A Race Between Capture and Decay: How the Elements Heavier than Iron Are Made — Epoche C1
Every chemical element, we are often told, was forged by nuclear fusion inside stars. The claim holds up to iron and fails beyond it, where fusion costs energy rather than releasing it; the heavier half of the periodic table — silver, gold, iodine, uranium — was assembled another way. The founding paper of the field, by Margaret Burbidge, Geoffrey Burbidge, William Fowler and Fred Hoyle in 1957, universally cited as B²FH, assigned essentially everything above iron to the capture of neutrons, and a single comparison — is capture faster than beta decay? — divides that territory into two families with different products and different birthplaces. The birthplace of the second was confirmed only in 2017. Why fusion stops paying at iron The relevant quantity is the binding energy per nucleon: the energy needed to pull a nucleus apart, divided by the number of protons and neutrons in it. It rises from about $7.1\ \mathrm{MeV}$ for helium-4 to a maximum near $8.8\ \mathrm{MeV}$ in the iron group, around mass number $A \approx 56$–$62$, and then declines towards uranium. Fusion releases energy only while the curve rises, since only there is the merged nucleus more tightly bound than its parts. Why the curve should turn over at all follows from two competing terms in the standard semi-empirical description of nuclear masses. Nuclear binding is short-ranged and saturates, so the bulk contribution grows in proportion to $A$; but nucleons at the surface have fewer neighbours, and the resulting deficit scales as the surface area, $A^{2/3}$. Dividing through by $A$, the surface penalty falls off as $A^{-1/3}$, which is what makes larger nuclei more tightly bound per nucleon and drives the rise. Working against it is the electrostatic repulsion of the protons, which is long-ranged and therefore does not saturate: it accumulates as $Z^{2}/A^{1/3}$, so per nucleon it grows as $Z^{2}/A^{4/3}$. The rising surface term wins early and the growing Coulomb term wins late, and the crossover sits in the iron group. (The curve actually peaks at nickel-62; silicon burning in a star ends at nickel-56, which decays to iron-56, which is why iron rather than nickel is the byword.) Electric charge seals the verdict a second time, through the reaction rate rather than the energy budget. Two nuclei must reach contact against their mutual repulsion, and the Coulomb barrier — the electrostatic energy at touching distance — is $E_C \approx Z_1Z_2e^{2}/R$. For two iron-56 nuclei, $Z_1 = Z_2 = 26$, with $e^{2} \approx 1.44\ \mathrm{MeV\,fm}$ in Gaussian units and touching radius $R \approx 9.2\ \mathrm{fm}$ (nuclear radii scale as $1.2\,A^{1/3}\ \mathrm{fm}$), this is $$E_C \approx \frac{26\times 26\times 1.44}{9.2}\ \mathrm{MeV} \approx 106\ \mathrm{MeV}.$$ Silicon burning at $T \approx 4\times10^{9}\ \mathrm{K}$ supplies a thermal energy $k_\mathrm{B}T \approx 0.34\ \mathrm{MeV}$, three hundred times too small. Quantum tunnelling rescues charged-particle reactions at lower charges, but the penetration probability falls off exponentially in the product $Z_1Z_2$ divided by the relative velocity, and raising that product from 1, as in hydrogen burning, to 676 suppresses the rate beyond any temperature a star can supply before it collapses. One inequality, two families The route past both obstacles is the neutron. Being uncharged, it feels no barrier and enters a heavy nucleus freely. Successive captures leave the nucleus increasingly neutron-rich until beta decay intervenes — a neutron inside the nucleus turns into a proton, emitting an electron and an antineutrino — which raises the atomic number by one and moves the material one element up the table. The waiting time between captures follows from an elementary rate argument. A nucleus sitting in a flux of neutrons of number density $n_n$ moving at relative speed $v$ presents an effective target area $\sigma$, the capture cross-section, so it captures at the rate $n_n\sigma v$ and waits, on average, $$\tau_{\mathrm{cap}} = \frac{1}{n_n\,\sigma\,v}$$ between captures — strictly $1/(n_n\langle\sigma v\rangle)$, with the product averaged over the thermal distribution of neutron speeds. B²FH's classification compares this with the beta-decay lifetime $\tau_\beta$ of the nucleus reached. If $\tau_{\mathrm{cap}} \gg \tau_\beta$, every unstable nucleus decays before the next neutron arrives, and the path creeps along the valley of stable nuclei one step at a time: the slow, or s-, process. If $\tau_{\mathrm{cap}} \ll \tau_\beta$, dozens of captures happen before any decay, and the path plunges far off the valley into extremely neutron-rich territory: the rapid, or r-, process. The two are not variations in degree. They visit different nuclei and therefore make different elements. The slow family: a capture every decade The s-process runs in asymptotic-giant-branch stars — old stars of one to eight solar masses in their final shell-burning phase, in which a thin helium-burning shell flashes intermittently and mixes material outwards. There the reaction $^{13}\mathrm{C}(\alpha,n)^{16}\mathrm{O}$, in which carbon-13 absorbs a helium nucleus and releases a neutron, sustains neutron densities of order $10^{7}$–$10^{8}\ \mathrm{cm^{-3}}$. Put representative numbers through the formula. With $k_\mathrm{B}T = 30\ \mathrm{keV}$, a neutron of rest energy $940\ \mathrm{MeV}$ moves at $v = \sqrt{2k_\mathrm{B}T/m_n} \approx 2.4\times10^{8}\ \mathrm{cm\,s^{-1}}$, and a typical heavy nucleus presents $\sigma \approx 100\ \mathrm{mb} = 10^{-25}\ \mathrm{cm^{2}}$. Then $$\tau_{\mathrm{cap}} \approx \frac{1}{10^{8}\times10^{-25}\times2.4\times10^{8}}\ \mathrm{s} \approx 4\times10^{8}\ \mathrm{s},$$ about thirteen years between captures. Unstable nuclei close to the valley of stability decay in minutes to months, so $\tau_{\mathrm{cap}} \gg \tau_\beta$ holds with room to spare, and over tens of thousands of years the process assembles strontium, barium and lead. The s-process is the better tested of th