Fusion and Nucleosynthesis

Introduction

Every atom of carbon in your body, every atom of oxygen you breathe and every atom of gold in a wedding ring was made by nuclear reactions. Hydrogen and helium date from the first few minutes after the Big Bang, but almost everything heavier was forged inside stars or in violent events such as supernovae and neutron-star mergers. This page uses the binding energy per nucleon curve to explain why fusion powers stars, why stellar burning stops at iron, and how neutron capture builds the heavy elements.

Core explanation

Why fusion releases energy. The binding energy per nucleon rises steeply from hydrogen to a broad maximum near iron-56 and nickel-62 (about 8.8 MeV per nucleon). When two light nuclei fuse, the product lies higher on this curve, so its nucleons are more tightly bound. The extra binding appears as released energy, equal to the mass defect multiplied by c². For fusion of four protons into one helium-4 nucleus, the overall release is about 26.7 MeV, roughly 0.7% of the original rest mass.

The Coulomb barrier. Nuclei are positively charged, so they repel each other strongly until they are close enough (about 10⁻¹⁵ m) for the short-range strong force to take over. For two protons the barrier is of the order of 1 MeV, yet the average thermal energy in the Sun's core (about 1.5 × 10⁷ K) is only about 1–2 keV. Fusion happens because a tiny fraction of nuclei in the high-energy tail of the Maxwell–Boltzmann distribution can tunnel through the barrier — the same quantum effect that allows alpha decay. The product of the falling energy distribution and the rising tunnelling probability gives a narrow window of effective energies called the Gamow peak .

Stellar burning stages. A star is a sequence of fusion stages, each needing a higher temperature because heavier nuclei carry more charge:

Stage Main fuel Main products Approximate core temperature --- --- --- --- Hydrogen burning ¹H ⁴He 1–3 × 10⁷ K Helium burning ⁴He ¹²C, ¹⁶O 1–2 × 10⁸ K Carbon burning ¹²C ²⁰Ne, ²³Na, ²⁴Mg about 8 × 10⁸ K Neon and oxygen burning ²⁰Ne, ¹⁶O ²⁸Si, ³²S 1.5–2 × 10⁹ K Silicon burning ²⁸Si ⁵⁶Ni, then ⁵⁶Fe about 3 × 10⁹ K

Hydrogen burns by the proton–proton chain in Sun-like stars and by the CNO cycle in hotter, more massive stars, where carbon, nitrogen and oxygen nuclei act as catalysts. Helium burns by the triple-alpha process : two ⁴He nuclei form unstable ⁸Be, which can capture a third ⁴He to give ¹²C before it falls apart.

The iron wall. Beyond the iron peak, fusion absorbs energy rather than releasing it, so a stellar core of iron-group nuclei has no further fuel. In massive stars this leads to core collapse and a supernova.

Beyond iron: neutron capture. Neutrons feel no Coulomb barrier, so heavy elements are built mainly by neutron capture followed by beta-minus decay. In the s-process (slow), neutron capture takes years, so an unstable nucleus usually beta-decays before the next capture, and the path hugs the valley of stability. It operates in asymptotic giant branch stars. In the r-process (rapid), neutron densities are so high that dozens of neutrons are captured in seconds, driving nuclei far to the neutron-rich side; they then beta-decay back towards stability. Neutron-star mergers are a confirmed r-process site.

Formulae

Energy released: Q = (Σm(reactants) − Σm(products)) × c², with 1 u × c² = 931.5 MeV.

Coulomb barrier (estimate): E C ≈ Z₁Z₂e² / (4πε₀ r), where r is the sum of the nuclear radii and r ≈ r₀A^(1/3) with r₀ ≈ 1.2 fm.

Step-by-step reasoning

To decide whether a fusion reaction can power a star:

1. Write the balanced nuclear equation, conserving mass number and charge. 2. Calculate Q from the atomic masses; a positive Q means energy is released. 3. Estimate the Coulomb barrier from Z₁Z₂ — larger charges need higher temperatures. 4. Compare the product's position on the binding energy curve with the reactants'. 5. If the product lies beyond the iron peak, fusion is endothermic and cannot sustain the star.

Visual explanation

Picture the binding energy per nucleon curve as a hill with its summit at iron. Light nuclei climb the left slope by fusion; heavy nuclei slide down the right slope by fission. Stars march up the left slope stage by stage until they reach the summit, where no further climb is possible. Neutron capture is a separate ladder leaning against the right side of the hill.

Real-world analogy

Fusion is like trying to push two strong magnets together with like poles facing. Most pushes fail, but if you push hard enough and get close enough, a hidden latch (the strong force) snaps them together, and the latch releases far more energy than you spent pushing.

Real-world example

The Sun converts about 6 × 10¹¹ kg of hydrogen into helium every second, turning roughly 4 × 10⁹ kg of mass into energy each second. Experimental fusion devices on Earth, such as tokamaks, aim to fuse deuterium and tritium, which has the highest reaction rate at achievable temperatures and releases 17.6 MeV per reaction.

Why?

Why do stars need ever-higher temperatures for each new stage? The Coulomb barrier grows with the product Z₁Z₂. Carbon nuclei (Z = 6) repel each other about 36 times more strongly than protons do, so only a much hotter core provides enough nuclei in the energetic tail to tunnel through at a useful rate.

Common misconception

"All the elements up to uranium were made by fusion in ordinary stars." Fusion stops releasing energy at the iron peak. Nearly all elements heavier than iron were built by neutron capture in the s-process and r-process, not by charged-particle fusion.

Worked example

Question: Calculate the energy released in the fusion ²H + ³H → ⁴He + n. Atomic masses: ²H = 2.014102 u, ³H = 3.016049 u, ⁴He = 4.002603 u, n = 1.008665 u.

Reasoning: Mass of reactants = 2.014102 + 3.016049 = 5.030151 u. Mass of products = 4.002603 + 1.008665 = 5.011268 u. Mass defect = 0.018883 u. Q = 0.018883 × 931.5 MeV ≈ 17.6 MeV.

Answer: About 17.6 MeV per reaction, most of it carried by the neutron.

Quick check

1. Why can neutron capture build elements beyond iron when charged-particle fusion cannot? Answer: Neutrons are uncharged, so there is no Coulomb barrier to overcome, and capture does not rely on a positive fusion Q-value.

Exam focus

Be ready to use the binding energy per nucleon curve to explain why both fusion of light nuclei and fission of heavy nuclei release energy. Calculate Q-values from atomic masses using 931.5 MeV per u, and clearly contrast the s-process and r-process in terms of capture rate compared with beta-decay rate.

Advanced insight

The triple-alpha process works only because ¹²C has an excited state at about 7.65 MeV that sits just above the combined energy of ⁸Be + ⁴He, making the capture resonant. Fred Hoyle predicted this state from the observed abundance of carbon before it was measured. Abundance patterns also carry fingerprints: the s-process creates peaks near the magic neutron numbers 50, 82 and 126 at stable nuclei such as barium and lead, while the r-process creates shifted peaks at lighter masses, such as those near tellurium and platinum.

Summary

Fusion of light nuclei releases energy because the products are more tightly bound, up to the iron peak. Quantum tunnelling through the Coulomb barrier allows fusion at stellar temperatures. Stars burn hydrogen, helium, carbon, neon, oxygen and silicon in turn, each stage hotter than the last, ending at iron. Heavier elements form by neutron capture: slowly in the s-process and rapidly in the r-process.

Practice questions

1. State two ways in which hydrogen is converted into helium in stars. Answer: The proton–proton chain, dominant in Sun-like stars, and the CNO cycle, dominant in hotter, more massive stars. 2. Explain why silicon burning is the last energy-releasing stage in a massive star. Answer: It produces iron-group nuclei at the peak of the binding energy per nucleon curve; further fusion would absorb energy rather than release it. 3. Explain the role of the unstable nucleus ⁸Be in making carbon. Answer: Two ⁴He nuclei briefly form ⁸Be, which can capture a third ⁴He to form ¹²C before it decays back into two alpha particles. 4. In which process does the reaction path lie far from the valley of stability, and why? Answer: The r-process, because neutrons are captured much faster than beta decay can occur, so very neutron-rich nuclei build up before decaying.