Deuterium quasi-equilibrium in proton-proton burning ($n_2\simeq {\lambda }_{11}{n}_1/(2{\lambda }_{21})$)

deuterium proton capture is much faster than the weak reaction that produces deuterium. Its abundance therefore adjusts so that ${\lambda }_{21}{n}_2{n}_1\simeq {\lambda }_{11}{n}_1^2/2$. Substituting this balance eliminates the fast intermediate from the slower proton and helium-3 evolution equations. This is a quasi-steady approximation, not an assumption that hydrogen itself is in nuclear equilibrium.