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Fission and fusion

A slow neutron can split the heaviest nuclei, and temperatures of millions of kelvin can fuse the lightest. Both routes move the products towards iron, both leave them lighter than the starting materials, and the mass difference, converted by E = mc², is the energy released.

Year 13AQA 3.8.1.6, 3.8.1.7 CIE 23.1.2, 23.1.5, 23.1.6, 23.1.7 OCR A 6.4.4(b), 6.4.4(g), 6.4.4(j), 6.4.4(k)EDEXCEL 164, 166, 167, 170

Builds on Mass-energy and binding energy and Radioactive decay and half-life.

IN THIS TOPIC

Describe induced fission by thermal neutrons and balance a fission equation.
Explain the chain reaction and the meaning of critical mass.
Calculate the energy released in fission and fusion reactions from nuclear masses.

COMMON MISCONCEPTION

Splitting any atom releases energy.

Splitting the heavyweight

Uranium-235 will not usually split on its own, but offer it a thermal neutron, one moving at everyday molecular speeds, and it captures it readily, becoming a violently excited uranium-236 that deforms and tears in two. The binding-energy curve from the last lesson sets the limits here. Only nuclei on the heavy side of iron sit low enough for their fragments to end up better bound, so only they release energy by splitting. Split a light nucleus and you would have to pay for the privilege.

FIG. 1Induced fission: a slow neutron in, two mid-mass fragments out, plus three fast neutrons and about 200 MeV.

A typical split turns uranium-235 plus a neutron into barium-141 and krypton-92 plus three neutrons. Check the books both ways. Proton number gives 92 = 56 + 36, and nucleon number gives 236 = 141 + 92 + 3. Around 200 MeV comes out per fission, most of it as kinetic energy of the fragments. Those fragments carry too many neutrons for their new size and sit above the stable band, so they decay by β⁻ in chains, a fact whose consequences fill the next lesson.

Energy from the masses

The energy bookkeeping follows one recipe every time. Total nuclear mass before, minus total after, times 931.5. The products of any release reaction weigh less than the ingredients, and the difference leaves as kinetic energy of the products. For fission the sum uses the fuel nucleus, the incoming neutron, both fragments and the freed neutrons. The question supplies the masses, and the method never changes.

GUIDED PRACTICE

A fission yield in familiar units

A uranium-235 fission event has a total mass difference of 0.186 u. Find the energy released in MeV and in joules.

Show the working

Multiply by 931.5, and 0.186 × 931.5 = 173 MeV.

In SI that becomes 173 × 10⁶ × 1.60 × 10⁻¹⁹ = 2.8 × 10⁻¹¹ J. Tiny per event, but the events come by the countless billion, as the next example shows.

The chain and the critical mass

Each fission's spare neutrons can induce further fissions, which is the door to a chain reaction. Left alone in a large enough mass of fuel, one fission becomes three, becomes nine, growing by powers. Held so that exactly one neutron per fission goes on to cause another, the chain ticks over at a steady rate, which is a reactor's whole art.

FIG. 2A schematic example, not a census: real fissions free two or three neutrons, 2.4 on average for uranium-235. Left, uncontrolled: in this cartoon each fission frees three and each finds a nucleus, so one becomes three becomes nine becomes twenty-seven, the frame filling faster each generation. Right, controlled: absorbers eat two of every three, holding the multiplication factor k_eff at about one, steady power instead of a bomb; in a real core that balance also counts neutrons lost to leakage and to capture without fission.

Whether a chain can sustain itself at all is a question of size. In a small lump, too many neutrons reach the surface and escape before meeting a nucleus; the chain fizzles. The critical mass is the minimum amount of fissile material in which, on average, one neutron per fission is retained to fission again, and it is not a constant of the material alone: shape, density, enrichment and any surrounding reflector all move it. Below it a chain is impossible; at and above it, a chain can be sustained.

WORKED EXAMPLE

Counting a power station's fissions

Each fission releases about 200 MeV. How many fissions per second sustain a reactor producing 1.0 GW of thermal power?

One fission in joules is 200 × 10⁶ × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹¹ J.

Rate = power/energy per event = 1.0 × 10⁹/(3.2 × 10⁻¹¹) = 3.1 × 10¹⁹ fissions per second.

Thirty billion billion splittings a second, every one of them microscopic. Nuclear power is small numbers multiplied by Avogadro-scale counts, the recurring arithmetic of this whole unit.

Fusion, the harder prize

At the curve's other end, light nuclei release energy by fusion, and the steep left slope makes each fused nucleon worth more than a fissioned one, so kilogram for kilogram fusion yields several times more. The standard reaction fuses deuterium and tritium into helium-4 plus a neutron, releasing 17.6 MeV.

FIG. 3Deuterium and tritium fuse to helium-4 and a neutron: 0.0189 u vanishes, worth 17.6 MeV.

The catch is the doorstep. Two nuclei have to come within a few femtometres for the strong force to act, and both of them are positive, so this is the closest-approach problem from earlier in the unit, now working against us. Temperatures of order 10⁷ K raise the collision energy and so cut the distance of closest approach, and the crushing pressure keeps the density, and with it the collision rate, enormous. That much is the answer an exam question is after.

The arithmetic does not close there, though, and it is worth seeing why. At 1.5 × 10⁷ K the average nucleus carries only about 2 keV, while the electrical hill between two protons runs to several hundred keV. Practically nothing in a star has the energy to get over it, so a purely classical treatment predicts a fusion rate hopelessly below what stars actually deliver. What rescues it is quantum tunnelling, the same barrier-penetration effect that lets alpha particles leave a nucleus.

So the full picture combines the two. The fastest nuclei in the tail of the speed distribution get close, and tunnelling carries them the rest of the way through a barrier they were never going to climb. Stars manage it, but not easily: the Sun's core releases only a few hundred watts per cubic metre, and it is sheer size that makes the total colossal. Building a reactor that sustains it remains one of engineering's great open problems, and this physics is what makes both the promise and the difficulty concrete.

ASSESSMENT FOCUS

Thermal means slow. Neutrons at speeds comparable to ordinary molecular motion, which uranium-235 captures far more readily than fast ones. That single word carries the mark.
Balance fission equations twice, nucleon numbers along the top and proton numbers along the bottom, and count the freed neutrons; two or three per fission is the expected answer.
Energy calculations are before minus after, in u, times 931.5 for MeV. Show the subtraction explicitly, since sign errors here are the commonest slip.
Critical mass answers need the surface argument. Below it, too many neutrons escape through the surface before they cause a fission, so the chain cannot sustain itself.
For why fusion needs extreme temperature, name the electrostatic repulsion between positive nuclei and the need to approach within strong-force range. That is the marked answer. Do not write that the nuclei have enough energy to get right over the barrier, because at stellar temperatures they do not, and tunnelling does the rest.
E = mc² is the only equation the booklet prints for this topic, with 931.5 MeV per u sitting among the constants. Everything else here is method, so the fission and fusion recipes have to come from memory.

CHECK YOURSELF

Deuterium (2.01355 u) and tritium (3.01550 u) fuse to form helium-4 (4.00150 u) and a neutron (1.00867 u), masses given as nuclear masses. Find the energy released.

Show a hint

Total mass before, total after, difference times 931.5.

Show the answer

Before, 2.01355 + 3.01550 = 5.02905 u. After, 4.00150 + 1.00867 = 5.01017 u.

Δm = 5.02905 − 5.01017 = 0.01888 u.

E = 0.01888 × 931.5 = 17.6 MeV, carried off as kinetic energy of the helium nucleus and, mostly, the neutron.

Both roads lead to iron, with heavy nuclei splitting and light nuclei fusing.

Sum the masses before and after; the missing u, times 931.5, is the MeV set free.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the fission and fusion questions page.

7 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

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Describe induced fission by thermal neutrons and balance a fission equation.
Explain the chain reaction and the meaning of critical mass.
Calculate the energy released in fission and fusion reactions from nuclear masses.

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