---
title: "Hydrogen Burning: pp Chains and the CNO Cycle"
draft: false
module: Nuclear Astrophysics
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  Four protons fuse into one helium-4 nucleus, releasing 26.7 MeV, through two
  competing networks. The pp chain begins with a weak-interaction bottleneck and
  branches three ways; the CNO cycle uses carbon, nitrogen, and oxygen as catalysts
  and is limited by nitrogen-14 proton capture. Their steep and gentle temperature
  dependences cross near 1.8e7 K, which divides pp-powered lower-main-sequence
  stars from CNO-powered upper-main-sequence stars.
topics: [Nuclear Astrophysics]
sources:
  - book: Carroll & Ostlie
    ref: "Ch. 10 — The Interiors of Stars; §10.3 Nuclear Reaction Rates; Ch. 11 — The Sun"
  - book: Maoz
    ref: "Ch. 3 — Stellar Physics"
---

Hydrogen burning converts four protons into one helium-4 nucleus. The net reaction,

$$
4\,{}^{1}\text{H} \longrightarrow {}^{4}\text{He} + 2e^{+} + 2\nu_e,
$$

releases $Q = 26.73\ \text{MeV}$ once the two positrons annihilate with ambient
electrons. No single collision accomplishes this; four protons meeting at one point
is vanishingly unlikely, and two of them must convert to neutrons by the weak
interaction. The transformation proceeds through a sequence of two-body reactions,
organized into two networks that dominate in different temperature regimes: the
proton–proton (pp) chain and the carbon–nitrogen–oxygen (CNO) cycle.

## The mass defect

The energy released is the binding energy locked up when four nucleons assemble into
a tightly bound helium nucleus. Using atomic masses,

$$
4\,m({}^{1}\text{H}) = 4 \times 1.007825\ \text{u} = 4.031300\ \text{u},
\qquad
m({}^{4}\text{He}) = 4.002602\ \text{u},
$$

so the mass defect is $\Delta m = 0.028698\ \text{u}$, which the mass–energy
relation converts to

$$
Q = \Delta m\,c^2 = 0.028698\ \text{u} \times 931.494\ \frac{\text{MeV}}{\text{u}}
= 26.73\ \text{MeV}.
$$

The fraction of the rest mass released is $\Delta m / (4 m_{{}^{1}\text{H}}) =
0.71\%$. This efficiency, an order of magnitude larger than any chemical process and
smaller than the several-percent yields of later burning stages and of accretion
onto compact objects, sets the nuclear timescale of the Sun: at
$L_\odot = 3.83\times 10^{26}\ \text{W}$, converting $0.71\%$ of the Sun's hydrogen
supplies power for roughly $10^{10}\ \text{yr}$.

$$
% caption: The rest mass of four hydrogen atoms exceeds that of one helium-4 atom by
% 0.0287 u; the deficit, 0.71 percent of the input mass, is carried off as 26.73 MeV
% of photons, positron annihilation, and neutrinos.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% left bar: four protons
\draw[very thick] (0.6,0) rectangle (2.2,4.03);
\node[anchor=south] at (1.4,4.05) {four H atoms};
\node[black!70, anchor=north] at (1.4,-0.05) {4.0313 u};
% right bar: helium
\draw[very thick] (4.4,0) rectangle (6.0,4.0026);
\node[anchor=south] at (5.2,4.28) {helium-4};
\node[black!70, anchor=north] at (5.2,-0.05) {4.0026 u};
% deficit strip on top of helium bar
\draw[acc, <->] (7.0,4.0026) -- (7.0,4.03);
\draw[black, densely dotted] (6.0,4.0026) -- (7.2,4.0026);
\draw[black, densely dotted] (2.2,4.03) -- (7.2,4.03);
\node[acc, anchor=west, align=left] at (7.15,3.6)
  {mass defect\\0.0287 u\\= 26.73 MeV};
\end{tikzpicture}
$$

## The proton–proton chain

The chain opens with the slowest reaction in all of stellar physics,

$$
p + p \longrightarrow d + e^{+} + \nu_e,
\qquad Q = 0.42\ \text{MeV}.
$$

Two protons must not only tunnel through the Coulomb barrier but simultaneously
undergo a $\beta^{+}$ conversion of one proton to a neutron, a weak-interaction
process, during the fleeting moment of contact. The resulting deuteron is the only
bound two-nucleon state; the diproton is unbound, so without the weak conversion the
collision produces nothing. The combined smallness of the barrier penetration and
the weak matrix element makes the mean lifetime of a proton against this reaction
about $10^{10}\ \text{yr}$ at the solar center. Every later step is faster by many
orders of magnitude, so this first reaction throttles the entire chain and fixes the
Sun's luminosity and lifetime.[^co-pp]

The deuteron captures a proton almost immediately,

$$
d + p \longrightarrow {}^{3}\text{He} + \gamma,
\qquad Q = 5.49\ \text{MeV},
$$

with a mean lifetime of order seconds, so deuterium never accumulates. The chain
then completes by one of three branches, distinguished by the fate of ${}^{3}\text{He}$.

**pp-I** closes when two helium-3 nuclei react:

$$
{}^{3}\text{He} + {}^{3}\text{He} \longrightarrow {}^{4}\text{He} + 2p,
\qquad Q = 12.86\ \text{MeV}.
$$

Reaching one ${}^{4}\text{He}$ this way requires the first two reactions to run
twice. At the solar center pp-I produces about $69\%$ of the helium.

**pp-II** and **pp-III** begin instead with a capture on a pre-existing ${}^{4}\text{He}$,

$$
{}^{3}\text{He} + {}^{4}\text{He} \longrightarrow {}^{7}\text{Be} + \gamma,
\qquad Q = 1.59\ \text{MeV}.
$$

Beryllium-7 then either captures an electron (pp-II) or a proton (pp-III):

$$
\begin{aligned}
\text{pp-II:}\quad
&{}^{7}\text{Be} + e^{-} \to {}^{7}\text{Li} + \nu_e, &
&{}^{7}\text{Li} + p \to 2\,{}^{4}\text{He}; \\
\text{pp-III:}\quad
&{}^{7}\text{Be} + p \to {}^{8}\text{B} + \gamma, &
&{}^{8}\text{B} \to {}^{8}\text{Be} + e^{+} + \nu_e \to 2\,{}^{4}\text{He}.
\end{aligned}
$$

At the solar center pp-II accounts for about $31\%$ of the helium and pp-III for
only about $0.3\%$. Small as it is, pp-III matters out of proportion to its rate: the
$\beta^{+}$ decay of ${}^{8}\text{B}$ emits neutrinos with energies up to
$\sim 15\ \text{MeV}$, the only solar neutrinos energetic enough for the early
chlorine and water-Cherenkov detectors, so this rare branch dominated the historical
[solar-neutrino
measurements](/astrophysics-cosmology/stellar-structure/the-standard-solar-model).

$$
% caption: The three pp branches share the first two reactions; helium-3 then either
% meets another helium-3 (pp-I) or captures on helium-4 to make beryllium-7, which
% splits into the electron-capture branch (pp-II) and the rare proton-capture branch
% (pp-III); solar-center branching fractions are shown.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >={Stealth[length=2mm]}]
\definecolor{acc}{HTML}{4A6FA5}
\tikzset{bx/.style={draw, very thick, inner sep=3pt, align=center}}
\node[bx] (pp) at (0,5) {p + p};
\node[bx] (d)  at (0,3.7) {deuteron + p};
\node[bx] (he3) at (0,2.4) {He-3};
\draw[->, thick] (pp) -- (d);
\draw[->, thick] (d) -- (he3);
% pp-I to the left
\node[bx] (ppone) at (-3.2,0.7) {He-3 + He-3};
\node[black!70, anchor=south] at (-4.7,1.7) {He-4 + 2p};
\draw[->, thick] (he3) -- node[black, above left, pos=0.55] {69\%} (ppone);
\draw[->, black, thick] (ppone) -- (-4.7,1.5);
\node[black, anchor=north] at (-3.2,0.55) {pp-I};
% branch to Be-7
\node[bx] (be7) at (2.8,0.9) {Be-7};
\draw[->, thick] (he3) -- node[black, above right, pos=0.55] {31\%} (be7);
% pp-II
\node[bx] (li7) at (1.4,-0.7) {Li-7 + p};
\draw[->, thick] (be7) -- node[black, left, pos=0.5] {add e} (li7);
\node[black, anchor=north] at (1.4,-1.0) {pp-II};
% pp-III
\node[bx] (b8) at (4.4,-0.7) {B-8};
\draw[->, black, thick] (be7) -- node[black, right, pos=0.5] {0.3\%} (b8);
\node[black, anchor=north] at (4.4,-1.0) {pp-III};
\end{tikzpicture}
$$

## Neutrino losses per branch

Each branch emits its neutrinos at different energies, and the neutrinos escape the
star immediately, carrying their energy away without contributing to the pressure or
luminosity. The effective heat deposited per helium nucleus is therefore
branch-dependent, always less than the full $26.73\ \text{MeV}$:

| Branch | neutrino source | typical $E_\nu$ | effective $Q$ |
| --- | --- | --- | --- |
| pp-I | $p+p$ (twice) | $\le 0.42\ \text{MeV}$ | $26.2\ \text{MeV}$ |
| pp-II | $p+p$, ${}^{7}\text{Be}$ capture | $0.86$/$0.38\ \text{MeV}$ line | $25.7\ \text{MeV}$ |
| pp-III | $p+p$, ${}^{8}\text{B}$ decay | $\le 15\ \text{MeV}$ | $19.3\ \text{MeV}$ |

The pp-I branch loses only about $2\%$ of $Q$ to neutrinos, while pp-III loses nearly
a third because the ${}^{8}\text{B}$ neutrino is so energetic. The neutrino spectrum
is a direct probe of which branches operate, and measuring it tests the
[standard solar
model](/astrophysics-cosmology/stellar-structure/the-standard-solar-model) at the
level of individual reactions.

## The CNO cycle

Where carbon, nitrogen, and oxygen are already present, a second network fuses
hydrogen using those nuclei as catalysts that are consumed and regenerated. The main
branch, CNO-I, is a closed loop of six reactions:

$$
\begin{aligned}
{}^{12}\text{C} + p &\to {}^{13}\text{N} + \gamma, &
{}^{13}\text{N} &\to {}^{13}\text{C} + e^{+} + \nu_e, \\
{}^{13}\text{C} + p &\to {}^{14}\text{N} + \gamma, &
{}^{14}\text{N} + p &\to {}^{15}\text{O} + \gamma, \\
{}^{15}\text{O} &\to {}^{15}\text{N} + e^{+} + \nu_e, &
{}^{15}\text{N} + p &\to {}^{12}\text{C} + {}^{4}\text{He}.
\end{aligned}
$$

Summing the loop, four protons enter and one ${}^{4}\text{He}$ leaves, with the same
net $Q = 26.73\ \text{MeV}$ minus the energy of two neutrinos from the
${}^{13}\text{N}$ and ${}^{15}\text{O}$ decays; the ${}^{12}\text{C}$ reappears
unchanged. The rate-limiting reaction is the proton capture on nitrogen-14,
${}^{14}\text{N}(p,\gamma){}^{15}\text{O}$, which has the highest Coulomb barrier of
the loop and a small cross section. Because it is the slowest link, the catalytic
material piles up as ${}^{14}\text{N}$: a star running the CNO cycle converts most of
its initial carbon and oxygen into nitrogen, the nucleosynthetic origin of much of
the galaxy's ${}^{14}\text{N}$.

$$
% caption: The CNO-I loop cycles a carbon-12 seed through nitrogen and oxygen
% isotopes and back, adding four protons and ejecting one helium-4 per turn; the
% nitrogen-14 proton capture (bold) is the slowest step and controls the rate.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >={Stealth[length=2mm]}]
\definecolor{acc}{HTML}{4A6FA5}
\tikzset{iso/.style={draw, very thick, circle, inner sep=2.5pt, minimum size=9mm}}
\node[iso] (c12) at (90:2.4) {C-12};
\node[iso] (n13) at (30:2.4) {N-13};
\node[iso] (c13) at (-30:2.4) {C-13};
\node[iso] (n14) at (-90:2.4) {N-14};
\node[iso] (o15) at (-150:2.4) {O-15};
\node[iso] (n15) at (150:2.4) {N-15};
\draw[->, thick] (c12) to[bend left=18] node[black, above right] {add p} (n13);
\draw[->, thick] (n13) to[bend left=18] node[black, right] {beta+} (c13);
\draw[->, thick] (c13) to[bend left=18] node[black, below right] {add p} (n14);
\draw[->, acc, very thick] (n14) to[bend left=18] node[black!70, below] {add p, slow} (o15);
\draw[->, thick] (o15) to[bend left=18] node[black, left] {beta+} (n15);
\draw[->, thick] (n15) to[bend left=18] node[black, above left] {add p} (c12);
\node[black!70, anchor=west] at (2.9,-1.2) {eject He-4};
\draw[->, black] (n15) .. controls (-1.6,1.8) and (-1.2,0.2) .. (2.8,-1.0);
\end{tikzpicture}
$$

About one proton capture on ${}^{15}\text{N}$ in a thousand takes the alternative
channel ${}^{15}\text{N}(p,\gamma){}^{16}\text{O}$ rather than releasing helium. This
opens the CNO-II branch, which threads through ${}^{16}\text{O}$, ${}^{17}\text{F}$,
and ${}^{17}\text{O}$ before returning to ${}^{14}\text{N}$, extending the catalytic
network to oxygen isotopes without changing the net hydrogen-to-helium conversion.

## The crossover and the main-sequence division

The two networks differ sharply in temperature sensitivity, for the reason derived
in [the Gamow-peak
analysis](/astrophysics-cosmology/nuclear-astrophysics/thermonuclear-reaction-rates-and-the-gamow-peak):
the CNO reactions run against a charge product $Z_1 Z_2 = 6$–$7$ rather than the
$Z_1 Z_2 = 1$ of the first pp step, so their Gamow peaks sit far higher on the
thermal tail. Near $1.5\times 10^7\ \text{K}$ the local power laws are

$$
\epsilon_{\rm pp} \propto \rho\,X^2\,T^{4},
\qquad
\epsilon_{\rm CNO} \propto \rho\,X\,X_{\rm CNO}\,T^{\,\sim 20},
$$

with $X$ the hydrogen mass fraction and $X_{\rm CNO}$ the catalyst abundance. The
gentle $T^{4}$ of the pp chain dominates at low temperature; the steep
$T^{\sim 20}$ of the CNO cycle overtakes it above a crossover near
$T \approx 1.8\times 10^7\ \text{K}$ for solar composition. The Sun's central
temperature, $1.57\times 10^7\ \text{K}$, sits just below the crossover, so the Sun
generates about $99\%$ of its power through the pp chain and only $\sim 1\%$ through
CNO.

$$
% caption: Energy generation rate versus central temperature; the shallow pp curve
% dominates below the crossover near 1.8e7 K and the steep CNO curve above it, so the
% Sun (marked, just below crossover) is pp-powered while more massive, hotter stars
% are CNO-powered.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.0,0) node[right, black!70] {temperature};
\draw[->, black] (0,0) -- (0,4.6) node[above, black!70] {log energy rate};
% pp curve (shallow slope)
\draw[acc, very thick] (0.5,0.9) .. controls (3.0,1.7) and (5.5,2.4) .. (8.5,3.0);
\node[acc, anchor=south] at (1.7,1.3) {pp chain};
% CNO curve (steep)
\draw[black, very thick, densely dashed] (2.6,0.2) .. controls (4.6,1.6) and (5.4,3.2) .. (6.6,4.4);
\node[black, anchor=east] at (6.4,4.1) {CNO cycle};
% crossover
\fill[black] (5.15,2.32) circle (2pt);
\draw[black, densely dotted] (5.15,0) -- (5.15,2.32);
\node[black, anchor=north] at (5.15,-0.05) {crossover};
% Sun marker
\draw[black, densely dotted] (4.2,0) -- (4.2,2.05);
\fill[acc] (4.2,2.05) circle (2pt);
\node[acc, anchor=south east] at (4.2,2.1) {Sun};
\end{tikzpicture}
$$

This crossover organizes the main sequence. Stars below about $1.2\,M_\odot$ have
central temperatures under the crossover and burn hydrogen by the pp chain with a
radiative core; stars above it run the CNO cycle, whose extreme temperature
sensitivity concentrates the energy generation in a small central region and drives
a **convective core**. The structural difference between the lower and upper main
sequence, developed in [the main sequence and its
structure](/astrophysics-cosmology/stellar-evolution/the-main-sequence-and-its-structure),
follows directly from which of these two networks supplies the star's luminosity.
The next lesson turns to the fuel that follows hydrogen, [helium and the
triple-alpha
process](/astrophysics-cosmology/nuclear-astrophysics/helium-burning-and-the-triple-alpha-process),
whose ignition requires the far higher temperatures the Gamow scaling demands for
charge-2 nuclei.

[^co-pp]: Carroll & Ostlie, §10.3 and Ch. 11 — the proton–proton chain and its branches, the CNO cycle, the branching ratios and neutrino losses, and the temperature dependence separating pp- and CNO-dominated stars.