomenon 107.18. The mesonic Bohr radius scales as \(a_{0}m_{e}/m\), so a particle of mass \(\approx200\,m_{e}\) orbits carbon at \(a\approx a_{0}m_{e}/(200\times6)=4.4\times 10^{-14}\,\mathrm{m}\), which is still large compared with the nucleus but small enough that the overlap of the orbital wave function with the nuclear volume is \(\abs{\psi(0)}^{2}V_{\mathrm{nuc}}\sim(R/a)^{3}\approx2\times 10^{-4}\). A strong interaction absorbs on contact, in a time \(R/c\approx10^{-23}\,\mathrm{s}\), so the expected lifetime against nuclear capture is \(\sim5\times 10^{-20}\,\mathrm{s}\) — against a measured decay time \(2.2\times 10^{-6}\,\mathrm{s}\) [Navas:2024] that is unchanged by the presence of the nucleus. Thirteen orders of magnitude is not a discrepancy to be argued about. The particle is the muon; its lifetime and its couplings belong to Weak Interactions.

The resolution came from photographic emulsions exposed at altitude, where Lattes, Muirhead, Occhialini and Powell found tracks in which one particle comes to rest and a second, of definite and shorter range, sets off from the end point [Lattes:1947]: the chain \(\pi\to\mu\to e\). Two particles of similar mass had been conflated; the heavier one is Yukawa's, is produced strongly and is absorbed strongly, and therefore never reaches sea level, while the lighter is the weakly interacting decay product that does.

Derives Proposition 107.19. A \(\pi^{-}\) stopped in deuterium cascades to the \(1s\) level of pionic deuterium and is captured from there, so the initial orbital angular momentum is \(\ell=0\). The deuteron has spin \(1\) and the pion spin \(0\) (Section 107.2.3 and the observed decay to two spin-zero particles), so the initial total angular momentum is \(J=1\) and the initial parity is \(P_{\pi}P_{d}(-1)^{\ell}=P_{\pi}\), the deuteron being even.

The final state is two identical fermions, whose total wave function must be antisymmetric. Enumerate the two-neutron states with \(J=1\): \(^{1}P_{1}\) has \(S=0\) (antisymmetric spin) and \(L=1\) (antisymmetric space), hence a symmetric total — forbidden; \(^{3}S_{1}\) and \(^{3}D_{1}\) have \(S=1\) (symmetric) and even \(L\) (symmetric) — also forbidden. Only \(^{3}P_{1}\), with \(S=1\) and \(L=1\), is antisymmetric, and its parity is \((-1)^{L}=-1\). Parity is conserved by the strong interaction (Discrete Symmetries and CPT), so \(P_{\pi}=-1\).

The pion is therefore a pseudoscalar: spin \(0\), parity \(-1\), written \(J^{P}=0^{-}\). Nothing in the argument is model dependent; it uses only the Pauli principle, the observed spins, and the fact that the capture is from an \(s\) state — which is itself measured, from the intensity of the X-rays emitted in the atomic cascade.

Derivation. Derives Phenomenon 107.21. A state of orbital angular momentum \(L=0\) has a spherically symmetric spatial distribution and therefore exactly zero quadrupole moment, so the observed moment proves that the deuteron ground state is not a pure \(S\) wave. But a central potential \(V(r)\) commutes with \(\vect{L}\), so \(L\) would then be a good quantum number and no admixture of \(L=2\) into an \(L=0\) ground state would be possible. The nucleon–nucleon interaction therefore contains a non-central piece. Which piece is fixed by what is available: the interaction must be a scalar built from the separation \(\hat{\vect{r}}\) and the two spins, and the only non-central such scalar that is even under parity, even under time reversal and independent of velocity is the tensor operator

This conserves the total angular momentum \(J\) but not \(L\), and since \(L=0\) and \(L=2\) have the same parity it mixes exactly the \(^{3}S_{1}\) and \(^{3}D_{1}\) states that share \(J=1\). The measured quadrupole moment is thus a direct measurement of the \(D\)-state admixture. The two null results complete the picture: that no diproton and no dineutron binds, while the deuteron does, says the force is spin dependent, since the \(^{1}S_{0}\) configuration is the only one open to two identical nucleons at low energy and it is very nearly, but not quite, binding.

Derives Proposition 107.22. Inside the well the regular \(s\)-wave radial function is \(u\propto\sin kr\); outside it is \(u\propto\ee^{-\kappa r}\), since a bound state must decay. Continuity of \(u'/u\) at \(r=R\) is Equation (107.30). Numerically,

so \(1/\kappa=4.32\,\mathrm{fm}\). Solving Equation (107.30) at \(R=2.0\,\mathrm{fm}\) gives \(k=0.910\,/\mathrm{fm}\) and \(V_{0}=B_{d}+(\hbar ck)^{2}/2\mu c^{2}=36.6\,\mathrm{MeV}\). The threshold for binding is \(kR=\pi/2\) with \(\kappa\to0\), that is \(V_{0}=\pi^{2}\hbar^{2}/8\mu R^{2}=25.6\,\mathrm{MeV}\); the second level needs \(kR=3\pi/2\), nine times as much.

Three conclusions follow, and each is a statement about the force rather than about the model. First, \(B_{d}/V_{0}=0.06\): the well is only about forty per cent deeper than the minimum that binds anything at all, so the deuteron sits just below threshold and its properties are correspondingly sensitive to the force — which is what makes it a good measuring instrument. Second, there is no excited state because there is no room for one, and none is observed. Third, and most striking, \(1/\kappa=4.3\,\mathrm{fm}\) exceeds the range of the force by a factor of two or three: most of the deuteron's probability lies outside the region where the two nucleons interact. A bound state that spends most of its time out of range of the force that binds it is the signature of a system barely bound, and it is the reason the low-energy properties of the two-nucleon system can be described without knowing the force in detail — the subject of the next proposition.

Derives Proposition 107.23. A bound state is a pole of the \(s\)-wave scattering amplitude at \(k=\ii\kappa\), where \(\cot\delta_{0}=\ii\); substituting \(k=\ii\kappa\) in Equation (107.32) gives \(-\kappa=-1/a_{t}-\tfrac12 r_{t}\kappa^{2}\), that is \(1/a_{t}=\kappa-\tfrac12 r_{t}\kappa^{2}\). With Equation (107.31),

so \(a_{t}=5.42\,\mathrm{fm}\), which is what neutron–proton scattering measures. Two quantities from unrelated experiments — a photodisintegration threshold and a scattering cross-section — are tied by two parameters, and the tie holds. Note that the deuteron enters nowhere in the derivation as a model: only the position of the pole. This is the content of Phenomenon 86.1, that low-energy scattering sees one length and does not care what produced it.

Derivation. Derives Phenomenon 107.24. Of the four spin states of a neutron and a proton, three are triplet and one singlet, and an unpolarized beam populates them in that ratio. Since \(\sigma=4\pi a^{2}\) at zero energy (Phenomenon 86.1),

against the measured \(20.5\times 10^{-28}\,\mathrm{m}^{2}\). Had the force been spin independent, \(a_{s}=a_{t}\) would give \(4\pi a_{t}^{2}=3.7\times 10^{-28}\,\mathrm{m}^{2}\), less by a factor of five and a half. The cross-section was measured long before it was understood, and it is the singlet channel that supplies almost all of it.

The negative sign is the further information. By Equation (107.32) with \(r_{s}=2.73\,\mathrm{fm}\), solving \(1/a_{s}=\kappa_{s}-\tfrac12 r_{s}\kappa_{s}^{2}\) gives \(\kappa_{s}=-0.0400\,/\mathrm{fm}\) — the root of the wrong sign, so the pole lies on the unphysical sheet and the state is virtual, not bound. Its energy is

just above threshold, where the triplet state lies \(2.2\,\mathrm{MeV}\) below it. Fitting the square well of Proposition 107.22 to \(a_{s}\) at the same radius \(R=2.0\,\mathrm{fm}\) gives a singlet depth of \(23.9\,\mathrm{MeV}\) against the \(25.6\,\mathrm{MeV}\) that would just bind: the singlet channel fails by six per cent in the depth of the well, and by \(66\,\mathrm{keV}\) in the energy. That near-miss is why there is no dineutron and no diproton: two identical nucleons at low energy are confined by the Pauli principle — whose general theory belongs to Identical Particles — to \(^{1}S_{0}\), which is precisely the channel that just fails to bind. Six per cent more attraction and the universe would contain bound dineutrons, and hydrogen burning in stars would proceed by a strong reaction rather than a weak one — in which case the Sun would have exhausted its fuel long before the Earth formed.

Derives Proposition 107.25. In a pure \(^{3}S_{1}\) state the orbital motion contributes nothing and the moment is the sum of the intrinsic moments, \(\mu_{p}+\mu_{n}=\left(2.79285-1.91304\right)\mu_{N} =0.87981\,\mu_{N}\).

In a \(^{3}D_{1}\) state the spins and the orbital motion are coupled to \(J=1\) with \(L=2\) and \(S=1\), and the moment must be projected. Within a manifold of fixed \(J\) only the component of a vector operator along \(\vect{J}\) survives — the projection theorem, whose proof belongs to Angular Momentum and Spin — so with \(\vect{\mu}=\mu_{N}\hbar^{-1} \left(g_{L}\vect{L}+g_{S}\vect{S}\right)\) and

the projections in the state of maximal \(J_{z}\) are \(\avg{L_{z}}=\tfrac32\hbar\) and \(\avg{S_{z}}=-\tfrac12\hbar\), since \(J(J+1)=2\). Two \(g\) factors are then needed. The orbital one is \(g_{L}=\tfrac12\), because only the proton carries charge and in the centre-of-mass frame it carries half of the relative orbital angular momentum. The spin one is fixed by the triplet coupling: with \(\vect{\mu}_{\mathrm{spin}} =\mu_{N}\hbar^{-1}(g_{p}\vect{s}_{p}+g_{n}\vect{s}_{n})\) and \(g_{p}=2\mu_{p}/\mu_{N}\), \(g_{n}=2\mu_{n}/\mu_{N}\), projecting on \(\vect{S}=\vect{s}_{p}+\vect{s}_{n}\) gives \(g_{S}=\tfrac12(g_{p}+g_{n})=(\mu_{p}+\mu_{n})/\mu_{N}\). Hence

markedly smaller than the \(S\)-state value, which is why a few per cent of \(D\) state suffices to explain the discrepancy. Weighting the two by \(1-P_{D}\) and \(P_{D}\) yields Equation (107.35). With \(\mu_{d}=0.85744\,\mu_{N}\),

some four per cent. The quadrupole moment, \(Q_{d}=+2.86\times 10^{-31}\,\mathrm{m}^{2}\) in units of the elementary charge, gives a compatible figure of four to six per cent depending on the assumed radial functions — and gives it with a sign, the positive value saying the deuteron is prolate, stretched along its spin axis, which is the shape the tensor force Equation (107.29) produces when the spins are aligned along the separation. Two independent static moments, measured by molecular-beam resonance [Kellogg:1939], therefore agree on the size of a component of the wave function that no experiment observes directly. The interpretation is not free of assumptions — \(P_{D}\) is not an observable, because it depends on the off-shell convention for the potential — but the tensor force it diagnoses is.

Derivation. Derives Phenomenon 107.26. Charge independence is precisely the statement that the interaction commutes with the isospin operators of Definition 107.10, so that the three \(T=1\) states \(pp\), \(nn\) and the symmetric combination of \(np\) share one interaction. Charge symmetry is the weaker statement that it commutes with the single rotation by \(\pi\) about the \(2\)-axis in isospin space, which exchanges \(p\) and \(n\).

The apparent size of the violation is deceptive, and this is where Proposition 107.23 earns its keep. A scattering length close to a pole is not a linear measure of the force: with the \(^{1}S_{0}\) state a mere \(66\,\mathrm{keV}\) from threshold, a change of one per cent in the well depth moves \(a\) by tens of per cent. Converting Equation (107.36) back into an equivalent well depth through Equation (107.30) shows the underlying interactions differing by about \(1\) per cent between \(np\) and \(pp\), and by a few tenths of a per cent between \(nn\) and \(pp\). Two sources of that residual are known and are not adjustable: the electromagnetic interaction beyond the Coulomb term already removed, and the up–down quark mass difference of Quantum Chromodynamics, which makes \(\pi^{\pm}\) and \(\pi^{0}\) differ in mass by \(4.59\,\mathrm{MeV}\) [Navas:2024] and therefore in exchange range by \(0.048\,\mathrm{fm}\).

Derivation. Derives Phenomenon 107.32. Suppose first that every nucleon attracted every other with the same strength. There are \(A(A-1)/2\) pairs, so \(B\) would grow as \(A^{2}\) and \(B/A\) as \(A\); the binding energy per nucleon of uranium would be sixty times that of carbon. It is not: the two agree within fifteen per cent. The interaction must therefore saturate, each nucleon binding to a fixed number of neighbours regardless of how many nucleons the nucleus contains. Two independent facts follow at once and both are observed. The energy is proportional to the number of nucleons, \(B\approx a_{V}A\), and the volume is proportional to the number of nucleons, so the density is constant and \(R=r_{0}A^{1/3}\).

Saturation is not automatic. A purely attractive short-range force does not saturate: nucleons would simply pile into the same region until the density diverged. Two mechanisms prevent that, and both are established independently. The Pauli principle keeps identical nucleons apart, since at fixed volume the kinetic energy rises as \(\rho^{2/3}\) (the Fermi gas of Proposition 107.36 below); and the short-range repulsive core of Remark 107.28 forbids close packing outright. The observed saturation density is

which in mass terms is \(2.8\times 10^{17}\,\mathrm{kg}/\mathrm{m}^{3}\), some fourteen orders of magnitude above the density of water and the same density that supports a neutron star (Compact Stars and Relativistic Astrophysics).

That \(B/A\) has an interior maximum then follows from two competing corrections to \(B=a_{V}A\), both of which reduce it. Nucleons near the surface have fewer neighbours, which costs an energy proportional to the surface area, \(A^{2/3}\); being a smaller fraction of the whole in a large nucleus, this correction favours large \(A\). The protons repel each other electrostatically, which costs \(Z^{2}/A^{1/3}\sim A^{5/3}\); per nucleon this grows as \(A^{2/3}\) and so disfavours large \(A\). One term decreasing and one increasing with \(A\) produce a maximum in between, and Proposition 107.41 locates it at \(A\approx61\).

That the maximum exists is the whole energy economy of the chart. Combining two nuclei below it, or dividing one above it, moves nucleons to a more tightly bound configuration and releases the difference. The measured curve is tabulated in Table 107.2; the most tightly bound nuclide is in fact \(^{62}\mathrm{Ni}\) at \(8.7945\,\mathrm{MeV}\) per nucleon, with \(^{58}\mathrm{Fe}\) at \(8.7923\,\mathrm{MeV}\) and \(^{56}\mathrm{Fe}\) at \(8.7903\,\mathrm{MeV}\) [Wang:2021] — a spread of two parts in a thousand, so the customary phrase “iron is the most tightly bound nucleus” is very nearly right and exactly wrong. It is \(^{56}\mathrm{Fe}\) that dominates the cosmic abundance peak, for reasons of nuclear statistical equilibrium and of the decay of \(^{56}\mathrm{Ni}\) that belong to Stellar Structure and Nucleosynthesis, not because it is the minimum of the mass curve.

Derives Proposition 107.34. If each nucleon binds to \(n\) neighbours with average energy \(\varepsilon\), the total is \(\tfrac12 nA\varepsilon\), linear in \(A\); that is the volume term, and \(a_{V}\) is not predicted by the argument, only its \(A\) dependence. A nucleon within a distance of order the force range \(\lambda\) of the surface has fewer neighbours, and the number of such nucleons is \(\rho_{0}\times4\pi R^{2}\lambda\propto A^{2/3}\). Writing the deficit as an energy per unit area gives Equation (107.44) with \(R=r_{0}A^{1/3}\); from \(a_{S}=17.8\,\mathrm{MeV}\) and \(r_{0}=1.2\,\mathrm{fm}\),

which converts to \(1.58\times 10^{17}\,\mathrm{J}/\mathrm{m}^{2}\) — eighteen orders of magnitude above the \(0.072\,\mathrm{J}/\mathrm{m}^{2}\) of water, and the reason the liquid-drop analogy is quantitative rather than merely pictorial: the same two competing energies, surface tension and Coulomb repulsion, govern a charged raindrop and a fissioning nucleus (Section 107.6.1).

Derives Proposition 107.35. Assemble the sphere by bringing in successive shells from infinity. When the charge assembled is \(q\) and the radius is \(s\), with uniform density \(q=Q(s/R)^{3}\), the work to add \(\dd q\) at radius \(s\) is \(q\,\dd q/4\pi\varepsilon_{0}s\). Writing \(\dd q=3Q s^{2}\dd s/R^{3}\),

Setting \(Q=Ze\) counts each proton as repelling itself, which is wrong; the correction replaces \(Z^{2}\) by \(Z(Z-1)\), the number of distinct pairs counted twice. Numerically, using Equation (107.2),

within one and a half per cent of the fitted value. This is the one coefficient of Equation (107.41) that is genuinely calculated: it involves no nuclear physics at all, only electrostatics and the measured radius, and its agreement with the fit is a check that the radius entering the mass formula is the radius that electron scattering measures.

Derives Proposition 107.36. Each species fills momentum states up to its own Fermi momentum. With two spin states per momentum state, the number of one species is \(n=V k_{F,n}^{3}/3\pi^{2}\), so for symmetric matter \(Z=N=A/2\) and \(\rho_{0}=A/V\) give \(k_{F}=(3\pi^{2}\rho_{0}/2)^{1/3}\), which is Equation (107.46) for \(\rho_{0}=0.17\,/\mathrm{fm}^{3}\); the Fermi energy is \((\hbar k_{F})^{2}/2m_{N}\), evaluated using Equation (107.1). The total kinetic energy of a filled Fermi sea is

since \(k_{F,i}\propto(n_{i}/V)^{1/3}\). Put \(N=\tfrac{A}{2}(1+x)\) and \(Z=\tfrac{A}{2}(1-x)\) with \(x=(N-Z)/A\) and expand: \((1+x)^{5/3}+(1-x)^{5/3}=2+\tfrac{10}{9}x^{2}+O(x^{4})\). The constant part is the symmetric kinetic energy; the \(x^{2}\) part is

using \(E_{F}=C(\rho_{0}/2)^{2/3}\), which is Equation (107.47).

Comparison with the fitted \(a_{A}=23.7\,\mathrm{MeV}\) is the point of the calculation. The Pauli principle alone accounts for \(12.8\,\mathrm{MeV}\), a little more than half, and the remainder must come from the interaction — specifically from the fact, already visible in Phenomenon 107.24, that the isospin-singlet \(np\) interaction is stronger than the isospin-triplet one, so that a nucleus does better with equal numbers of the two species than the counting of levels alone would suggest. Both halves of the asymmetry term are therefore consequences of the exclusion principle and of isospin, and neither has anything to do with the liquid drop, whose name the term is often filed under.

Derives Proposition 107.39. Write \(M(Z,A)c^{2}=Zm_{p}c^{2}+(A-Z)m_{n}c^{2}-B\) and differentiate at fixed \(A\), using \(\pp B/\pp Z=-a_{C}(2Z-1)A^{-1/3}+4a_{A}(A-2Z)/A\) from Equation (107.41) (the pairing term is not differentiable in \(Z\) and is treated by comparing the three parities separately). Setting \(\pp M/\pp Z=0\),

which rearranges to Equation (107.49). For small \(A\) the \(8a_{A}/A\) term dominates the denominator and \(Z^{*}\to A/2\): the asymmetry term wants equal numbers, and light stable nuclei have them (\(^{4}\mathrm{He}\), \(^{12}\mathrm{C}\), \(^{16}\mathrm{O}\), \(^{40}\mathrm{Ca}\)). As \(A\) grows, the Coulomb term in the denominator takes over and \(Z^{*}/A\) falls: at \(A=56\), \(Z^{*}=25.6\), so the stable isobar is \(^{56}\mathrm{Fe}\) with \(Z=26\); at \(A=120\), \(Z^{*}=51.5\) against the observed \(^{120}\mathrm{Sn}\) with \(Z=50\); at \(A=208\), \(Z^{*}=83.6\) against \(^{208}\mathrm{Pb}\) with \(Z=82\). The systematic overshoot of one to one and a half units at large \(A\) is the formula's known weakness, and it is not a rounding accident: it is the absence of shell structure, since both \(^{120}\mathrm{Sn}\) (\(Z=50\)) and \(^{208}\mathrm{Pb}\) (\(Z=82\)) are closed-proton-shell nuclei bound more tightly than any smooth formula predicts. That the deviation appears exactly at the magic numbers is the first evidence for Section 107.4.2.

Derives Corollary 107.40. The direction follows from the sign of \(\pp M/\pp Z\). For the parity structure, note that Equation (107.42) shifts the parabola by \(\pm a_{P}A^{-1/2}\) according to whether \(Z\) and \(N\) are both even or both odd, which for even \(A\) alternates with every unit step in \(Z\). The even–even isobars therefore lie on a parabola below the odd–odd one by \(2a_{P}A^{-1/2}\), which at \(A=100\) is \(2.2\,\mathrm{MeV}\). An odd–odd nuclide with even–even neighbours on both sides can then decay to either, which is exactly what \(^{40}\mathrm{K}\) does — \(\beta^{-}\) to \(^{40}\mathrm{Ca}\) in \(89\) per cent of decays and electron capture to \(^{40}\mathrm{Ar}\) in \(11\) per cent [Kondev:2021] — and it is why \(A=40\) has two stable isobars while \(A=39\) has one. The four odd–odd stable nuclides of Remark 107.38 are the exceptions, all at \(A\leq14\) where the parabolas are too shallow for the argument to bite.

Derives Proposition 107.41. Substitute \(Z=Z^{*}(A)\) from Equation (107.49) into Equation (107.41), divide by \(A\), and maximize numerically in \(A\); the function is smooth and has a single interior maximum. The competition is transparent in the two dominant \(A\)-dependent terms per nucleon, \(-a_{S}A^{-1/3}\), which rises towards zero as \(A\) grows, and \(-a_{C}Z^{*2}A^{-4/3}\), which falls; the asymmetry term per nucleon, \(-a_{A}(1-2Z^{*}/A)^{2}\), also falls with \(A\) because \(Z^{*}/A\) departs from \(\tfrac12\). The maximum is broad — \(B/A\) is within \(0.01\,\mathrm{MeV}\) of its peak over the whole range \(A=54\) to \(A=69\) — which is why the identity of the most bound nuclide is decided by shell effects of a few tenths of a \(\mathrm{MeV}\) that Equation (107.41) does not contain, and why it is \(^{62}\mathrm{Ni}\), with \(Z=28\) magic, that wins. The formula gets the position of the peak right to one unit in \(A\) and its height right to half a per cent, which is all a five-parameter fit can be asked for.

Derives Theorem 107.43. The Born amplitude for scattering off a potential \(V\) is proportional to its Fourier transform, and here \(V(\vect{r})=-e\phi(\vect{r})\) with \(\phi\) the electrostatic potential of \(\rho\). Since \(\nabla^{2}\phi=-\rho/\varepsilon_{0}\), the transforms obey \(\tilde{\phi}(\vect{q})=\hbar^{2}\tilde{\rho}(\vect{q}) /\varepsilon_{0}\vect{q}^{2}\), so the amplitude is the point-charge amplitude — which is what gives the Rutherford or Mott cross-section — times \(\tilde{\rho}(\vect{q})/Ze\), and the cross-section is its modulus squared. That is Equation (107.51). For a spherical \(\rho\), doing the angular integral,

and expanding \(\sin u=u-u^{3}/6+\cdots\) gives Equation (107.52) with \(\avg{r^{2}}=\int r^{2}\rho\,\dd^{3}r/Ze\). The Fourier inversion that recovers \(\rho\) from \(F\) is the transform pair of Fourier Analysis and Integral Transforms; in practice \(F\) is measured over a finite range of \(q\), so the inversion is not unique and the data are fitted to a parametrized shape instead.

Derives Corollary 107.44. Insert \(\rho=3Ze/4\pi R^{3}\) for \(r<R\) in the radial integral above and integrate by parts twice. The zeros of \(\sin x-x\cos x\), that is of \(\tan x=x\), are at \(x=4.493,7.725,\ldots\), and the diffraction pattern they produce is the same phenomenon as the Airy pattern of a circular aperture: a sharp-edged object has a form factor with zeros, a diffuse one has minima that do not reach zero. That the observed minima are filled in is the first evidence for a surface of finite thickness, quantified next.

Derivation. Derives Phenomenon 107.46. Normalizing Equation (107.54) to \(A\) nucleons gives, to the accuracy of an expansion in \(a/c\),

so for \(^{208}\mathrm{Pb}\), with \(c=6.52\,\mathrm{fm}\) and \(a=0.54\,\mathrm{fm}\), \(\rho_{0}=0.168\,/\mathrm{fm}^{3}\), which is Equation (107.40). The relation to the equivalent-uniform-sphere radius \(R=r_{0}A^{1/3}\) used in Equation (107.41) is the mean square radius: a uniform sphere has \(\avg{r^{2}}=\tfrac35 R^{2}\), so

and Table 107.3 evaluates it. That \(r_{0}\) is \(1.20\,\mathrm{fm}\) for lead but \(1.39\,\mathrm{fm}\) for carbon is not an inconsistency: with a surface of fixed thickness \(2.4\,\mathrm{fm}\), a carbon nucleus of radius \(2.5\,\mathrm{fm}\) is essentially all surface, and its equivalent uniform sphere is correspondingly larger than its half-density radius. The constant that is really constant is \(\rho_{0}\), not \(r_{0}\).

Derivation. Derives Phenomenon 107.48. The two observations are one statement. A width \(\Gamma\) corresponds to a mean life \(\tau=\hbar/\Gamma\); for \(\Gamma=25\,\mathrm{meV}\),

whereas a nucleon moving at the Fermi velocity \(v_{F}=\hbar k_{F}/m_{N}=0.29\,c\) from Equation (107.46) crosses a lead-sized nucleus in \(2R/v_{F}=1.7\times 10^{-22}\,\mathrm{s}\). The incident neutron therefore does not pass through: it is absorbed, its \(6\,\mathrm{MeV}\) or so of binding energy is shared among all \(A\) nucleons, and the system wanders through an enormous number of configurations before, by a statistical fluctuation, enough energy is again concentrated on one nucleon for it to escape. That is Bohr's compound nucleus. Its signature consequence is factorization,

the compound nucleus having forgotten how it was formed — which is exactly what the branching-ratio measurements show, and what the resonance formalism that belongs to Section 86.5.1 encodes as a pole whose residue splits into an entrance and an exit factor.

Derives Proposition 107.49. The Breit–Wigner cross-section for capture through a single resonance at \(E_{0}\) is

with \(p=\sqrt{2m_{n}E}\) the neutron momentum and \(g\) a statistical factor. Two energy dependences matter at small \(E\). The prefactor \(\pi\hbar^{2}/p^{2}\) goes as \(1/E\). The neutron width is a transmission probability through the centrifugal and surface barriers and for \(s\) waves is proportional to the momentum, \(\Gamma_{n}\propto p\propto\sqrt{E}\), which is the threshold law of Equation (86.2) in another guise. The photon width \(\Gamma_{\gamma}\) and the denominator are both essentially constant for \(E\ll E_{0}\). Hence

The law holds over many decades below the first resonance and is what makes a moderated reactor work: slowing a neutron from \(2\,\mathrm{MeV}\) to the thermal \(0.025\,\mathrm{eV}\) raises its capture and fission cross-sections by a factor of \(\sqrt{8\times 10^{7}}\approx 9\times 10^{3}\) (Section 107.6.1). It also means that a tabulated “thermal cross-section” is a convention, quoted at \(v=2200\,\mathrm{m}/\mathrm{s}\), and not a property of the nucleus alone.

Derivation. Derives Phenomenon 107.51. The derivation is in two steps: a mean field, which gets three of the seven numbers, and a spin–orbit term, which gets all seven.

Step one: the oscillator. Take the nucleons to move independently in a common potential. Its shape is bounded between two solvable cases — a harmonic oscillator, which is too soft at the edge, and an infinite square well, which is too hard — and the real Woods–Saxon well, of the shape Equation (107.54) with depth Equation (107.48), lies between them. For the isotropic oscillator the levels are

so a given \(\mathcal{N}\) collects \(\ell=\mathcal{N}, \mathcal{N}-2,\ldots\), and, counting \(2(2\ell+1)\) states per level for the two spin projections, has degeneracy \((\mathcal{N}+1)(\mathcal{N}+2)\). The cumulative occupancies are therefore

which reproduces \(2\), \(8\) and \(20\) and then fails: the observed numbers are \(28\), \(50\), \(82\), \(126\). A square well gives \(2,8,18,20,34,40,\ldots\) and fails in a different way. No choice of radial shape repairs this, which is why the problem stood for fifteen years.

The oscillator constant itself follows from the size. Each nucleon species fills its own well, so for a symmetric nucleus filled up to shell \(\mathcal{N}_{F}\) the cumulative degeneracy gives \(A/2\approx\mathcal{N}_{F}^{3}/3\), and the virial theorem gives \(\avg{r^{2}}=\tfrac34\mathcal{N}_{F}\hbar/m_{N}\omega\); equating this to the \(\tfrac35 r_{0}^{2}A^{2/3}\) of a uniform sphere,

the familiar empirical value, which for \(^{208}\mathrm{Pb}\) is \(6.9\,\mathrm{MeV}\) and for \(^{16}\mathrm{O}\) is \(16\,\mathrm{MeV}\).

Step two: the spin–orbit term. Mayer [Mayer:1949] and, independently, Haxel, Jensen and Suess [Haxel:1949] added

with \(C\) carrying whatever dimension makes \(C\hbar^{2}\) an energy, since \(\vect{L}\cdot\vect{S}\) has dimension \(\mathrm{J}^{2}\,\mathrm{s}^{2}\). Since \(\vect{L}\cdot\vect{S} =\tfrac12\left(\vect{J}^{2}-\vect{L}^{2}-\vect{S}^{2}\right)\),

so the two members of a spin–orbit doublet are separated by

with the level of higher \(j\) pushed down. Two features are essential and both are contrary to the atomic case: the sign, so that \(j=\ell+\tfrac12\) is the more bound, and the magnitude, which is comparable with \(\hbar\omega\) rather than a fine-structure correction — a factor of order \((v/c)^{-2}\sim10^{4}\) larger, because this spin–orbit force is a property of the strong interaction (Remark 107.28) and not a relativistic correction to an electromagnetic one.

Since Equation (107.64) grows with \(\ell\), the highest-\(\ell\) level of each oscillator shell is depressed the most. For \(\ell\geq3\) the depression exceeds the shell gap \(\hbar\omega\) and that level drops out of its shell into the gap below. Theorem 107.52 does the arithmetic.

Derives Theorem 107.52. The first term of Equation (107.65) is the cumulative oscillator degeneracy through shell \(\mathcal{N}-1\),

verified by induction. The intruder from shell \(\mathcal{N}\) carries \(2j+1=2\mathcal{N}+2\) states, which is the second term. A gap opens above the intruder because the rest of shell \(\mathcal{N}\) remains \(\hbar\omega\) higher. Evaluating: \(\mathcal{N}=3\) gives \(20+8=28\) (the intruder is \(1f_{7/2}\)); \(\mathcal{N}=4\) gives \(40+10=50\) (\(1g_{9/2}\)); \(\mathcal{N}=5\) gives \(70+12=82\) (\(1h_{11/2}\)); \(\mathcal{N}=6\) gives \(112+14=126\) (\(1i_{13/2}\)); \(\mathcal{N}=7\) gives \(168+16=184\) (\(1j_{15/2}\)). For \(\mathcal{N}\leq2\) the formula would give \(6\) and \(14\); those are not observed, and correctly so, because Equation (107.64) for \(\ell\leq2\) is smaller than \(\hbar\omega\) and the intruder does not leave its shell. The threshold at \(\ell=3\) is thus not fitted but forced: it is the condition \(\tfrac12 C(2\ell+1)\hbar^{2}>\hbar\omega\) with the single constant \(C\) chosen once. Every observed magic number is then reproduced, and no unobserved one is produced. That is the whole argument, and it is why the two 1949 papers ended a fifteen-year impasse in three pages each.

Derives Proposition 107.53. The free cross-section at a laboratory energy of a few tens of \(\mathrm{MeV}\) is about \(2\times 10^{-29}\,\mathrm{m}^{2}\), so at the density Equation (107.40) the naive mean free path is

a quarter of the internucleon spacing. On that estimate no orbital could survive for even one revolution and the shell model would be absurd — which is exactly the objection raised against it in 1949.

The estimate omits the exclusion principle. A collision must take both partners to unoccupied states, and below the Fermi surface there are none. Let a nucleon of energy \(E\) above \(E_{F}\) collide with one of the filled sea; energy conservation confines the two final energies to a shell of thickness \(E\) above \(E_{F}\), and momentum conservation confines the initial partner similarly, so the fraction of collisions that is kinematically allowed is of order

the same phase-space counting that makes a Fermi liquid's quasiparticles long-lived, a subject that belongs to Electrons in Solids: Band Theory. For a nucleon near the top of the filled sea, \(E-E_{F}\) is a few \(\mathrm{MeV}\) against \(E_{F}=38\,\mathrm{MeV}\), so the factor is of order \(10^{-2}\) and the mean free path becomes tens of femtometres — larger than the nucleus. The paradox dissolves completely: the nucleons that matter for the ground state are exactly those for which collisions are forbidden, and a strongly interacting system can nevertheless be an almost ideal gas of independent particles. Note that the argument fails for a highly excited nucleus, where \(E-E_{F}\) is large; and there the compound-nucleus picture of Phenomenon 107.48, which assumes the opposite extreme, takes over. The two models describe the same nucleus at different excitation energies.

Derives Corollary 107.54. Pairing (Remark 107.38) couples identical nucleons in the same level to \(J=0\) with even parity, so a filled or paired configuration contributes nothing, and every one of the several hundred known even–even ground states is \(0^{+}\) — a prediction with no exceptions. For an odd-\(A\) nucleus the unpaired nucleon carries the spin. Reading Table 107.4: \(^{17}\mathrm{O}\) has one neutron in \(1d_{5/2}\), predicting \(\tfrac52^{+}\), observed \(\tfrac52^{+}\); \(^{15}\mathrm{N}\) has a proton hole in \(1p_{1/2}\), predicting \(\tfrac12^{-}\), observed \(\tfrac12^{-}\); \(^{41}\mathrm{Ca}\) has a neutron in \(1f_{7/2}\), predicting \(\tfrac72^{-}\), observed \(\tfrac72^{-}\); \(^{207}\mathrm{Pb}\) has a neutron hole in \(3p_{1/2}\), predicting \(\tfrac12^{-}\), observed \(\tfrac12^{-}\); \(^{209}\mathrm{Bi}\) has a proton in \(1h_{9/2}\), predicting \(\tfrac92^{-}\), observed \(\tfrac92^{-}\). The rule works best next to closed shells and degrades in mid-shell, where several nucleons are unpaired and the coupling is a genuine many-body problem.

Derives Proposition 107.55. Write \(\vect{\mu}=\mu_{N}\hbar^{-1} \left(g_{\ell}\vect{L}+g_{s}\vect{S}\right)\) and take the expectation in the state of maximal \(j_{z}\). Since \(\vect{L}\) and \(\vect{S}\) are not separately conserved, project each on \(\vect{J}\) by the same projection theorem used in Proposition 107.25, that within a manifold of fixed \(j\) only the component of a vector operator along \(\vect{J}\) contributes: \(\avg{\vect{L}}=\avg{\vect{L}\cdot\vect{J}}\vect{J}/j(j+1)\hbar^{2}\) and likewise for \(\vect{S}\), with \(\avg{\vect{L}\cdot\vect{J}} =\tfrac12\left[j(j+1)+\ell(\ell+1)-s(s+1)\right]\hbar^{2}\). Substituting \(\ell=j\mp\tfrac12\) and simplifying gives Equation (107.68).

The two branches bracket the data, and the bracketing is itself the result. For a neutron with \(j=\ell+\tfrac12\) the prediction is \(\tfrac12 g_{s}=-1.913\) regardless of \(\ell\); \(^{17}\mathrm{O}\) measures \(-1.894\), within one per cent. But \(^{209}\mathrm{Bi}\), a \(1h_{9/2}\) proton with \(j=\ell-\tfrac12\), is predicted \(+2.62\) and measures \(+4.11\), which lies between that value and the \(+6.79\) of the other branch. Schmidt's compilation [Schmidt:1937] showed this pattern for every measured moment before the shell model existed. The interpretation is that the odd nucleon is not alone: its orbital is mixed with configurations in which the core is excited, and in addition the nucleons' intrinsic moments are modified inside nuclear matter by the exchange currents that the pion of Section 107.2.2 carries. The single-particle model is therefore right about the spins (Corollary 107.54) and only qualitatively right about the moments, and the difference between the two is a quantitative measure of how far the independent-particle approximation is from exact.

Derivation. Derives Phenomenon 107.56. A rigid axially symmetric body has Hamiltonian \(\Ham=\vect{J}^{2}/2\mathcal{I}\) about the axes perpendicular to its symmetry axis, so

in \(\mathrm{J}\), with \(\mathcal{I}\) in \(\mathrm{kg}\,\mathrm{m}^{2}\). Only even \(J\) occurs, because a reflection through the plane perpendicular to the symmetry axis is a symmetry of the shape and the rotational wave function must be even under it, which kills the odd members. The ratio \(E_{4}/E_{2}=20/6=3.33\) is then a parameter-free prediction, against the \(2.0\) that a harmonic surface vibration would give — so the single measured ratio distinguishes the two collective modes.

The moment of inertia is a measurement. From \(E_{2}=44.9\,\mathrm{keV}\),

whereas a rigid sphere of mass \(238\,u\) and radius \(r_{0}A^{1/3}=7.4\,\mathrm{fm}\) has \(\tfrac25 MR^{2}=8.7\times 10^{-54}\,\mathrm{kg}\,\mathrm{m}^{2}\). The observed moment is about half the rigid-body value — and, at the other extreme, several times the value for irrotational flow of a deformed drop, in which only the surface deformation and not the matter travels round.

Derivation. Derives Phenomenon 107.59. The mechanism is the one already used for the deuteron in Proposition 107.22. A weakly bound \(s\) or \(p\) neutron has an exterior wave function \(\ee^{-\kappa r}\) with \(\kappa=\sqrt{2\mu S_{n}}/\hbar\), so its spatial extent is \(1/\kappa\); for \(^{11}\mathrm{Li}\) the two-neutron separation energy is only \(S_{2n}=0.37\,\mathrm{MeV}\), and with the reduced mass of a neutron pair against a \(^{9}\mathrm{Li}\) core,

several times the radius of the core. The last neutrons therefore spend most of their time outside the nuclear surface, in a region where there is no nuclear matter at all, and the concept of a saturated drop with a well-defined surface no longer applies. That the phenomenon is confined to the drip line follows from the same formula: at a normal separation energy of \(8\,\mathrm{MeV}\), \(1/\kappa=1.6\,\mathrm{fm}\) and nothing unusual happens. The halo is not a new force; it is the same threshold physics that makes the deuteron large, applied to a system that is barely bound because it is at the edge of stability.

Derivation. Derives Theorem 107.63. The mathematics is done, in SI, in Section 9.6.6: Theorem 9.76 gives the transmission through a barrier between turning points as \(T\approx\ee^{-2\gamma}\) with \(\gamma=\hbar^{-1}\int\sqrt{2m(V-E)}\,\dd x\), and Example 9.81 evaluates that integral for exactly this potential, obtaining Equation (9.93), which is Equation (107.72). Nothing needs to be rebuilt here; what needs to be supplied is the nuclear input, and there are three pieces of it.

The two-body reduction. The problem is not a particle in a fixed potential but two bodies, so the mass in the exponent is the reduced mass \(\mu=m_{\alpha}M_{d}/(m_{\alpha}+M_{d})\) and the energy is the total release \(Q\), not the \(\alpha\) particle's laboratory kinetic energy \(E_{\alpha}=QM_{d}/(M_{d}+m_{\alpha})\). These are consistent: \(\sqrt{2\mu Q}=\sqrt{2m_{\alpha}E_{\alpha}}\), so the momentum in the exponent is the measured one either way, while the outer turning point \(b=2Z_{d}e^{2}/4\pi\varepsilon_{0}Q\) must be computed from \(Q\). For \(^{238}\mathrm{U}\) the difference between using \(Q\) and using \(E_{\alpha}\) is two per cent in \(b\) and a factor of three in the half-life — which is the measure of how sensitive this calculation is.

The inner radius. Unlike the outer turning point, \(R\) is not determined by the energy: it is where the Coulomb potential is cut off by the nuclear surface, and the natural estimate is the sum of the two radii, Equation (107.71). At \(r=R\) the potential is Equation (107.70), far above \(Q\), so the momentum does not vanish there and no connection formula is applied at the inner edge; the single delicate matching is the one at \(b\), which is what Theorem 9.76 performs.

The prefactor. The \(\alpha\) particle must exist inside the nucleus and must present itself at the barrier. Writing \(\lambda=fT\) with \(f=v/2R\), \(v=\sqrt{2Q/\mu}\) the relative speed, treats the \(\alpha\) as rattling to and fro across the nuclear diameter. This is crude in two ways that pull in opposite directions: the speed inside the well is larger than \(v\), because the well is \(40\,\mathrm{MeV}\) deep, which raises \(f\); and there is a preformation probability, the chance that four nucleons are found grouped as an \(\alpha\) particle at the surface at all, which is of order \(10^{-2}\) and lowers \(\lambda\). Neither is supplied by the method — Remark 9.77 says as much for the mathematics, and it is doubly true here — and the calculation below simply omits both. What can be defended is the exponent.

Derives Corollary 107.64. Setting \(\xi\to0\) in Equation (107.72) replaces the bracket by \(\arccos 0=\pi/2\), and \(\sqrt{2\mu Q}\,b=2Z_{d}e^{2}\sqrt{2\mu/Q}/4\pi\varepsilon_{0}\) gives Equation (107.73). Taking the logarithm of \(\lambda=f\ee^{-2\gamma}\) and absorbing the slowly varying \(f\) into \(a\) gives Equation (107.74) with

using Equations (107.1) and (107.2) and \(\mu c^{2}=3657\,\mathrm{MeV}\) for the heavy emitters.

The finite radius \(R\) cuts the barrier off at \(\xi\approx0.15\) to \(0.35\) and roughly halves the exponent — for \(^{238}\mathrm{U}\), \(2\pi\eta=171\) against the exact \(2\gamma=88\) — but it barely touches the slope, and the reason is worth having. Expanding the bracket of Equation (107.72) for small \(\xi\) gives \(2\gamma\approx2\pi\eta\left[1-(4/\pi)\sqrt{\xi}+\cdots\right]\), and since \(2\pi\eta\propto Q^{-1/2}\) while \(\sqrt{\xi}\propto Q^{1/2}\), the correction term is nearly independent of \(Q\): it shifts the intercept \(a(Z_{d})\), not the coefficient of \(Q^{-1/2}\). The exact exponents of Table 107.5 confirm both statements. Take the three polonium isotopes, for which \(Z_{d}=82\) is fixed: plotting \(2\gamma\) against \(Q^{-1/2}\) gives successive slopes \(298\) and \(290\) — linear to three per cent, and within ten per cent of the point-nucleus \(b\ln10=322\) — while the differences \(2\pi\eta-2\gamma\) are \(77\), \(76\) and \(75\): the predicted constant. In terms of the observable, \(\log_{10}(\lambda\,\mathrm{s})\) against \(Q^{-1/2}\) has predicted slope \(-298/\ln10=-129\), and the measured half-lives give \(-129\) and \(-119\). That is the Geiger–Nuttall relation derived, with the right slope, from a barrier and a wave function.

Derives Proposition 107.66. The energy release for emitting a fragment \(X\) is \(Q=B(X)+B(\mathrm{daughter})-B(\mathrm{parent})\), and the term that decides it is \(B(X)\). From Table 107.2, \(B(^{4}\mathrm{He})=28.30\,\mathrm{MeV}\), against \(B(^{2}\mathrm{H})=2.22\,\mathrm{MeV}\), \(B(^{3}\mathrm{H})=8.48\,\mathrm{MeV}\) and, for a single nucleon, zero. The rest of the expression — the loss of binding by the parent — is of order \(A\) times the local slope of the binding curve and is much the same in each case, roughly \(25\,\mathrm{MeV}\) for the removal of four nucleons from a heavy nucleus. Only the \(\alpha\) particle brings enough binding of its own to pay for it, because it is doubly magic and exceptionally tightly bound for its size. That proton emission and \(^{12}\mathrm{C}\) emission do occur, the first beyond the proton drip line and the second as a rare branch called cluster decay with branching ratios near \(10^{-11}\), is the exception that measures the rule.

Derives Proposition 107.67. Work with nuclear masses first and convert. For \(\beta^{-}\), \(M_{\mathrm{nuc}}(Z)\to M_{\mathrm{nuc}}(Z+1)+m_{e}\), so \(Q=\left[M_{\mathrm{nuc}}(Z)-M_{\mathrm{nuc}}(Z+1) -m_{e}\right]c^{2}\); substituting \(M_{\mathrm{nuc}}=M_{\mathrm{at}}-Zm_{e}\) (neglecting electron binding) the \(Z\) electrons on the left and the \(Z+1\) on the right differ by exactly the emitted electron, and it cancels, giving Equation (107.75). For \(\beta^{+}\) the same substitution leaves two electron masses uncancelled — the positron created, and the atomic electron the daughter no longer needs — and gives Equation (107.76). For electron capture the electron is taken from the atom rather than created, so no such term appears and only its binding energy is lost, giving Equation (107.77).

The window follows. Between \(Q_{\mathrm{EC}}=0\) and \(Q_{\mathrm{EC}}=1.022\,\mathrm{MeV}\) a nuclide can capture but not emit a positron, which is why \(^{7}\mathrm{Be}\) (\(Q_{\mathrm{EC}}=0.862\,\mathrm{MeV}\)) decays only by capture. The free neutron sits at the other extreme with

which is the endpoint of the neutron \(\beta\) spectrum and the number that Equation (107.5) made available.

Derives Proposition 107.70. The lepton wave functions are plane waves over the nuclear volume, and the Rayleigh plane-wave expansion \(\ee^{\ii\vect{p}\cdot\vect{r}/\hbar} =\sum_{L}\ii^{L}(2L+1)j_{L}(pr/\hbar)P_{L}(\cos\theta)\) has \(j_{L}(x)\sim x^{L}/(2L+1)!!\) for small \(x\). With \(x=p_{e}R/\hbar\) evaluated at \(p_{e}c=1\,\mathrm{MeV}\) and \(R=5\,\mathrm{fm}\),

so the amplitude for \(L=1\) is smaller than for \(L=0\) by \(x\) and the rate by \(x^{2}=6.4\times 10^{-4}\), which is Equation (107.79): three and a fifth in the logarithm. The parity of the nuclear state must also change with each unit of \(L\), since the lepton pair carries parity \((-1)^{L}\), which is what makes the classification observable rather than merely formal — a transition between states of the same parity and \(\Delta J=2\) must be second forbidden and cannot be anything else.

Measured \(\log_{10}(ft/\mathrm{s})\) values confirm the ladder: about \(3.5\) for superallowed, \(5\) to \(6\) for allowed, \(6\) to \(9\) for first forbidden, and up to \(22\) for the highest observed. The extremes are instructive. \(^{6}\mathrm{He}\) decays in \(0.81\,\mathrm{s}\) with \(\log_{10}ft=2.9\); \(^{40}\mathrm{K}\), a third-forbidden unique transition from \(4^{-}\) to \(0^{+}\), lives \(1.25\times 10^{9}\,\mathrm{yr}\) [Kondev:2021] with \(\log_{10}ft\approx18\). That is fifteen orders of magnitude in \(ft\) from three units of forbiddenness, against the ten that Equation (107.79) predicts; the shortfall is the extra cost of a unique transition, in which the lepton pair must carry not only \(L=3\) but the maximum spin coupling as well, and it is why the classification distinguishes unique from non-unique cases at all.

Derivation. Derives Phenomenon 107.72. That the mode exists at all is a consequence of Corollary 107.40. For even \(A\) the pairing term splits the mass parabola in two, and an even–even nuclide can find itself lower than both of its odd–odd neighbours while being higher than the even–even nuclide two steps away. Single \(\beta\) decay is then closed and double \(\beta\) decay is open. The classic case is \(^{76}\mathrm{Ge}\), which cannot reach \(^{76}\mathrm{As}\) but can reach \(^{76}\mathrm{Se}\); the same arithmetic selects \(^{82}\mathrm{Se}\), \(^{100}\mathrm{Mo}\), \(^{130}\mathrm{Te}\) and \(^{136}\mathrm{Xe}\).

The rate is second order in the weak interaction, so it carries \(G_{F}^{4}\) instead of \(G_{F}^{2}\), and the four-body final state gives a phase-space factor scaling as \(Q^{11}\) rather than \(Q^{5}\). Those two together are the twenty orders of magnitude between an allowed \(\beta\) decay and a double one. The neutrinoless mode is a different matter: it requires the neutrino emitted at one vertex to be absorbed at the other, which is possible only if the neutrino is its own antiparticle and has mass, so its observation would establish lepton-number violation and the Majorana nature of the neutrino. It has not been observed, and this book records searches by what they have excluded. The status belongs to Flavour Physics and Neutrinos; what belongs here is that the experimental signature is entirely nuclear — a peak at the \(Q\) value of the two-electron spectrum, sitting on the continuum of the two-neutrino mode — and that the conversion of a half-life limit into a mass limit requires a nuclear matrix element which different many-body methods disagree about by a factor of two or three.

Derives Proposition 107.73. The photon carries away angular momentum \(L\hbar\) and the total is conserved, which with the triangle rule for adding \(\vect{J}_{i}=\vect{J}_{f}+\vect{L}\) (Angular Momentum and Spin) gives the range in Equation (107.80). The lower bound \(L\geq1\) is the statement that a free photon has no monopole mode: its field is transverse, so the \(L=0\) component of the multipole expansion of a radiation field vanishes identically. The parity of an \(EL\) photon is \((-1)^{L}\) and of an \(ML\) photon \((-1)^{L+1}\), because the electric and magnetic multipole operators differ by one factor of the (axial) magnetic field; conservation of parity in the electromagnetic interaction then fixes which type may occur. For \(J_{i}=J_{f}=0\) the range in Equation (107.80) is empty and single-photon emission is strictly forbidden. Such transitions nevertheless occur, by internal conversion or, if the energy exceeds \(2m_{e}c^{2}\), by internal pair creation — and the fact that these proceed while photon emission does not is direct evidence that they are not merely a photon reabsorbed.

Derivation. Derives Theorem 107.74. Equation (107.81) is the multipole expansion of the radiation field applied to Fermi's golden rule; its structure is what matters here and it is fixed by dimensional analysis together with the small parameter of the problem. The photon field at the nucleus varies as \(\ee^{\ii\vect{k}\cdot\vect{r}}\) with \(k=E_{\gamma}/\hbar c\), and \(kR\ll1\) — for \(E_{\gamma}=1\,\mathrm{MeV}\) and \(R=6\,\mathrm{fm}\), \(kR=0.030\) — so each successive term of the expansion of Proposition 107.70 costs one power of \(kR\) in the amplitude. Two powers of \(kR\) per unit of \(L\) in the rate, and one factor of \(E_{\gamma}\) from the density of photon states, give the \((E_{\gamma}/\hbar c)^{2L+1}\); the numerical coefficient comes from the normalization of the vector spherical harmonics and is quoted.

Substituting Equation (107.82) and using \(e^{2}/4\pi\varepsilon_{0}\hbar=\alpha c\) collapses the prefactor to a pure number times \(\alpha c\):

For \(L=1\) the two numerical factors are \(16\pi/9\) and \((1/4\pi)(3/4)^{2}\), whose product is exactly \(\tfrac14\), so

and inserting \(\alpha=7.2974\times 10^{-3}\), \(c=3.0\times 10^{23}\,\mathrm{fm}/\mathrm{s}\), \(R^{2}=1.44A^{2/3}\,\mathrm{fm}^{2}\) and \(\hbar c=197.33\,\mathrm{MeV}\,\mathrm{fm}\) gives Equation (107.83). The same substitution at \(L=2,3,4\) gives the rest. These are the Weisskopf units, and they are estimates of an order of magnitude, not predictions: a measured rate is quoted as a multiple of them, and the multiple is the physics.

Magnetic multipoles are suppressed relative to electric ones of the same order by roughly

about \(2\times 10^{-3}\) for a medium-mass nucleus, since the magnetic moment of a nucleon carries the extra factor \(\hbar/m_{p}c\) that a charge does not.

Derives Corollary 107.75. From Equations (107.83) and (107.86), for \(A=125\) and \(E_{\gamma}=1\,\mathrm{MeV}\) the rates are \(2.6\times 10^{15}\), \(4.6\times 10^{10}\), \(5.3\times 10^{5}\) and \(4.2\times 10^{0}\,/\mathrm{s}\) for \(L=1,2,3,4\): a factor near \(10^{5}\) per unit of \(L\), which is \((kR)^{2}\) combined with the falling numerical coefficient. A \(4^{+}\) state lying just above a \(0^{+}\) ground state therefore lives of order a second rather than a femtosecond. The islands of isomerism sit just below the magic numbers, where Table 107.4 places levels of very different \(j\) next to one another — \(1g_{9/2}\) beside \(2p_{1/2}\) below \(Z\) or \(N=50\), \(1h_{11/2}\) beside \(3s_{1/2}\) below \(82\) — so that the lowest available transition has \(\Delta J=4\) or \(5\). That the isomer islands are found exactly where the shell model puts high-\(j\) intruders is a further confirmation of Theorem 107.52 that uses no binding energies at all.

Derivation. Derives Phenomenon 107.77. A free nucleus of mass \(M\) emitting a photon of energy \(E_{\gamma}\) recoils with momentum \(E_{\gamma}/c\) and so carries away

to first order in \(E_{R}/E_{\gamma}\). The emitted quantum is thus short of the transition energy by \(E_{R}\), while an absorbing nucleus needs \(E_{R}\) more than the transition energy in order to recoil in turn: emission and absorption lines are displaced from one another by \(2E_{R}\). Resonance survives only if that displacement is small against the line width \(\Gamma\). For the \(14.4\,\mathrm{keV}\) line of \(^{57}\mathrm{Fe}\), \(Mc^{2}=53.0\,\mathrm{GeV}\) gives \(E_{R}=1.96\times 10^{-3}\,\mathrm{eV}\), whereas the mean life of \(141\,\mathrm{ns}\) gives \(\Gamma=\hbar/\tau=4.67\times 10^{-9}\,\mathrm{eV}\): the displacement exceeds the width by a factor of \(8.4\times 10^{5}\), and no resonance between free nuclei is possible. Bound in a lattice the nucleus cannot recoil by itself. A finite fraction of the transitions leave the crystal in its original vibrational state, and for those the mass appearing in Equation (107.88) is the mass of the whole crystal, which makes \(E_{R}\) utterly negligible. The line is then as narrow as the lifetime permits, and it is that narrowness, not any large signal, which gives the method its resolution:

Derives Proposition 107.78. The probability of a zero-phonon transition is \(f=\exp(-k^{2}\avg{x^{2}})\) with \(k=E_{\gamma}/\hbar c\) the photon wave number and \(\avg{x^{2}}\) the mean square displacement of the emitting nucleus along the photon direction — the same Debye–Waller factor that governs the intensity of a Bragg peak, a subject that belongs to Phonons and Lattice Dynamics. Both formulas are used here as stated. In the Debye model at \(T=0\), \(\avg{x^{2}}=3\hbar^{2}/(4Mk_{B}\Theta_{D})\), so \(k^{2}\avg{x^{2}}=3E_{\gamma}^{2}/(4Mc^{2}k_{B}\Theta_{D}) =3E_{R}/2k_{B}\Theta_{D}\), which is Equation (107.90). With \(k_{B}\Theta_{D}=40.5\,\mathrm{meV}\) and \(E_{R}=1.96\,\mathrm{meV}\) the exponent is \(0.073\) and \(f=0.93\). Two conditions for a usable Mössbauer isotope follow at once: a low transition energy, since \(E_{R}\propto E_{\gamma}^{2}\), and a stiff lattice, since \(f\) falls with \(\Theta_{D}\). That \(^{57}\mathrm{Fe}\) satisfies both is why essentially all Mössbauer spectroscopy uses it.

Derives Corollary 107.79. The first-order Doppler shift gives Equation (107.91) directly from Equation (107.89); the millimetre-per-second scale of a Mössbauer spectrometer's drive is therefore not a convenience but a consequence of the line width. For the redshift,

so Pound and Rebka had to locate the centre of a line to better than one per cent of its width, which they did by modulating the source velocity and measuring the asymmetry of the counting rate [Pound:1960]. The experiment is a test of the equivalence principle and belongs equally to The Equivalence Principle and Classical Tests; what belongs here is that the instrument is a nuclear energy level, and that its precision comes from a lifetime.

Derivation. Derives Phenomenon 107.80. The energy released is the increase in total binding energy, and it can be read straight off the curve of Phenomenon 107.32. At \(A\approx236\) the binding is about \(7.6\,\mathrm{MeV}\) per nucleon; near the fragment masses, \(A\approx100\) to \(140\), it is about \(8.5\,\mathrm{MeV}\) per nucleon [Wang:2021]. The gain is therefore

the observed order of magnitude, and — since two numbers were read off a curve — as much precision as the argument is entitled to. The same estimate says why fission is confined to the heavy end of the chart: the gain is positive only where the curve is falling, and it is worth a substantial fraction of a nucleon's binding only where it has fallen far. It says nothing about the rate, which is governed by the barrier and not by the energy release, and that is why \(^{235}\mathrm{U}\) and \(^{238}\mathrm{U}\) behave so differently under slow neutrons although their release is nearly the same.

Derives Theorem 107.81. Volume conservation fixes \(R_{0}\) in terms of the equivalent spherical radius \(R\): expanding \(V=\tfrac{4\pi}{3}R_{0}^{3}\left(1+\tfrac35\alpha_{2}^{2}\right)\) to second order gives \(R_{0}=R\left(1-\tfrac15\alpha_{2}^{2}\right)\). The surface area of the deformed shape, expanded to the same order, is \(4\pi R^{2}\left(1+\tfrac25\alpha_{2}^{2}\right)\), which with \(E_{S}=4\pi R^{2}\sigma\) from Equation (107.44) gives Equation (107.93); the electrostatic energy of a uniformly charged spheroid, expanded likewise, gives Equation (107.94). The first coefficient is positive and the second negative because stretching increases the surface but moves charge apart. Subtracting the spherical values leaves Equation (107.95), and the sphere is stable if and only if the bracket is positive, which rearranges to Equation (107.96) with the coefficients Equation (107.43).

The parameter is customarily normalized as \(x=(Z^{2}/A)/(Z^{2}/A)_{\mathrm{crit}}\), and for the actinides: \(^{232}\mathrm{Th}\) has \(x=0.70\), \(^{238}\mathrm{U}\) \(x=0.71\), \(^{239}\mathrm{Pu}\) \(x=0.74\). All are stable against small deformations and all can fission, because a stable sphere is only a local minimum: the ultimate configuration, two separated fragments, lies \(200\,\mathrm{MeV}\) lower (Equation (107.92)), and between them stands a barrier that Equation (107.95) describes only near \(\alpha_{2}=0\). The full barrier requires the whole deformation energy surface, which is the calculation of Bohr and Wheeler [Bohr:1939b]; its height for the actinides is \(5\,\mathrm{MeV}\) to \(6\,\mathrm{MeV}\), small compared with the release and comparable with the neutron separation energy — which is the coincidence on which the whole subject turns.

Derivation. Derives Phenomenon 107.82. Capture of a neutron leaves a compound nucleus excited by exactly the neutron separation energy of the product, and it is that number, not the fission \(Q\) value, which must be compared with the barrier. From the mass evaluation [Wang:2021],

against barriers of about \(5.7\,\mathrm{MeV}\) and \(6.2\,\mathrm{MeV}\) respectively. In the first case the compound nucleus is created above its barrier and fissions promptly, with no help from the neutron's kinetic energy; in the second it is created more than a \(\mathrm{MeV}\) below, and the neutron must supply the difference.

The \(1.7\,\mathrm{MeV}\) between the two separation energies is pairing. \(^{235}\mathrm{U}\) has \(N=143\), odd, so the captured neutron completes a pair and the system gains the pairing energy of Equation (107.42); \(^{238}\mathrm{U}\) has \(N=146\), even, so the captured neutron is unpaired and gains nothing. The predicted difference is \(2a_{P}A^{-1/2}=2\times11.18/\sqrt{237} =1.45\,\mathrm{MeV}\), against the \(1.7\,\mathrm{MeV}\) observed. A term worth \(1\,\mathrm{MeV}\) in a formula whose leading term is \(3700\,\mathrm{MeV}\) therefore decides which isotope of uranium is a fuel, which is as sharp an illustration as the subject affords of why the small terms of Equation (107.41) were worth deriving.

Derivation. Derives Theorem 107.85. The exponent in Equation (107.100) is the point-Coulomb Gamow factor Equation (107.73) rewritten: with \(\eta=Z_{1}Z_{2}e^{2}/(4\pi\varepsilon_{0}\hbar v)\) and \(v=\sqrt{2E/\mu}\),

using \(e^{2}/4\pi\varepsilon_{0}\hbar=\alpha c\); the factor \(1/E\) outside is the geometric area \(\pi(\hbar/p)^{2}\propto1/E\), and everything not exponential and not geometric is collected into the astrophysical \(S\) factor \(S(E)\), which is what a laboratory measurement extrapolates to stellar energies. Here the point-Coulomb limit is the right one — unlike in Corollary 107.64, where the finite nuclear radius mattered — because the classical turning point \(b=Z_{1}Z_{2}e^{2}/4\pi\varepsilon_{0}E\) at \(E=6\,\mathrm{keV}\) is \(240\,\mathrm{fm}\), so \(\xi=R/b\approx0.01\) and the correction is a per cent.

The rate is the cross-section averaged over the Maxwell–Boltzmann distribution — whose derivation belongs to Kinetic Theory of Gases — which contributes \(\ee^{-E/k_{B}T}\) and a factor \(E\) that varies slowly. So

The two exponents pull in opposite directions — one kills large \(E\), the other kills small \(E\) — so the integrand is sharply peaked. Setting the derivative of the bracket to zero,

which is Equation (107.101). Expanding the bracket about \(E_{0}\) to second order gives a Gaussian of variance \(1/\abs{f''(E_{0})}\) with \(f''=-\tfrac34\sqrt{E_{G}}E_{0}^{-5/2}=-3/(2E_{0}k_{B}T)\), whence Equation (107.102). The construction is an evaluation by steepest descent, carried out on the real axis.

Derivation. Derives Phenomenon 107.87. The energy release is the mass difference. Using atomic masses, so that the two positrons annihilate with two of the electrons already counted,

equivalently \(4m_{p}c^{2}-M_{\mathrm{nuc}} (^{4}\mathrm{He})c^{2}=25.71\,\mathrm{MeV}\) of nuclear release plus \(4m_{e}c^{2}=2.04\,\mathrm{MeV}\) from the two annihilations, minus the \(2m_{e}c^{2}\) carried into existence as positrons — the two bookkeepings agreeing, as they must.

The rate-limiting step is the first, \(p+p\to d+e^{+}+\nu_{e}\), which is a weak process: two protons cannot form a bound diproton (Phenomenon 107.24), so one of them must convert to a neutron during the collision. Its cross-section has never been measured and cannot be, being some twenty orders of magnitude below anything a laboratory can reach; it is computed from the weak coupling of Weak Interactions and the two-nucleon wave function of Proposition 107.23. That a weak interaction gates the whole chain is why the Sun burns for \(10^{10}\) years instead of exploding, and it is a rare case in which the slowness of a rate is the reason there is anybody to measure it.

The flux at the Earth follows from the luminosity alone, with no solar model. Each helium nucleus releases \(Q\) and two neutrinos, so

some sixty billion neutrinos through every square centimetre of the reader every second. That this number is right, and that the number of electron neutrinos observed was for thirty years only a third of it, is the subject of Experiment: Neutrino Oscillations.

Derives Proposition 107.90. Setting \(A=2.5Z\) in Equation (107.96) gives \(Z/2.5=50.1\), that is \(Z=125\): beyond that, a nucleus is unstable against an infinitesimal deformation and cannot exist even momentarily. Below it the barrier does not vanish but shrinks rapidly, since Equation (107.95) is the leading term of a surface that flattens as \(x\to1\); the height falls from \(6\,\mathrm{MeV}\) at \(x=0.7\) (uranium) to a fraction of a \(\mathrm{MeV}\) at \(x\approx0.9\) (\(Z\approx106\) at the same neutron-to-proton ratio). A barrier that low is transparent, and the liquid drop therefore predicts that no nuclide beyond about \(Z=106\) survives long enough to have a chemical identity.