Nuclear Fusion and Stellar Reactions

Nuclear fusion is the joining of light nuclei into a heavier one with the release of energy, and it powers the Sun and the stars: in the Sun, four hydrogen nuclei are converted to one helium-4 nucleus, releasing 26.7 MeV. In brief: the deuterium–tritium reaction, 2H + 3H → 4He + n + 17.6 MeV, releases about 340 TJ per kilogram of fuel, roughly four times as much as fission per kilogram.

Fusion is the source of nearly all of the energy that reaches the Earth, and it is the process in which almost all elements heavier than hydrogen and helium were made. It is also the goal of research for a nearly limitless energy source on Earth. The difficulty is that nuclei repel each other, and fusion needs temperatures of tens of millions of degrees. This article explains how fusion releases energy, describes the reactions in stars and in the laboratory, shows the calculations and gives the status of fusion research.

Why does fusion release energy?

The binding energy per nucleon rises from 1.1 MeV for deuterium to 7.07 MeV for helium-4, and reaches a maximum near iron. When light nuclei fuse into a more tightly bound nucleus, the difference in binding energy is released, for example 28.3 − (2 × 2.22) = 23.9 MeV for two deuterons forming helium-4. The released energy appears as the kinetic energy of the products. Fusion of nuclei lighter than iron releases energy, and fusion of heavier ones consumes it.

Why is fusion difficult?

Nuclei are positively charged and repel each other. To fuse, they must come within about 1–2 fm, where the strong force takes over. The Coulomb barrier for two protons is about 0.5–1 MeV, and for deuterium and tritium about 0.4 MeV. The thermal energy of a nucleus is (3/2)kT, which corresponds to only 1.3 keV at a temperature of 107 K, the temperature at the center of the Sun, much less than the barrier. Fusion takes place in the Sun because of two effects: a small fraction of the nuclei have energies much higher than the average (the tail of the Maxwell–Boltzmann distribution), and nuclei can penetrate the barrier by quantum tunneling. In the Sun the reaction is very slow: the average proton waits about 1010 years before it fuses, which is why the Sun lives so long.

On Earth, to get a useful rate in a plasma, the temperature must be about 100–150 million K for deuterium–tritium, and the plasma must be held long enough and dense enough: the Lawson criterion requires a triple product of density, temperature and confinement time of about 3 × 1021 keV·s/m3.

How does the Sun produce energy?

The Sun, and stars of similar or smaller mass, fuse hydrogen to helium mainly through the proton–proton chain:

  1. 1H + 1H → 2H + e+ + νe (very slow, as it involves the weak interaction)
  2. 2H + 1H → 3He + γ
  3. 3He + 3He → 4He + 2 1H

The net result is 4 1H → 4He + 2e+ + 2νe + energy. The positrons annihilate with electrons. The total energy released is 26.7 MeV per helium nucleus, of which about 0.5 MeV is carried off by neutrinos. In stars more massive than about 1.3 solar masses, the CNO cycle, in which carbon, nitrogen and oxygen act as catalysts, dominates. The Sun’s power output is 3.83 × 1026 W.

Important fusion reactions

Fusion reactions and the energy released
Reaction Energy released (MeV) Where it occurs
4 1H → 4He + 2e+ + 2ν (proton–proton chain)26.7The Sun
2H + 3H → 4He + n17.6Fusion reactors (ITER, NIF)
2H + 2H → 3He + n3.27Experiments, deuterium plasmas
2H + 2H → 3H + 1H4.03Experiments
2H + 3He → 4He + 1H18.3Future “aneutronic” concepts
3 4He → 12C (triple-alpha process)7.27Red giant stars

How are the elements made in stars?

1 1 H + 1 1 H → 2 1 H + e+​ + νe​ 2 1 H + 1 1 H → 3 2 He + γ 3 2 He + 3 2 He → 4 2 He + 2 1 1 H

The carbon, oxygen, iron and calcium in our bodies were made in stars: we are made of “stardust”.

How is fusion done on Earth?

The advantages of fusion are abundant fuel, no carbon dioxide and no risk of a runaway chain reaction, and less long-lived waste than fission, although neutrons make the reactor structure radioactive. The difficulties are materials that can withstand the neutrons and the heat, tritium handling and the cost. A commercial fusion power plant is not expected before the 2040s or later.

Worked examples

Example 1: energy of the net hydrogen-burning reaction

Using atomic masses, 4 × 1.007825 = 4.031300 u (four hydrogen atoms) and m(4He) = 4.002603 u. Δm = 4.031300 − 4.002603 = 0.028697 u. E = 0.028697 × 931.494 = 26.73 MeV. The mass fraction that is converted is 0.028697 / 4.0313 = 0.71 %.

4 × 1.007825 = 4.031300 u → 4 2 He = 4.002603 u Δm = 0.028697 u × 931.494 = 26.73 MeV

Example 2: the mass the Sun converts into energy

The power is L = 3.83 × 1026 W, so the mass converted per second is m = L/c2 = 3.83 × 1026 / (3.00 × 108)2 = 4.26 × 109 kg/s, about 4.3 million tonnes per second. Since 0.71 % of the hydrogen mass is converted, the Sun fuses 4.26 × 109 / 0.0071 = 6.0 × 1011 kg, 600 million tonnes of hydrogen, every second. The Sun has a mass of 2.0 × 1030 kg and is not in danger of running out: it has used up about half of the hydrogen in its core in 4.6 billion years.

Example 3: energy of 1 kg of deuterium–tritium fuel

One D–T reaction releases 17.6 MeV = 2.82 × 10−12 J and consumes 5.03 u = 8.35 × 10−27 kg of fuel. The energy per kilogram is 2.82 × 10−12 / 8.35 × 10−27 = 3.4 × 1014 J/kg. This is about four times the value for uranium-235 fission (8.2 × 1013 J/kg), and about ten million times the energy from coal (2.9 × 107 J/kg).

2 1 H + 3 1 H → 4 2 He + 1 0 n + 17.6 MeV

Example 4: the energy of the products

In D–T fusion, 17.6 MeV is shared between the helium nucleus (mass 4) and the neutron (mass 1) according to conservation of momentum, in the inverse ratio of their masses: the neutron gets 4/5 × 17.6 = 14.1 MeV and the alpha particle 3.5 MeV. The alpha particles heat the plasma, and the neutrons leave and heat the blanket, and breed tritium.

4 2 He 3.5 MeV    1 0 n 14.1 MeV 3.5 + 14.1 = 17.6 MeV

Example 5: the thermal energy at the center of the Sun

At 1.5 × 107 K, the mean thermal energy is (3/2)kT = 1.5 × 1.381 × 10−23 × 1.5 × 107 = 3.1 × 10−16 J = 1.9 keV. This is only about 0.2 % of the Coulomb barrier for two protons (about 1 MeV), and shows why tunneling and the high-energy tail of the distribution are needed.

How was fusion discovered?

Arthur Eddington suggested in 1920 that stars are powered by the fusion of hydrogen into helium, using the measurements by Aston showing that helium is lighter than four hydrogen atoms. George Gamow showed in 1928 how tunneling makes it possible, and Hans Bethe worked out the proton–proton chain and the CNO cycle in 1938–1939, for which he received the Nobel Prize in Physics in 1967. The first thermonuclear explosion, a hydrogen bomb, took place in 1952. The first tokamak was built in the Soviet Union in the 1950s, and Margaret Burbidge, Geoffrey Burbidge, William Fowler and Fred Hoyle explained the formation of the elements in stars in a classic paper of 1957.

Which mistakes are common?