Fusion conditions
Explain Coulomb repulsion, temperature, density and confinement, and distinguish fusion energy per event from reaction rate and nuclear power.
On this page
This optional supporting lesson separates the energy of one fusion reaction from the conditions needed for many reactions. First study Nuclear fusion to calculate its released energy Q. Positive Q does not mean that cold fuel will readily fuse.
Bring positively charged nuclei close enough
Two nuclei repel electrically because both are positively charged. This Coulomb repulsion is a barrier to close approach. The attractive strong nuclear force acts over nuclear distances, so the nuclei must approach very closely before a bound product can form.
High temperature means high typical kinetic energy of the nuclei. It changes the distribution of encounter energies; it does not give every nucleus the same energy. Quantum tunnelling also permits fusion when the collision energy is below the classical barrier. You do not need a tunnelling calculation here.
Guided practice 1
Why high temperature?
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.
Show solution step by step
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
Use thermal kinetic energy
Method
High temperature raises typical nuclear kinetic energies.
Reason
Higher encounter energies can increase the chance of sufficiently close approach and tunnelling through the repulsive barrier.
Working
higher T → higher typical kinetic energy
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
Temperature, density and confinement work together
For practical fusion, nuclei must encounter one another sufficiently often while energy losses are limited. Temperature affects encounter energies; density affects how many nuclei are available; confinement keeps the reacting plasma together long enough. These are connected conditions, not three interchangeable labels. See ITER’s explanation of fusion conditions.
| Quantity | Question it answers |
|---|---|
| Released energy Q | How much energy does one reaction release? |
| Reaction rate R | How many reactions occur per second? |
| Nuclear power P = RQ | How much nuclear energy is released per second? |
| Useful net output | How much useful energy remains after losses and energy supplied to the system? |
In P = RQ, use Q in joules per reaction and R in reactions per second to obtain watts. This is nuclear power, not a guarantee of net electrical output.
Use a stated model rather than a universal temperature rule
Heating a given plasma can improve fusion conditions over a suitable range. Do not infer that increasing temperature always increases reaction rate. A prediction needs the fuel, the relevant range and the encounter model.
Try it yourself 2
Reaction rate and energy per event
Independent transfer
Try this before viewing the solution
Hints
Hint 1: Separate rate from energy per event
Show solution step by step
Calculate rates
Method
The expected rates are 200 and 500 reactions per second.Reason
Multiply encounters per second by the stated probability of fusion per encounter.Working
R₁ = (1.0 × 10⁸)(2.0 × 10⁻⁶) = 200 s⁻¹, R₂ = 500 s⁻¹.Compare nuclear power
Method
Power increases by a factor of 2.5.Reason
The same energy is released per event, but there are more events per second.Working
P₂/P₁ = R₂/R₁ = 500/200 = 2.5Check the boundary
Method
Q stays at 17.6 MeV per reaction.Reason
It comes from the reaction’s mass difference, not the number of reactions. The probability trend was supplied for this limited model.Working
Q₂ = Q₁
Mind stretcher 1: How can stellar fusion occur below the classical barrier?Extension
Not every nuclear encounter in a star has enough kinetic energy to cross the repulsive barrier classically. How can fusion still occur?
Compare your explanation
The nuclei have a distribution of kinetic energies, including a high-energy tail. Quantum tunnelling also gives some encounters below the classical barrier a non-zero fusion probability. With many nuclei and continued confinement in the star, rare successful encounters can supply energy. This does not mean every collision fuses or that the Sun mainly uses the deuterium–tritium reaction studied in the energy example.
Try a different power comparison
Two illustrative conditions have reaction rates R and 3R. Each event releases the same Q. A learner says the second condition has three times the nuclear power and three times the energy per event. Which claim follows?
Check the distinction
Only the power claim follows: P₂ = 3RQ = 3P₁. Energy per event stays at Q. More successful events per second changes power without changing the rest-mass difference of the reaction.
Summary
Use binding energy or mass differences to find the energy of one event. Use a stated encounter model to predict rate. Temperature, density and confinement influence whether useful fusion power is possible; a positive reaction energy alone is insufficient.
Optional video
Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30