8 min readPathway sheet1-mark MCQ, 2-mark fission versus fusion, 3-mark energy calculation; also a NEET favourite

Nuclear fusion

In fusion, light nuclei join to form a heavier nucleus with higher binding energy per nucleon. It needs temperatures of millions of kelvin and powers the Sun.

Example reaction²₁H + ²₁H → ³₂He + ¹₀n + 3.27 MeV

ConditionVery high temperature (about 1.5 × 10⁷ K in the Sun's core), so nuclei can overcome their electric repulsion: thermonuclear fusion

Energy of the Sun4 ¹₁H + 2e⁻ → ⁴₂He + 2ν + 6γ + 26.7 MeV (proton-proton cycle)

Energy per nucleonFusion gives more energy per kilogram of fuel than fission

Why fusion needs a very high temperature

Both nuclei in a fusion reaction are positively charged, so they repel each other. To fuse, they must come within a few femtometres, the range of the nuclear force. Only nuclei moving very fast can push through this electric repulsion. Such speeds come from very high temperatures, millions of kelvin. So fusion driven by heat is called thermonuclear fusion. The core of the Sun is at about 1.5 × 10⁷ K. This is below the temperature an average proton would need, so fusion there comes from the few protons moving much faster than average.

Some fusion reactions

Check each one: mass numbers and charges balance.

1
¹₁H + ¹₁H → ²₁H + e⁺ + ν + 0.42 MeV
Mass numbers 2 = 2; charges 2 = 1 + 1. e⁺ is a positron and ν a neutrino.
2
²₁H + ²₁H → ³₂He + ¹₀n + 3.27 MeV
Mass numbers 4 = 3 + 1; charges 2 = 2 + 0.
3
²₁H + ²₁H → ³₁H + ¹₁H + 4.03 MeV
Mass numbers 4 = 3 + 1; charges 2 = 1 + 1.

Energy in the Sun: the proton-proton cycle

Overall, four hydrogen nuclei make one helium-4 nucleus and release 26.7 MeV.

Worked example: fusing two deuterium nuclei into helium-4

Problem

Find the energy released when two ²₁H nuclei fuse to form ⁴₂He. (Atomic masses: ²₁H = 2.0141 u, ⁴₂He = 4.0026 u; 1 u = 931.5 MeV)

Working

Mass before = 2 × 2.0141 u = 4.0282 u.
Mass after = 4.0026 u.
Mass lost ΔM = 4.0282 − 4.0026 = 0.0256 u.
Energy = 0.0256 × 931.5 MeV ≈ 23.8 MeV.
Per nucleon: 23.8 ÷ 4 ≈ 6 MeV, compared with about 200 ÷ 236 ≈ 0.85 MeV per nucleon in uranium fission.

Answer

About 23.8 MeV, roughly 6 MeV per nucleon

Fission and fusion

Nuclear fissionA heavy nucleus splits into two medium nuclei. Started by a slow neutron; works at ordinary temperatures. About 200 MeV per event, about 0.85 MeV per nucleon. Used in nuclear power stations.

Nuclear fusionLight nuclei join into a heavier one. Needs millions of kelvin. More energy per nucleon (about 6 MeV for two ²H forming ⁴He). Powers the Sun and stars. Controlled fusion for electricity is still being developed.

In real life

Every bit of sunlight that grows our rice, wheat and sugarcane began as fusion energy in the Sun's core. India is also a partner in ITER, an international project in France to build an experimental fusion reactor.

Exam angle

2 marks: “Distinguish between nuclear fission and nuclear fusion (any two points).” Fission: a heavy nucleus splits; needs a slow neutron. Fusion: light nuclei join; needs very high temperature. Fusion gives more energy per nucleon.

3 marks: “Calculate the energy released when two deuterium nuclei fuse to form helium-4. (²₁H = 2.0141 u, ⁴₂He = 4.0026 u, 1 u = 931.5 MeV)” ΔM = 2 × 2.0141 − 4.0026 = 0.0256 u. Energy = 0.0256 × 931.5 ≈ 23.8 MeV.

1 marks: “Why is a very high temperature needed for nuclear fusion?” To give the nuclei enough kinetic energy to overcome their electrostatic repulsion and come within range of the nuclear force.

Remember it

Fission splits big, fusion joins small; both climb towards iron.