Thermonuclear Fusion in the Sun and Other Stars — Question 7

Question 7

The CNO cycle returns its catalyst while converting four protons to helium

Use this cycle: ${}^{12}\mathrm C+p\to{}^{13}\mathrm N+\gamma$; ${}^{13}\mathrm N\to{}^{13}\mathrm C+e^++\nu_e$; ${}^{13}\mathrm C+p\to{}^{14}\mathrm N+\gamma$; ${}^{14}\mathrm N+p\to{}^{15}\mathrm O+\gamma$; ${}^{15}\mathrm O\to{}^{15}\mathrm N+e^++\nu_e$; ${}^{15}\mathrm N+p\to{}^{12}\mathrm C+{}^4\mathrm{He}$.

1. Cancel intermediate species and identify the net nuclear transformation and catalyst.

2. Including positron annihilation, calculate total Q using $m({}^1\mathrm H)=1.007825$ u, $m({}^4\mathrm{He})=4.002603$ u and $1\,\mathrm u c^2=931.5$ MeV.

Question 7 - Solution

The CNO cycle returns its catalyst while converting four protons to helium

(a) Method and calculation

$${}^{12}\mathrm C+4p\to{}^{12}\mathrm C+{}^4\mathrm{He}+2e^++2\nu_e+\text{photons}.$$$$\boxed{4p\to{}^4\mathrm{He}+2e^++2\nu_e+\text{energy}};\quad{}^{12}\mathrm C\text{ is regenerated}.$$

(b) Interpretation and further calculation

$$\Delta m=4m({}^1\mathrm H)-m({}^4\mathrm{He})=4(1.007825)-4.002603=.028697\,\mathrm u.$$$$\boxed{Q=.028697(931.5)=26.7313\,\mathrm{MeV}}.$$

Physical check

The catalyst changes reaction pathways but not the net mass difference. This atomic-mass Q includes positron annihilation and neutrino energy.