Nuclear fusion

In fusion, two light nuclei, such as isotopes of hydrogen, join to make a heavier nucleus, releasing energy. Nuclei are positive and repel each other, so fusion only happens at the huge temperatures and pressures inside stars. Fusion of hydrogen into helium powers the Sun.

Key facts

  • Fusion: light nuclei (e.g. hydrogen) join to make heavier ones; it needs extreme temperature and pressure.
  • The Sun fuses hydrogen into helium in its core: it is not burning like a fire.
  • Mass numbers and atomic numbers must balance on both sides of a nuclear equation.
  • E = mc² (c = 3.0 × 10⁸ m/s); 1 u of mass ≡ 931.5 MeV; 1 MeV = 1.6 × 10⁻¹³ J.
  • Fusing light nuclei releases more energy per kilogram of fuel than fission does.

Worked examples

Example 1

What process releases energy in the Sun?

  1. The Sun is not burning like a fire, and it has no uranium fuel.
  2. In its core, small hydrogen nuclei join together.
  3. Hydrogen fuses into helium, releasing energy.

Answer: fusion of hydrogen nuclei into helium

Example 2

Why does fusion need extremely high temperatures?

  1. Both nuclei are positive, so they repel each other.
  2. They must get very close for the strong nuclear force to join them.
  3. Only very hot, fast-moving nuclei can get that close.

Answer: Nuclei must move fast enough to overcome their electrostatic repulsion

Example 3

In this fusion, ²₁H + ³₁H → X + ¹₀n. What is the mass number (top number) of X?

  1. Left side: mass numbers add to 5.
  2. Right side without X: 1.
  3. X: 5 − 1 = 4

Answer: 4

Example 4

In this fusion, ²₁H + ²₁H → X + ¹₀n. What is the atomic number (bottom number) of X?

  1. Left side: atomic numbers add to 2.
  2. Right side without X: 0.
  3. X: 2 − 0 = 2

Answer: 2

Example 5

In a fusion reaction the mass decreases by 0.0189 u. How much energy is released, in MeV and in J?

  1. Multiply the mass lost by 931.5 MeV per u. E = 0.0189 × 931.5 = 17.6 MeV
  2. Change MeV to joules. 17.6 × 1.6 × 10⁻¹³ = 2.8 × 10⁻¹² J

Answer: 17.6 MeV; 2.8 × 10⁻¹² J

Example 6

In the reaction four ¹H → ⁴He (overall, in the Sun), the mass lost is 0.02866 u. How much energy is released, in MeV, to 3 significant figures? (1 u = 931.5 MeV)

  1. The mass lost is 0.02866 u.
  2. Each u of mass is 931.5 MeV of energy.
  3. 0.02866 × 931.5 = 26.7 MeV

Answer: 26.7 MeV

Example 7

In the reaction ²H + ³H → ⁴He + ¹n (deuterium–tritium fusion), the mass lost is 0.01888 u. How much energy is released, in MeV, to 3 significant figures? (1 u = 931.5 MeV)

  1. The mass lost is 0.01888 u.
  2. Each u of mass is 931.5 MeV of energy.
  3. 0.01888 × 931.5 = 17.59 MeV

Answer: 17.6 MeV

Example 8

In the reaction ²H + ²H → ³He + ¹n, the mass lost is 0.0035 u. How much energy is released, in MeV, to 3 significant figures? (1 u = 931.5 MeV)

  1. The mass lost is 0.0035 u.
  2. Each u of mass is 931.5 MeV of energy.
  3. 0.0035 × 931.5 = 3.26 MeV

Answer: 3.26 MeV