Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 1 / ii / Solution

Along the pp-chain homologous sequence, $R$ is constant and therefore $T_c\propto M$. Normalizing to the Sun gives the transition mass

For the CNO cycle law $ϵ\propto \rho T^{23/2}$, the nuclear scaling becomesThe opacity law is unchanged, so $L_{\mathrm{rad}}\propto M^{11/2}{R}^{-1/2}$. Equating them givesand then

The effective temperature follows from the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$. Thus the pp branch has $T_e\propto M^{11/8}$ and $L\propto T_e^4$, whereas the CNO branch hasOn a Hertzsprung-Russell diagram, both branches rise toward larger luminosity and, conventionally, leftward toward larger temperature. They join near $4M_{\odot }/3$; the CNO branch has the steeper $L$--$T_e$ logarithmic slope and lies to the cooler side of the extrapolated constant-radius branch at fixed luminosity.

If $X$ is the hydrogen mass fraction and $X_{\mathrm{CNO}}$ the combined carbon-nitrogen-oxygen mass fraction, the reaction-pair and catalyst abundances give approximatelyThe CNO nuclei are catalysts, so their abundance multiplies the CNO-cycle rate rather than being consumed by the completed cycle.