= Nickel–iron nuclear equilibrium ratio
{title2=$n_{56}/n_{54}=K n_p^2T^{-3}e^{\Delta B/(kT)}$}
Nickel-56 and iron-54 have the same <neutron> number and differ by two <protons>. Dividing their <nuclear statistical equilibrium> abundance formulas cancels the <neutron> density and leaves the displayed ratio, with $\Delta B=B_{56}-B_{54}$ and $K$ containing temperature-independent statistical and <mass> factors in the simplified model. For material with equal total <proton> and <neutron> numbers and only these nuclei plus free <protons>, charge balance gives $n_p=2n_{54}$. Thus $n_{56}/n_{54}^3\propto T^{-3}e^{\Delta B/(kT)}$. This has a finite <temperature> maximum only if the signed energy difference is negative; conventional positive total nuclear binding energies for these two actual isotopes instead give a positive difference.
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