---
title: Stellar Nucleosynthesis
module: Fusion and Nucleosynthesis
moduleNumber: 10
lessonNumber: 2
order: 1002
summary: >
  Main-sequence stars burn hydrogen to helium through the proton-proton chain and the
  CNO cycle, both releasing 26.7 MeV per helium nucleus. Helium burning bridges the
  mass-5 and mass-8 gaps by the triple-alpha process through the Beryllium-8 and Hoyle
  resonances, and successive carbon-to-silicon burning stages climb to the iron peak,
  where fusion stops. The elements beyond iron are built by slow and rapid neutron
  capture, and the solar neutrino flux confirms the reactions directly.
topics: [Fusion and Nucleosynthesis]
sources:
  - book: Krane
    ref: "Ch. 14 — Nuclear Fusion; §14.3 Thermonuclear Reactions in Stars, Stellar Nucleosynthesis"
  - book: Tipler & Llewellyn
    ref: "Ch. 13 — Astrophysics and Cosmology; solar fusion and stellar energy generation"
draft: false
---

Stars are self-regulating fusion reactors held together by gravity. A main-sequence star
sits in hydrostatic equilibrium: the outward pressure of a hot plasma balances its own
weight, and the temperature at the center is whatever fusion rate replaces the energy
radiated from the surface. Because the thermonuclear rate depends so steeply on temperature,
this balance is stable — a small contraction raises $T$, raises the rate, and restores the
pressure. The sequence of nuclei a star builds is set by which Coulomb barriers its central
temperature can overcome, and the reactions run in a definite order from hydrogen up to the
iron peak, past which fusion no longer releases energy.[^krane-stars]

## Hydrogen burning: the proton-proton chain

The net result of hydrogen burning is $4p \to {}^{4}\mathrm{He} + 2e^+ + 2\nu_e$, releasing
$Q = 26.73\ \mathrm{MeV}$ once the positrons annihilate. The barrier problem is severe: the
first step must fuse two protons, and there is no bound diproton, so the reaction proceeds
only if one proton converts to a neutron by the weak interaction during the brief collision,

$$
p + p \to {}^{2}\mathrm{H} + e^+ + \nu_e, \qquad Q = 0.42\ \mathrm{MeV}.
$$

This is a weak process happening inside a Coulomb-suppressed collision, so its rate is
minute — the mean time for a given proton in the solar core to undergo it is billions of
years. That slowness is why the Sun burns for ten billion years rather than exploding. The
deuteron then captures a proton and two helium-3 nuclei combine, in the branch called
**pp-I**:

$$
{}^{2}\mathrm{H} + p \to {}^{3}\mathrm{He} + \gamma\;(5.49\ \mathrm{MeV}),
\qquad
{}^{3}\mathrm{He} + {}^{3}\mathrm{He} \to {}^{4}\mathrm{He} + 2p\;(12.86\ \mathrm{MeV}).
$$

The two neutrinos of pp-I carry away about $0.6\ \mathrm{MeV}$, leaving roughly
$26.2\ \mathrm{MeV}$ to heat the star. When helium has accumulated, the ${}^{3}\mathrm{He}$
can instead capture a ${}^{4}\mathrm{He}$, opening the **pp-II** and **pp-III** branches
through ${}^{7}\mathrm{Be}$, which either captures an electron to ${}^{7}\mathrm{Li}$ or a
proton to ${}^{8}\mathrm{B}$. The ${}^{8}\mathrm{B}$ positron decay produces the highest-energy
solar neutrinos, up to $15\ \mathrm{MeV}$, and although this branch is rare it dominates the
neutrino signal that terrestrial detectors can see.[^krane-stars]

$$
% caption: The proton-proton chain fuses four protons into helium-4; after the slow
% weak-interaction first step it splits into three branches, pp-I completing through two
% helium-3 nuclei and pp-II/pp-III proceeding through beryllium-7.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0, node distance=6mm]
  \definecolor{acc}{HTML}{4A6FA5}
  \tikzset{nuc/.style={draw=acc, fill=acc!10, thick, minimum width=8mm, minimum height=6mm, inner sep=2pt, font=\footnotesize}}
  \node[nuc] (dd) at (0,0) {d};
  \node[nuc] (he3) at (2.6,0) {He3};
  \node[nuc] (he4a) at (5.5,1.3) {He4};
  \node[nuc] (be7) at (5.5,-1.3) {Be7};
  \node[nuc] (li7) at (8.0,-0.2) {Li7};
  \node[nuc] (b8) at (8.0,-2.4) {B8};
  \draw[->, black, thick] (-1.5,0) node[left, black!70, font=\scriptsize] {$p + p$} -- (dd);
  \draw[->, black, thick] (dd) -- (he3) node[midway, above, black, font=\scriptsize] {$+p$};
  \draw[->, acc, thick] (he3) -- (he4a) node[midway, above left, black, font=\scriptsize] {pp-I};
  \draw[->, acc, thick] (he3) -- (be7) node[midway, below left, black, font=\scriptsize] {$+$ He4};
  \draw[->, black, thick] (be7) -- (li7) node[midway, above, black, font=\scriptsize] {pp-II};
  \draw[->, black, thick] (be7) -- (b8) node[midway, left, black, font=\scriptsize] {pp-III};
\end{tikzpicture}
$$

## The CNO cycle

In stars more massive than about $1.3$ solar masses the central temperature exceeds
$1.7\times10^7\ \mathrm{K}$, and hydrogen burns faster through a catalytic cycle that uses
pre-existing carbon, nitrogen, and oxygen. The **CNO cycle** captures four protons onto a
carbon-12 seed and returns the carbon at the end, ejecting one helium-4:

$$
{}^{12}\mathrm{C}(p,\gamma){}^{13}\mathrm{N}(\beta^+\nu){}^{13}\mathrm{C}(p,\gamma)
{}^{14}\mathrm{N}(p,\gamma){}^{15}\mathrm{O}(\beta^+\nu){}^{15}\mathrm{N}(p,\alpha){}^{12}\mathrm{C}.
$$

The net reaction and energy release are identical to the pp chain, $4p \to {}^{4}\mathrm{He}
+ 2e^+ + 2\nu_e$ with $Q = 26.73\ \mathrm{MeV}$, but the temperature dependence is far steeper
because the rate-limiting proton captures must penetrate the $Z = 6$–$8$ Coulomb barriers.
The slowest step is ${}^{14}\mathrm{N}(p,\gamma){}^{15}\mathrm{O}$, so the cycle drives most
of the catalyst into ${}^{14}\mathrm{N}$, a signature of CNO processing. The pp chain scales
roughly as $T^4$ near solar conditions and the CNO cycle as $T^{16}$–$T^{20}$, so the two
mechanisms cross at about $1.7\times10^7\ \mathrm{K}$: the Sun runs on pp (CNO contributes
about $1\%$), while more massive, hotter stars run on CNO.

$$
% caption: The CNO cycle is catalytic, threading four proton captures and two positron
% decays around a loop that consumes four protons, emits one helium-4, and regenerates the
% carbon-12 seed; the nitrogen-14 proton capture is the bottleneck.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \tikzset{nuc/.style={draw=acc, fill=acc!10, thick, minimum width=8mm, minimum height=6mm, inner sep=2pt, font=\footnotesize}}
  \node[nuc] (c12) at (90:2.3) {C12};
  \node[nuc] (n13) at (30:2.3) {N13};
  \node[nuc] (c13) at (-30:2.3) {C13};
  \node[nuc] (n14) at (-90:2.3) {N14};
  \node[nuc] (o15) at (210:2.3) {O15};
  \node[nuc] (n15) at (150:2.3) {N15};
  \draw[->, black, thick] (c12) to[bend left=18] (n13);
  \draw[->, black, thick] (n13) to[bend left=18] (c13);
  \draw[->, black, thick] (c13) to[bend left=18] (n14);
  \draw[->, acc, very thick] (n14) to[bend left=18] (o15);
  \draw[->, black, thick] (o15) to[bend left=18] (n15);
  \draw[->, black, thick] (n15) to[bend left=18] (c12);
  \node[acc, font=\scriptsize] at (0,0) {catalytic loop};
  \node[black, anchor=north, font=\scriptsize] at (-90:3.3) {slowest step};
\end{tikzpicture}
$$

## Helium burning and the triple-alpha bottleneck

When core hydrogen is exhausted the star contracts and heats until helium can fuse near
$10^8\ \mathrm{K}$. Building carbon from helium is blocked by two gaps: there is no stable
nucleus at mass $5$ or at mass $8$. Two alphas make ${}^{8}\mathrm{Be}$, which is unbound by
$92\ \mathrm{keV}$ and flies apart in $10^{-16}\ \mathrm{s}$. The path to carbon exists only
because of two resonances stacked in coincidence. A tiny equilibrium population of
${}^{8}\mathrm{Be}$ survives long enough for a third alpha to be captured, and that capture
is resonant with an excited $0^+$ state of ${}^{12}\mathrm{C}$ at $7.65\ \mathrm{MeV}$ — the
**Hoyle state**, predicted from the requirement that carbon be produced at all before it was
found experimentally:

$$
\alpha + \alpha \rightleftharpoons {}^{8}\mathrm{Be},
\qquad
{}^{8}\mathrm{Be} + \alpha \to {}^{12}\mathrm{C}^{\ast}(7.65) \to {}^{12}\mathrm{C} + \gamma.
$$

The overall **triple-alpha** reaction $3\,\alpha \to {}^{12}\mathrm{C}$ releases
$Q = 7.275\ \mathrm{MeV}$. Because it needs three bodies, the rate scales as the cube of the
alpha density and, through the two resonances, as roughly $T^{40}$ near $10^8\ \mathrm{K}$ —
the most temperature-sensitive reaction in a star. Once carbon is present it captures
another alpha, ${}^{12}\mathrm{C}(\alpha,\gamma){}^{16}\mathrm{O}$, and some oxygen captures a
further alpha to ${}^{20}\mathrm{Ne}$, setting the carbon-to-oxygen ratio that later burning
stages inherit.[^krane-stars]

$$
% caption: The triple-alpha process crosses the mass-8 gap through the short-lived
% beryllium-8 resonance and the Hoyle resonance of carbon-12; the equilibrium of the first
% step and the resonant capture of the second make carbon production possible.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \tikzset{lev/.style={acc, very thick}}
  % He4 + He4 level (left)
  \draw[lev] (0,0.6) -- (1.6,0.6);
  \node[black!70, anchor=south, font=\scriptsize] at (0.8,0.62) {He4 $+$ He4};
  % Be8 resonance (dashed, slightly above)
  \draw[acc, very thick, dashed] (2.6,0.85) -- (4.2,0.85);
  \node[black!70, anchor=south, font=\scriptsize] at (3.4,0.87) {Be8 resonance};
  % Hoyle state (higher, right)
  \draw[lev] (5.2,2.4) -- (6.8,2.4);
  \node[black!70, anchor=south, font=\scriptsize] at (6.0,2.42) {Hoyle state};
  % C12 ground state (bottom right)
  \draw[lev] (5.2,-0.9) -- (6.8,-0.9);
  \node[black!70, anchor=north, font=\scriptsize] at (6.0,-0.92) {C12 ground};
  % arrows
  \draw[->, black, thick] (1.6,0.6) -- (2.6,0.85);
  \draw[<-, black, thick, dashed] (2.55,0.72) .. controls (2.2,0.4) .. (1.7,0.45);
  \draw[->, black, thick] (4.2,0.85) -- (5.2,2.35) node[midway, above left, black, font=\scriptsize] {$+$ He4};
  \draw[->, acc, thick] (6.0,2.4) -- (6.0,-0.9) node[midway, right, black, font=\scriptsize] {gamma};
\end{tikzpicture}
$$

## Advanced burning to the iron peak

A massive star burns through successive fuels, each igniting when the ash of the previous
stage contracts and heats enough to overcome the next barrier. The stages run faster as they
proceed, because each releases less energy per nucleon while radiating from an ever hotter,
denser core.[^krane-stars]

- **Carbon burning** ($\sim 5\times10^8\ \mathrm{K}$):
  ${}^{12}\mathrm{C} + {}^{12}\mathrm{C} \to {}^{20}\mathrm{Ne} + \alpha$,
  ${}^{23}\mathrm{Na} + p$, ${}^{24}\mathrm{Mg} + \gamma$.
- **Neon burning** ($\sim 1.2\times10^9\ \mathrm{K}$): photodisintegration
  ${}^{20}\mathrm{Ne}(\gamma,\alpha){}^{16}\mathrm{O}$ frees alphas that recombine onto other
  neon, building ${}^{24}\mathrm{Mg}$.
- **Oxygen burning** ($\sim 2\times10^9\ \mathrm{K}$):
  ${}^{16}\mathrm{O} + {}^{16}\mathrm{O} \to {}^{28}\mathrm{Si} + \alpha$, ${}^{31}\mathrm{P}
  + p$, ${}^{31}\mathrm{S} + n$.
- **Silicon burning** ($\sim 3\times10^9\ \mathrm{K}$): photodisintegration and re-capture
  reach a quasi-equilibrium that funnels nuclei toward the most bound species near
  ${}^{56}\mathrm{Ni}$, which decays to ${}^{56}\mathrm{Fe}$.

Each stage lasts far less time than the last — hydrogen burning of a $25$-solar-mass star
takes millions of years, silicon burning about a day — and the star develops an onion-shell
structure with the heaviest ash at the center. Fusion halts at the iron peak because $B/A$
is maximal there: fusing iron would consume energy rather than release it, so no further
exothermic fusion can support the core. When the inert iron core exceeds the Chandrasekhar
mass it collapses, and the star's fusion history ends.

$$
% caption: A massive star near the end of its life has concentric burning shells, each
% fusing the ash of the layer outside it, with an inert iron core at the center where
% exothermic fusion stops.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) circle (3.2);
  \draw[black, thick] (0,0) circle (2.55);
  \draw[black, thick] (0,0) circle (1.95);
  \draw[black, thick] (0,0) circle (1.4);
  \draw[black, thick] (0,0) circle (0.9);
  \draw[acc, very thick] (0,0) circle (0.5);
  \fill[acc!14] (0,0) circle (0.5);
  \node[acc, font=\scriptsize] at (0,0) {Fe};
  \node[black!70, font=\scriptsize] at (0,0.7) {Si};
  \node[black!70, font=\scriptsize] at (0,1.15) {O, Ne};
  \node[black!70, font=\scriptsize] at (0,1.68) {C};
  \node[black!70, font=\scriptsize] at (0,2.25) {He};
  \node[black!70, font=\scriptsize] at (0,2.85) {H};
  \node[black, anchor=west, font=\scriptsize] at (3.3,0) {envelope};
\end{tikzpicture}
$$

## Building the heavy elements by neutron capture

Charged-particle fusion cannot climb past iron: the Coulomb barrier grows with $Z$ while the
energy return turns negative. Elements heavier than the iron peak are assembled instead by
neutron capture, which has no barrier. A nucleus captures a neutron, $(n,\gamma)$, moving one
step to the right in $A$; if the product is unstable it $\beta^-$-decays, raising $Z$ and
moving up the chart. Two regimes are distinguished by whether capture or decay is faster.

- **The s-process** (slow): the neutron flux is low, so between captures an unstable nucleus
  has time to $\beta^-$-decay. The path hugs the valley of stability, stepping up along it
  one mass unit at a time. It runs in thermally pulsing AGB stars, with neutrons supplied by
  ${}^{13}\mathrm{C}(\alpha,n){}^{16}\mathrm{O}$ and ${}^{22}\mathrm{Ne}(\alpha,n){}^{25}\mathrm{Mg}$,
  and terminates in a cycle at ${}^{208}\mathrm{Pb}$–${}^{209}\mathrm{Bi}$. Abundance peaks
  appear at the magic neutron numbers $N = 50, 82, 126$ ($A \approx 88, 138, 208$), where the
  small capture cross section makes nuclei pile up.
- **The r-process** (rapid): the neutron flux is enormous, so many captures occur before any
  $\beta^-$-decay. The path runs far out on the neutron-rich side, along a contour of nearly
  constant neutron separation energy, pausing at **waiting points** where a magic neutron
  number lowers the capture rate until a $\beta^-$-decay lets it advance. It requires an
  explosive, neutron-rich site — neutron-star mergers, confirmed by the kilonova of a merger
  event, and possibly core-collapse supernovae. After the flux ceases, the neutron-rich
  isotopes decay back to stability, producing abundance peaks at $A \approx 80, 130, 195$,
  displaced below the s-process peaks because the freeze-out happens at fixed neutron number.

$$
% caption: On the chart of nuclides the s-process climbs in a staircase along the valley of
% stability, while the r-process runs horizontally into neutron-rich territory and then
% decays diagonally back to stability after the neutron flux stops.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (8.4,0) node[right, black!70] {neutron number $N$};
  \draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {proton number $Z$};
  % stability band (diagonal)
  \draw[black, thick] (0.6,0.4) -- (6.6,4.0);
  \node[black, anchor=north west, font=\scriptsize, rotate=31] at (4.0,2.3) {stability};
  % s-process staircase along stability (solid acc)
  \draw[acc, very thick]
    (1.2,0.9) -- (1.8,0.9) -- (1.8,1.35) -- (2.6,1.35) -- (2.6,1.8)
    -- (3.4,1.8) -- (3.4,2.25) -- (4.2,2.25) -- (4.2,2.7);
  \node[acc, anchor=south east, font=\scriptsize] at (2.4,1.8) {s-process};
  % r-process horizontal then decay diagonal (dashed)
  \draw[black, very thick, dashed] (1.4,0.5) -- (5.6,0.5);
  \draw[black, thick, dashed] (5.6,0.5) -- (7.0,1.9);
  \node[black, anchor=north west, font=\scriptsize] at (3.2,0.5) {r-process};
  \node[black, anchor=west, font=\scriptsize] at (5.7,1.3) {decay to stability};
  % waiting point marker
  \fill[black!70] (5.6,0.5) circle (1.7pt);
  \node[black!70, anchor=south, font=\scriptsize] at (5.6,0.6) {waiting point};
\end{tikzpicture}
$$

## The solar neutrino confirmation

The reactions of the pp chain and CNO cycle can be checked directly because each weak step
emits a neutrino that leaves the star immediately, carrying a fingerprint of the reaction
that made it. The Sun's photons take $10^5$ years to random-walk out of the core; its
neutrinos arrive in $8$ minutes. Ray Davis's chlorine experiment in the 1960s measured a
flux of ${}^{8}\mathrm{B}$ neutrinos about a third of the standard solar model prediction,
the **solar neutrino problem**. The deficit was not a failure of the fusion model but of
particle physics: electron neutrinos oscillate into muon and tau flavors on the way out,
which the chlorine detector could not see. The Sudbury Neutrino Observatory settled it by
measuring both the electron-flavor flux and the total flux over all flavors; the total
matched the solar model while the electron flux was suppressed, confirming both the pp-chain
energy generation and neutrino flavor mixing.[^krane-nu] The same weak process that starts
hydrogen burning also set the neutron-to-proton ratio of the early universe, the subject of
the [next lesson](/nuclear-physics/fusion-nucleosynthesis/big-bang-nucleosynthesis).

[^krane-stars]: **Krane**, _Introductory Nuclear Physics_, §14.3 (Fusion in Stars): the
proton-proton chain with its three branches, the CNO catalytic cycle, the triple-alpha
process through the ${}^{8}\mathrm{Be}$ and Hoyle resonances ($Q = 7.275\ \mathrm{MeV}$), the
advanced burning stages to the iron peak, and the s- and r-process neutron-capture paths.
The synthesis framework is that of Burbidge, Burbidge, Fowler, and Hoyle. Reaction $Q$-values
and cross sections are compiled by the NNDC, [https://www.nndc.bnl.gov/](https://www.nndc.bnl.gov/).

[^krane-nu]: **Krane**, §14.3, and **Tipler & Llewellyn**, _Modern Physics_, Ch. 13 (solar
fusion). The ${}^{8}\mathrm{B}$ solar neutrinos, the Davis chlorine deficit, and its
resolution by neutrino oscillation as established by the Sudbury Neutrino Observatory.
Neutrino-oscillation parameters are reviewed by the Particle Data Group,
[https://pdg.lbl.gov/](https://pdg.lbl.gov/).