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E.4

Fission

Theme E · Nuclear and quantum physics · SL and HL

🎯What you need to be able to do

  • Use mass defect and mass–energy equivalence to find binding energy.
  • Interpret the binding energy per nucleon curve and use it to predict energy release.
  • Describe induced fission and the chain reaction, including critical mass.
  • Describe the components of a nuclear reactor and the function of each.

📚The physics

Mass defect. Weigh a nucleus and it comes out lighter than the sum of its separate nucleons. The missing mass \(\Delta m\) is the mass defect, and it corresponds to the energy released when the nucleus formed:

\[ E = \Delta m\,c^{2} \]

Turned round, that same energy — the binding energy — is what you would have to supply to pull the nucleus completely apart. Mass and energy are not being converted into one another so much as being two accounts of the same thing.

The convenient unit is the unified atomic mass unit, u, where 1 u corresponds to 931.5 MeV. Working in u and MeV avoids a great deal of scientific notation.

The binding energy per nucleon curve is the single most useful graph in nuclear physics, and it is worth being able to sketch from memory. It climbs steeply from hydrogen, peaks at about 8.8 MeV per nucleon around \( A = 56 \) — iron and nickel — then declines slowly to about 7.6 MeV per nucleon at uranium.

Everything about nuclear energy follows from that shape. Higher on the curve means more tightly bound, which means more stable. So energy is released by any process that moves nucleons towards the peak: fusing light nuclei from the left, or splitting heavy nuclei from the right. Iron sits at the top, which is why it is the end of the line for stellar fusion — a point that returns in E.5.

Induced fission. A slow neutron absorbed by uranium-235 makes it unstable enough to split into two lighter fragments, releasing typically two or three further neutrons and around 200 MeV. Those neutrons can induce further fissions — a chain reaction. Whether it sustains depends on whether, on average, at least one neutron from each fission goes on to cause another. The critical mass is the minimum mass for which that happens; below it, too many neutrons escape the surface before being absorbed.

A reactor is a chain reaction held at exactly self-sustaining. Four components, each with one job:

  • Fuel — uranium enriched to a few per cent U-235, since natural uranium is over 99% U-238.
  • Moderator — water or graphite, which slows the fast neutrons produced by fission down to the low speeds at which U-235 absorbs them efficiently. Without a moderator the chain reaction stalls.
  • Control rods — boron or cadmium, which absorb neutrons. Pushing them in reduces the reaction rate; pulling them out increases it.
  • Coolant — carries the thermal energy away to raise steam, which drives a turbine. From that point on it is an ordinary heat engine, subject to the Carnot limit from B.4.

The waste problem is real and worth stating plainly. Fission fragments are neutron-rich and therefore radioactive, some with half-lives of thousands of years. Against that, fission produces no carbon dioxide during operation and the fuel is extraordinarily energy-dense. Both halves belong in a balanced answer.

✏️Worked example

Calculate the binding energy per nucleon of helium-4. Take the mass of a proton as 1.00728 u, a neutron as 1.00867 u, and the helium-4 nucleus as 4.00151 u, with 1 u = 931.5 MeV.

Total mass of the separate nucleons: \( 2 \times 1.00728 + 2 \times 1.00867 = 2.01456 + 2.01734 = 4.03190 \) u.

Mass defect: \( \Delta m = 4.03190 - 4.00151 = 0.03039 \) u.

Binding energy: \( 0.03039 \times 931.5 = 28.3 \) MeV.

\[ \frac{E_B}{A} = \frac{28.3}{4} = 7.07\ \text{MeV per nucleon} \]
What that number tells you. Helium-4 sits high for such a light nucleus — unusually tightly bound for its size, which is exactly why alpha particles exist as a unit and why alpha decay happens at all. But 7.07 is still below the 8.8 MeV peak, so fusing helium into heavier elements still releases energy. Stars know this, and E.5 follows it upward.
The trap. Rounding the masses. The mass defect is a difference of numbers that agree to three decimal places, so rounding to three figures early destroys the answer entirely. Carry all five decimals until the subtraction is done.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A nuclear reaction has a mass defect of 0.20 u. Find the energy released, taking 1 u = 931.5 MeV.
\( E = 0.20 \times 931.5 = 186 \) MeV.
2. The total binding energy of an iron-56 nucleus is 492 MeV. Find the binding energy per nucleon.
\( 492/56 = 8.8 \) MeV per nucleon — the peak of the curve, which is why iron is the most tightly bound nucleus and the end point of stellar fusion.
3. Explain how the shape of the binding energy per nucleon curve allows energy to be released by both fission and fusion.
Higher on the curve means more tightly bound, hence more stable. The curve rises steeply from hydrogen to a peak near \( A = 56 \), then declines slowly to uranium. Any process that moves nucleons towards the peak releases energy: fusing light nuclei from the left-hand rise, or splitting heavy nuclei from the right-hand decline. Both run downhill in energy terms towards the same maximum.
4. State the function of the moderator in a nuclear reactor, and say what happens without it.
It slows the fast neutrons produced by fission down to the low speeds at which uranium-235 absorbs them efficiently, typically using water or graphite. Without a moderator too few neutrons are captured to induce further fissions, and the chain reaction stalls.
5. State the function of the control rods, and explain how the reaction rate is increased.
They are made of a neutron-absorbing material such as boron or cadmium. Pushing them further in absorbs more neutrons and reduces the reaction rate; withdrawing them lets more neutrons go on to cause further fissions, increasing the rate.
6. One fission of uranium-235 releases about 200 MeV. Express this in joules, and comment on why nuclear fuel is described as energy-dense.
\( 200 \times 10^{6} \times 1.60 \times 10^{-19} = 3.2 \times 10^{-11} \) J per fission. That is around a hundred million times the few electronvolts released per molecule in a chemical reaction such as burning, which is why a very small mass of fuel supplies an enormous amount of energy.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — nuclear binding energy and fission
  • The Physics Hypertextbook — nuclear energy