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Nuclear Fusion

Key idea: Explain nuclear fusion and why it releases energy for light nuclei using the binding energy per nucleon curve; describe the Coulomb barrier and conditions needed (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Relate binding energy per nucleon to fission, fusion, applications and hazards.

1. Definitions (Must Know)

A. Nuclear fusion

Nuclear fusion is the combining of two light nuclei to form a heavier nucleus.

B. Coulomb barrier

The Coulomb barrier is the electrostatic repulsion between two positively charged nuclei that must be overcome (or tunnelled through) for them to get close enough for the strong nuclear force to bind them.

C. Why energy can be released

Fusion can release energy because the product nucleus often has a higher binding energy per nucleon than the reactants (for light nuclei).

2. Key Ideas (What Earns Marks)

  • Use the binding energy per nucleon curve: light nuclei sit on the rising part, so fusion moves products towards higher E_b/A.
  • Fusion is difficult because nuclei repel electrically; high temperature (high kinetic energy) increases the chance of getting close enough.
  • Nuclear equations must conserve A and Z.
One-line explanation

“Energy is released because the fusion product has a higher binding energy per nucleon than the reactants.”

3. Detailed Explanations

For small A, the curve rises steeply. When two light nuclei fuse to form a heavier nucleus, the average binding energy per nucleon can increase, so total binding energy increases and the difference is released.

B. Why high temperature is needed

Higher temperature means higher typical kinetic energy, so a larger fraction of nuclei can get close enough to feel the attractive strong nuclear force (or to tunnel through the Coulomb barrier).

4. Common Mistakes

  • Saying fusion always releases energy (it releases energy mainly for light nuclei moving toward the peak near iron).
  • Forgetting the role of Coulomb repulsion.
  • Writing unbalanced nuclear equations (check A and Z).

5. Exam Tips

  • For “why energy released?”: binding energy per nucleon argument.
  • For “why difficult?”: Coulomb repulsion + need for very high temperature/pressure.
  • Mention where energy goes: kinetic energy of products and radiation.

6. Worked Examples

Modelled example 1

Balancing a fusion equation

Core

Problem

Complete ²₁H + ³₁H → ⁴₂He + X.
Study the worked solution

  1. Conserve nucleon number

    Method

    The missing particle has A = 1.

    Reason

    The reactants contain five nucleons and helium-4 accounts for four.

    Working

    2 + 3 = 4 + A_X ⇒ A_X = 1

  2. Conserve charge

    Method

    The missing particle has Z = 0.

    Reason

    The reactant charge total and helium product charge are both two.

    Working

    1 + 1 = 2 + Z_X ⇒ Z_X = 0

  3. Identify the particle

    Method

    X is a neutron, ¹₀n.

    Reason

    A particle with A = 1 and Z = 0 is a neutron.

    Working

    ²₁H + ³₁H → ⁴₂He + ¹₀n

Guided practice 2

Why high temperature?

About 4 min

Problem

Why do fusion reactors require extremely high temperatures?

Try this before viewing the solution

Hints

Hint 1: identify the barrier
Both nuclei are positively charged.
Hint 2: reach the short-range force
Explain how increased kinetic energy helps nuclei approach the range of the strong nuclear force.
View solution step by step

  1. Identify Coulomb repulsion

    Method

    The positively charged nuclei repel electrostatically.

    Reason

    They must approach very closely before the short-range strong nuclear force can bind them.

    Working

    positive nucleus ↔ positive nucleus: repulsion

  2. Use thermal kinetic energy

    Method

    High temperature raises typical nuclear kinetic energies.

    Reason

    A larger fraction of encounters can approach the Coulomb barrier closely enough for fusion or tunnelling.

    Working

    higher T → higher typical kinetic energy

  3. Complete the force argument

    Method

    At sufficiently small separation, the attractive strong nuclear force can form a bound product.

    Reason

    The strong force is effective only over nuclear distances.

    Working

    close approach → strong-force binding

Common misconception 3

Energy release reasoning from E_b/A

Find and correct the mistake

Learner claim

Deuterium has E_b/A ≈ 1.1 MeV and helium-4 has E_b/A ≈ 7.1 MeV. A learner says forming helium must absorb energy because its binding energy is larger. Diagnose the claim.

Try this before viewing the solution

Fusion energy outcome

View solution step by step

  1. Compare average binding

    Method

    Helium-4’s nucleons are much more tightly bound.

    Reason

    Its binding energy per nucleon is higher.

    Working

    7.1 MeV > 1.1 MeV

  2. Interpret mass-energy

    Method

    The more tightly bound product has lower total mass-energy than the separated reactants.

    Reason

    Binding energy is the energy required to separate the product, not extra positive energy stored in it.

    Working

    greater total binding → lower bound-system mass

  3. State the release

    Method

    The mass-energy difference is released as product kinetic energy and radiation.

    Reason

    Fusion moves these light nuclei upward toward the binding-curve peak.

    Working

    reactants → bound helium + released energy

Examiner practice 4

Energy released from mass defect (D–T fusion, given masses)

4 marks

Examination question

For ²₁H + ³₁H → ⁴₂He + ¹₀n, the masses are 2.014 u, 3.016 u, 4.002 u and 1.009 u, respectively. Find the energy released using 1u c² = 931 MeV. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Total reactant mass

    1 mark

    Method

    mᵣ = 5.030 u.

    Reason

    Add the deuterium and tritium masses.

    Working

    mᵣ = 2.014 + 3.016 = 5.030 u

  2. Total product mass

    1 mark

    Method

    mₚ = 5.011 u.

    Reason

    Add the helium-4 and neutron masses.

    Working

    mₚ = 4.002 + 1.009 = 5.011 u

  3. Find mass decrease

    1 mark

    Method

    Δ m = 0.019 u.

    Reason

    Released energy corresponds to reactant mass minus product mass.

    Working

    Δ m = 5.030-5.011 = 0.019 u

  4. Convert to energy

    1 mark

    Method

    E = 17.7 MeV.

    Reason

    Multiply the mass decrease by its energy equivalent.

    Working

    E = 0.019(931) = 17.7 MeV

Challenge 5

Fusion power and reaction rate (qualitative)

Minimal support

Independent transfer

In a fusion plasma, temperature increases while density stays the same. Predict the qualitative change in fusion reaction rate and explain it using the collision-energy distribution and Coulomb barrier.

Try this before viewing the solution

Hints

Hint 1: focus on the energetic fraction
The number density is fixed; consider how temperature changes the fraction of nuclei with useful encounter energies.
View solution step by step

  1. Predict the rate

    Method

    The fusion reaction rate increases qualitatively.

    Reason

    The same density does not mean the same distribution of collision energies.

    Working

    higher T at fixed density → higher rate

  2. Explain the energy distribution

    Method

    Higher temperature raises average kinetic energy and enlarges the high-energy fraction.

    Reason

    More nuclear encounters reach small separations relevant to the barrier.

    Working

    more energetic encounters per unit time

  3. Connect to fusion probability

    Method

    More encounters can overcome the Coulomb barrier classically or have greater tunnelling probability.

    Reason

    That increases the chance of strong-force capture and fusion.

    Working

    closer approach → greater fusion probability

7. Mind Stretchers

Mind stretcher 1: Fusion vs fission energy per unit massExtension

Fusion releases less energy per reaction than fission for many common examples, but why is it still attractive as an energy source?

Show Answer

Because fuel can be abundant and the products can be less long-lived radioactive compared to fission waste (depending on the fuel cycle). Also, energy density can still be extremely high compared to chemical fuels.

Mind stretcher 2: Why do stars fuse at “only” millions of kelvin?Extension

Explain how fusion can occur in the Sun even though not all collisions have enough energy to overcome the Coulomb barrier classically.

Show Answer

Not all nuclei need enough classical energy to climb over the Coulomb barrier because quantum tunnelling allows some nuclei to penetrate the barrier.

Also, in a very large population of particles, the high-energy tail of the distribution means a small fraction have unusually high energies, contributing to fusion.

8. Optional (Enrichment)

A. Video (optional)

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027