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Lesson 1 of 4

Nuclear fission.

About 25 minutes · 6 practice questions

A uranium nucleus that tears in two: why a neutron splits it, where the energy comes from by E = mc², the multiplication factor k, control rods that hold a chain steady, and why a growing chain multiplies itself.

Step 1 of 6 · A nucleus that tears

Step 1 of 6

A nucleus that tears

Uranium is the heaviest element found in quantity in nature, and its nuclei are barely holding together: their 92 protons push on each other, held in only by the short-range strong force. In 1938 Otto Hahn and Fritz Strassmann found barium, a much lighter element, in uranium they had bombarded with neutrons. Lise Meitner and Otto Frisch worked out what had happened: the nucleus had split in two. They called it fission.

The Scene has a reactor core: the orange nuclei are fissile, like uranium-235, and the grey ones swallow a neutron without splitting, like uranium-238.

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Step 2 of 6

Where the energy comes from

The fragments after a split weigh about a tenth of a per cent less than the uranium nucleus and neutron that went in. The missing mass has become energy, by Einstein’s formula:

E=mc2E = mc^2⁠

The speed of light squared is an enormous number, so a little mass is a great deal of energy: about 200 MeV a split, millions of times more than any chemical reaction between two atoms.

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Step 3 of 6

The multiplication factor

The new neutrons can split more nuclei, and those release more neutrons. What matters is how many of the new neutrons, on average, cause another split: the multiplication factor kk⁠. Here each split frees 2.5 neutrons on average, and each must miss the Control rods and meet a fissile nucleus:

k=2.5×(fissile share)×(1−absorbed share)k = 2.5 \times (\text{fissile share}) \times (1 - \text{absorbed share})⁠

If k<1k < 1⁠, every chain dies out. If k=1k = 1⁠, each split leads to exactly one more, and the reaction holds steady. If k>1k > 1⁠, the number of splits grows with every generation. Fissile fuel and Control rods are fixed now.

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Step 4 of 6

A steady chain

Natural uranium is only about 0.7% uranium-235, so fuel is enriched, to a few per cent in most reactors. Control rods, of boron or cadmium, soak up neutrons: pushing them in lowers kk⁠, pulling them out raises it. Count the neutrons is On: the graph shows how many neutrons are in flight, on a log scale. The Fissile fuel is fixed at 90%.

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Step 5 of 6

Pulling the rods out

Fissile fuel at 50% and Control rods at 20% absorbed: k=2.5×0.5×0.8=1k = 2.5 \times 0.5 \times 0.8 = 1⁠. After you predict, Count the neutrons will switch On.

Try it: after answering, push Control rods in to 60% absorbed, and watch the chains die out.

Back to the experiment

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Step 6 of 6

Ten generations

A generation in a reactor lasts much less than a thousandth of a second. A real reactor can be controlled at human speed only because a small share of the neutrons, under one in a hundred, arrive seconds late from decaying fragments. Fissile fuel and Control rods are fixed: k=2.5×0.5×0.88=1.1k = 2.5 \times 0.5 \times 0.88 = 1.1⁠.

To the question

Still of the Nuclear fission 3D scene
Interactive scene

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Each step sets up the Scene for you, and a control marked “fixed” stays put for that step.

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Practice · Choose

Why is it a neutron, not a proton, that splits a uranium-235 nucleus?

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Practice · Calculate

When one kilogram of uranium-235 is split completely, the fragments together weigh about 0.9 g less. How much energy is released? The speed of light is 3 × 10⁸ m/s.

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Practice · Calculate

Each split frees 2.5 neutrons on average. The Fissile fuel is at 90%, and the Control rods absorb 20% of the neutrons. What is the multiplication factor kk⁠?

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Practice · Reach the target

The Fissile fuel is fixed at 90%. Push the Control rods in until the chain holds steady: k=1k = 1⁠.

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Practice · Predict, then look

The Fissile fuel is at 50% and the Control rods at 20% absorbed, so k=1k = 1⁠: the number of neutrons holds steady. The rods are now pulled all the way out, to 0%. Predict what will happen.

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Practice · Calculate

The Control rods absorb 12% of the neutrons, and k=1.1k = 1.1⁠. For every neutron now, how many will there be ten generations on?