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Fission and fusion

A slow neutron can split the heaviest nuclei; a hundred million kelvin can weld the lightest. Both routes end nearer iron, both leave the products lighter than the ingredients, and the missing mass, times 931.5, is the energy that lights cities and stars.

Year 13AQA 3.8.1.6, 3.8.1.7CIE 23.1OCR A 6.4.4

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.

WHAT YOU PROBABLY THINK

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 lie above dies on the curve from last lesson: only nuclei on the heavy side of iron sit low enough for their fragments to be better bound, so only they release energy by splitting. Fission a light nucleus and you would have to pay.

Induced fission: a slow neutron is captured by uranium-235 and the excited nucleus splits into two mid-mass fragments and three fast neutronsslow nuranium-235barium-141krypton-923 fastneutronscharge and nucleon number balance across the splitabout 200 MeV a fission, mostly fragment kinetic energy
FIG. 1Induced fission: a slow neutron in, two mid-mass fragments out, plus three fast neutrons and about 200 MeV.

A typical split is uranium-235 plus a neutron becoming barium-141 and krypton-92 plus three neutrons: check the books, 92 = 56 + 36 for proton number and 236 = 141 + 92 + 3 for nucleon number. Around 200 MeV is released per fission, most of it as fragment kinetic energy. The fragments carry too many neutrons for their new size, sitting 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 is the same 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 sums use the masses of the fuel nucleus, the incoming neutron, both fragments and the freed neutrons; the question supplies the masses, the method never changes.

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.

the chain reaction, wild and tamedwild: threefold each generationtamed: two of three absorbedhydrogencarbonuranium
FIG. 2Left, uncontrolled: each fission frees three neutrons and each finds a nucleus, so one becomes three becomes nine becomes twenty-seven, the frame filling faster each generation. Right, controlled: absorber rods eat two of every three, and the chain ticks over at exactly one, steady power instead of a bomb. Below, why a moderator is made of light atoms: a head-on neutron stops dead on hydrogen, bounces back off carbon a little slower, and rebounds off uranium with almost everything it arrived with.

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. Below it a chain is impossible; at and above it, a chain can be sustained.

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.

Deuterium and tritium fuse into helium-4 and a neutron: the products weigh less, and the missing mass leaves as 17.6 MeVdeuteriumtritiumfusehelium-4neutron17.6 MeVreleased0.0189 u vanishes from the ledgerthe climb up the curve's steep left side
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 must touch for the strong force to act, and both are positive: the closest approach problem from earlier in the unit, now working against us. Only at temperatures of order 107 K and enormous pressures do nuclei arrive fast enough to climb the electrical hill, which is why fusion powers stars and why building a reactor that sustains it remains one of engineering's great open problems. Knowing this physics is what lets societies weigh their energy choices with open eyes.

THE EXAM BIT

  • Thermal means slow: neutrons at speeds comparable to 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; sign errors here are the commonest slip.
  • Critical mass answers need the surface argument: below it, too many neutrons escape through the surface before causing 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.

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: heavy nuclei split, light nuclei fuse.

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

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.