Astronomy · How stars sustain their light

Nuclear fusion and stellar energy

Follow hydrogen-to-helium reaction bookkeeping, distinguish total released energy from local heating and connect energy per reaction to a conditional power.

Continue in the approved chapter order. Necessary vocabulary, quantities and assumptions are explained in the lesson.

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After reading, you can

Explain fusion energy using consistent rest-mass accounting.
Count the repeated stages and net conversion in the proton–proton I chain.
Separate energy release, escaping neutrinos and power under a stated reaction-rate model.

Account for the whole reaction system

An atomic nucleus contains protons and, except for ordinary hydrogen, neutrons. Isotopes of an element have the same proton count but different neutron counts. Nuclear fusion combines nuclei through a sequence of reactions; its energy budget is not the same as ordinary chemical bonding among atoms.

If the complete specified products have less total rest mass than the complete specified initial system, the difference can appear as released energy. The relation Q = Δmc² connects that mass difference with energy, where c is the speed of light. A consistent calculation includes all particles required by the reaction.

Atomic masses include electrons, while bare nuclear masses do not. Mixing those conventions can count an electron or its annihilation energy incorrectly. A quoted hydrogen-to-helium energy therefore needs a clear net reaction and treatment of emitted particles.

A positive Q means energy is available per completed reaction. It does not give the reaction rate. A process can release energy when it occurs yet proceed too slowly under particular conditions to supply a significant power.

The first two pp-I stages occur twice

In the proton–proton I, or pp-I, branch, two protons first form deuterium, a positron (the electron’s positively charged antiparticle) and an electron neutrino. Deuterium is a hydrogen nucleus containing one proton and one neutron. This reaction includes a weak-interaction conversion within the reacting system; it is not a claim that an isolated proton spontaneously decays in that way.

The deuterium nucleus then captures another proton to form helium-3 and a photon. Helium-3 has two protons and one neutron. Two helium-3 nuclei are needed for the final stage, so each of the first two reactions must occur twice per completed pp-I conversion.

The final helium-3 pair forms a helium-4 nucleus and returns two protons. Summing the three stages therefore consumes four protons overall and produces one helium-4 nucleus, two positrons and two electron neutrinos, together with released energy.

The positrons can annihilate with ambient electrons. When using the familiar rounded total of about 26.73 MeV for converting hydrogen to helium-4, include that annihilation accounting consistently, as in an atomic-mass comparison. Do not attach that total unchanged to a differently defined bare-particle mass balance.

Bookkeeping for one completed pp-I branch conversion. p denotes a proton, D a deuterium nucleus, e⁺ a positron and ν_e an electron neutrino; γ denotes a photon.
Multiplicity	Reaction	Bookkeeping role
2	p + p → D + e⁺ + ν_e	Make two deuterium nuclei
2	D + p → ³He + γ	Make two helium-3 nuclei
1	³He + ³He → ⁴He + 2p	Make helium-4 and return two protons

Escaping neutrinos carry part of the energy away

Neutrinos interact weakly with stellar matter and can carry energy out of the star rather than depositing it locally as heat. Charged particles and photons transfer much of their energy to the surrounding material. The total nuclear release and the local deposited energy are therefore different quantities.

Construct an original energy-accounting case using a rounded total release Q_total = 26.73 MeV per completed conversion. Stipulate a mean escaping-neutrino energy of 0.53 MeV per completion. Then deposited energy is 26.20 MeV. The stipulated loss is not an observed average over all branches operating in the Sun.

An electronvolt is exactly 1.602176634 × 10⁻¹⁹ J, so 1 MeV is 1.602176634 × 10⁻¹³ J. If the hypothetical completion rate is 10³⁸ s⁻¹, multiplying energy per completion by that rate gives the powers in the table.

The total is about 4.28262 × 10²⁶ W, split into about 4.19770 × 10²⁶ W deposited and 8.49154 × 10²⁴ W carried away by neutrinos. These are powers for supplied teaching inputs. They are not a new determination of the Sun’s luminosity or a complete stellar energy budget.

Original conditional nuclear-energy accounting at 10³⁸ completed conversions per second. Total 26.73 MeV is rounded; the 0.53 MeV neutrino loss is stipulated. Displayed powers are rounded.
Energy channel	MeV per completion	Power (W)
Total release	26.73	4.28262 × 10²⁶
Deposited locally	26.20	4.19770 × 10²⁶
Escaping neutrinos	0.53	8.49154 × 10²⁴

A pathway is not a universal ignition switch

Charged nuclei repel electrically. Stellar reaction rates depend on their thermal motions, quantum tunnelling, density, abundances and the relevant nuclear or weak interactions. Temperature changes can strongly affect a rate, but no single sharp temperature switch describes all reactions and compositions.

Other branches of the proton–proton network produce different intermediate nuclei and neutrino energy distributions. The pp-I bookkeeping table is a specific branch description; its stipulated heating example should not be treated as a measured mixture of every branch.

Carbon–nitrogen–oxygen, or CNO, cycles provide another hydrogen-to-helium route. Their nuclei participate as catalysts through a reaction network, while the net fuel conversion is hydrogen to helium with associated energy release and neutrinos. Intermediate abundances and the contribution of the cycles depend on the physical state and composition.

The Borexino collaboration’s 2020 report of CNO solar-neutrino evidence illustrates how reaction pathways can be tested observationally. The measurements and their interpretation have detector and model assumptions; the later interior-probe lesson examines why a neutrino signal adds information rather than simply reproducing a photograph of the core.

Thermal evolution connects nuclear heating with the escaping light

Deposited nuclear energy contributes to the material’s internal-energy budget. That energy can be transported, stored or involved in mechanical work as the structure changes. It is not an extra direct outward thrust to add to the hydrostatic force equation.

A sufficiently steady model can connect its integrated net energy source with the luminosity passing through its outer layers. During contraction, expansion or other evolution, changes of internal and gravitational energy must also be included. Hydrostatic balance alone does not assert thermal equilibrium.

Energy travelling through opaque stellar matter undergoes repeated interactions. Its transport is not described by following one named core photon unchanged all the way to the photosphere. The next lesson introduces the circumstances in which radiation or bulk fluid motion transports energy.

A convincing stellar-energy argument therefore separates the reaction network, the rate under specified conditions, neutrino losses, local heating, transport and structural evolution. Conservation of energy links those stages, while each stage requires its own physical description.

Worked examples

Sum repeated reactions before naming the net fuel

Why does one completed pp-I conversion consume four protons rather than six?

The first reaction occurs twice and consumes four protons.
The second reaction occurs twice and consumes two additional protons.
The final reaction returns two protons, so net consumption is 6 − 2 = 4.

Four protons are consumed overall. The two returned in the final stage cancel two of the six that entered earlier stages.

Convert a supplied reaction rate into deposited power

What deposited power follows from 26.20 MeV per completion at 10³⁸ completions per second?

Convert 26.20 MeV to 26.20 × 1.602176634 × 10⁻¹³ J per completion.
The deposited energy is 4.19770278108 × 10⁻¹² J per completion.
Multiply by 10³⁸ s⁻¹ to obtain 4.19770278108 × 10²⁶ W.

About 4.19770 × 10²⁶ W for the stipulated rate and energy partition, without treating it as a complete measured solar luminosity.

Try the reasoning

How many helium-3 nuclei must be supplied to the final reaction in one completed pp-I conversion?
Show the answer and reasoning
Two.
That is why each of the first two listed stages occurs twice.
With total release 26.73 MeV and stipulated escaping-neutrino energy 0.53 MeV, how much is deposited locally per completion?
Show the answer and reasoning
26.20 MeV.
Subtract the escaping energy from the total under the stated bookkeeping: 26.73 − 0.53 = 26.20.
Keep the energy per completion fixed but halve the hypothetical completion rate. What happens to nuclear power?
Show the answer and reasoning
It halves.
Power is energy per completion multiplied by completions per second. This rate scaling does not determine how a real star’s structure would adjust.
Does a positive reaction energy Q by itself show that the reaction is fast enough to support a star?
Show the answer and reasoning
No.
The rate also depends on the reacting particles, physical state and reaction physics.

What to carry forward

Use a complete and consistent mass convention when describing nuclear energy release.
The pp-I branch repeats its first two stages twice and consumes four protons overall.
Total release, deposited heating, neutrino losses and stellar luminosity are distinct parts of an energy argument.

References

Teaching resources and research supporting this lesson’s explanations and examples.