= Solution
Write $n_{56}$ and $n_{54}$ for the number densities of nickel-56 and iron-54. Both nuclei contain 28 <neutrons>, while their <proton> numbers differ by two. Dividing the positive-exponential <nuclear statistical equilibrium> formulas printed in the PDF therefore gives the <nickel–iron nuclear equilibrium ratio>
$$
\boxed{\frac{n_{56}}{n_{54}}=K n_p^2T^{-3}
\exp\left(\frac{\Delta Q}{kT}\right),\qquad
\Delta Q=Q(56,28)-Q(54,26).}
$$
The neutron-density factors cancel. $K$ contains the isotope-dependent statistical and translational <mass> factors suppressed in the supplied proportionality; it is constant in this simplified calculation. The TeX conversion incorrectly reverses the exponential sign, so the PDF sign is essential here.
In the specified restricted composition, equality of total <proton> and <neutron> numbers means
$$
28n_{56}+26n_{54}+n_p=28n_{56}+28n_{54},
\qquad\boxed{n_p=2n_{54}.}
$$
Consequently
$$
F(T)=\frac{n_{56}}{n_{54}^3}=4K T^{-3}e^{\Delta Q/(kT)}.
$$
Differentiation, with the suppressed prefactors held constant, gives
$$
\frac{d\log F}{dT}=-\frac3T-\frac{\Delta Q}{kT^2}.
$$
Using the paper's supplied signed energy difference produces
$$
\boxed{T_{\max}=-\frac{\Delta Q}{3k}=9.0\times10^9\ \mathrm K.}
$$
With that negative value, $F$ tends to zero at both <temperature> endpoints, increases below this <temperature> and decreases above it. At the stationary point $d^2\log F/dT^2=-3/T^2<0$, so it is indeed a maximum for the printed toy data.
There is a genuine physical-data qualification: the stated negative energy difference is inconsistent with interpreting $Q$ as the conventional positive total <nuclear binding energy> of these actual isotopes. Using atomic masses, whose electron masses cancel in this reaction,
$$
\Delta B\simeq[2m({}^1\mathrm H)+m({}^{54}\mathrm{Fe})-m({}^{56}\mathrm{Ni})]c^2
\simeq+12.23\ \mathrm{MeV}>0.
$$
The masses used are $1.00782503223\,\mathrm u$, $53.93960899\,\mathrm u$ and $55.94212855\,\mathrm u$, from https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=H&isotype=all[NIST hydrogen masses], https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Fe&isotype=all[NIST iron masses] and https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Ni&isotype=all[NIST nickel masses]. Small electronic binding corrections do not change the sign. With a positive $\Delta B$, the simplified $F(T)$ above is strictly decreasing for $T>0$ and has no finite maximum. Thus the numerical maximum follows from the supplied negative parameter, not from consistent physical binding energies; changing the TeX sign to recover a maximum would not repair the PDF's data.
<Nuclear statistical equilibrium> is appropriate when the relevant forward and inverse strong or electromagnetic reactions are fast compared with changes of <temperature>, density and composition. It is useful in sufficiently hot late silicon-burning material and in the hottest explosive burning regions of massive stars. Some burning zones reach only quasi-equilibrium, and cooling or expansion eventually freezes the abundances out; an equilibrium expression cannot by itself predict the full ashes when these timescale conditions fail. <Temperature>, density, conserved <baryon number> and the <electron fraction in stellar matter> must all be specified. Negligible free-particle contributions in bulk bookkeeping are not the same as exactly zero free densities in the equilibrium formula.
Strong and electromagnetic reactions preserve total charge and <baryon number>, so they conserve the proton-to-neutron ratio. Weak reactions need not: <electron capture> and <beta-plus decay> convert <protons> into <neutrons> and lower it. Such conversions occur in hydrogen-burning networks and in the nitrogen-to-neon chain during <helium> burning. At high density in advanced stellar cores and core collapse, <electron capture> becomes particularly important; <neutrino> emission also affects energy and lepton loss. <Beta-minus decay> changes the ratio in the opposite direction and can compete during later relaxation. Nuclear equilibrium for strong reactions does not automatically imply weak equilibrium or a fixed proton-to-neutron ratio throughout all <stellar evolution>.
Back to article page
