Helium-3 equilibrium abundance
= Helium-3 equilibrium abundance
{title2=$n_{3e}=(-B+\sqrt{B^2+4AC})/(2A)$}
With $A=\lambda_{33}$, $B=\lambda_{34}n_4$ and $C=\lambda_{11}n_1^2/2$, <helium-3> evolves as $\dot n_3=C-An_3^2-Bn_3$ after <deuterium> elimination. The positive root is its unique attracting equilibrium for a fixed background. In the pp-I limit it reduces to $n_1\sqrt{\lambda_{11}/(2\lambda_{33})}$, making equilibrium abundance strongly <temperature> dependent.