Key Idea: This topic is where mass turns into energy. Weigh a nucleus and it is always a little lighter than its separate protons and neutrons — that missing mass defect, through E = mc², is the binding energy that holds the nucleus together. Plot the binding energy per nucleon against nucleon number and the curve peaks near iron: any reaction that moves toward that peak releases energy, which is why both fusion (joining light nuclei) and fission (splitting heavy ones) give out energy. From there the topic builds out to fission as a power source — induced fission, the chain reaction, the neutron economy that keeps it steady, and the four parts of a nuclear reactor. It is examined on both papers. Paper 1A is quick multiple-choice — read off the most stable nucleus, compare energy released per unit mass in fusion vs fission, pick the neutron loss that keeps a chain steady, or match a reactor component to its job. Paper 2 is longer structured work — find a mass defect, the binding energy and the binding energy per nucleon, calculate the energy released in a fission, or outline how a reactor works.
📋 Key formulas
Only one equation in this topic carries the data-booklet badge — E = mc² (look for it). The mass-defect, per-nucleon and neutron-economy relations below are bookkeeping rules, not printed equations, so you remember those.
- energy equivalent of the mass — the binding energy, or the energy released in a reaction (J, or MeV)
- mass converted — the mass defect Δm of the nucleus or reaction (kg, or u)
- speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
- mass defect — mass that goes missing when the nucleus forms (u or kg)
- number of protons (each of mass m_p)
- number of neutrons (each of mass m_n)
- actual mass of the bound nucleus (always the lighter total)
- total binding energy of the nucleus (MeV)
- nucleon number — the number of protons + neutrons
- number of neutrons released per fission (about 2–3)
- number that must be lost or absorbed so exactly ONE continues the chain (steady)
⚖️ Mass defect, binding energy & 'per nucleon'
☢️ Fusion vs fission on the curve
Higher on the binding-energy-per-nucleon curve = more tightly bound = more stable, and the peak sits near iron (A ≈ 56). Any reaction that moves a nucleus toward that peak ends up more tightly bound, so it releases energy — that is why BOTH fusion (from the light side) and fission (from the heavy side) give out energy. The classic trap is thinking only fission releases energy.
🔁 The chain reaction & the reactor
Important: Both the moderator and the control rods act on neutrons, but oppositely: - moderator → slows neutrons (helps fission) - control rods → absorb neutrons (limit fission) Swapping these two is the most common reactor mistake.
✏️ Worked exam-style questions
IB-style question — binding energy and binding energy per nucleon
A boron-11 nucleus has 5 protons and 6 neutrons. Proton mass = 1.007276 u, neutron mass = 1.008665 u, and the boron-11 nucleus has mass 11.00931 u. (a) Find the mass defect. (b) Find the total binding energy in MeV. (c) Find the binding energy per nucleon. (1 u = 931.5 MeV c⁻².)
Solution:
(a) Add up the masses of the separate nucleons (5 protons + 6 neutrons):
(a) Mass defect = separate nucleons − the actual nucleus:
(b) Turn that mass into energy with the given E = mc²; in MeV just multiply Δm by 931.5:
(c) Binding energy per nucleon = total binding energy ÷ nucleon number (A = 11):
(a) Δm = 0.07906 u. (b) Eb ≈ 73.6 MeV. (c) ≈ 6.7 MeV per nucleon. Keep all the decimals until the subtraction — the defect is a tiny difference of two large masses. ×931.5 gives MeV; ÷A gives 'per nucleon'.
IB-style question — energy per unit mass: fusion vs fission
A fusion reaction releases 18 MeV from a total of 5 nucleons of light fuel. A fission reaction releases 195 MeV from a total of 235 nucleons of heavy fuel. Find the ratio of the energy released per unit mass of the fusion fuel to that of the fission fuel.
Solution:
Energy per unit mass scales with energy per nucleon (every nucleon has almost the same mass). Write the rule first:
Fusion — 18 MeV over 5 nucleons:
Fission — 195 MeV over 235 nucleons:
Take the ratio fusion : fission:
Fusion releases about 3.6 MeV per nucleon and fission about 0.83 MeV per nucleon, so fusion gives roughly 4 times more energy per unit mass of fuel. Energy per nucleon is the fair per-kilogram comparison.
IB-style question — neutron economy keeps the chain steady
Each fission of a uranium-235 nucleus releases on average 3 neutrons. (a) For a steady, self-sustaining chain reaction, how many neutrons per fission must be lost or absorbed? (b) If instead only 0.9 neutrons per fission go on to cause the next fission, state whether the reaction grows, stays steady or dies out.
Solution:
(a) For a steady (critical) chain, exactly one neutron per fission must trigger the next one. Use the counting rule:
(b) Compare the number continuing with the steady value of 1:
(b) Fewer than one neutron per fission continues, so each generation produces fewer fissions than the last.
(a) 2 neutrons per fission must be lost or absorbed (3 − 1), leaving exactly one to keep it steady. (b) The reaction DIES OUT (subcritical), because 0.9 < 1. More than one continuing would make it grow (supercritical).
IB-style question — energy released in a single fission
In one fission event the total mass of the products is 0.198 u less than the mass of the original nucleus plus the absorbed neutron. (a) Using 1 u ≡ 931.5 MeV, find the energy released in this fission, in MeV. (b) A reactor produces 8.0 × 10⁸ W of thermal power and each fission releases 3.2 × 10⁻¹¹ J. Find the number of fissions per second.
Solution:
(a) The mass defect Δm = 0.198 u is converted to energy by the given E = mc². Using the 931.5 MeV/u shortcut:
(b) Power is energy per second, so the number of fissions per second is the power ÷ the energy per fission:
(b) Evaluate — watch the negative power of ten:
(a) E ≈ 184 MeV per fission. (b) n ≈ 2.5 × 10¹⁹ fissions per second. With a mass defect in u, use the ×931.5 shortcut rather than converting to kilograms; for the rate, divide reactor power by the energy of one fission.
🧠 Quick self-check
Tap each card to reveal the answer.
🎯 Exam tips
Exam Tips
- Mass defect FIRST: add the separate protons and neutrons, then subtract the nucleus mass. Keep every decimal place — Δm is a tiny difference of two large numbers, so early rounding ruins it.
- Binding energy: E = mc². With Δm in u, just multiply by 931.5 to get MeV; for joules, put Δm in kilograms (×1.661 × 10⁻²⁷) first, then ×c². For 'per nucleon', divide by A.
- Higher binding energy per nucleon = more stable. The curve peaks near iron (A ≈ 56). 'Most stable' means the highest PER-NUCLEON value, never the highest total.
- BOTH fusion (light nuclei) and fission (heavy nuclei) release energy by moving toward iron. Fusion releases more energy per unit mass of fuel — energy per nucleon is the fair per-kilogram comparison.
- Steady (critical) chain reaction = exactly ONE neutron per fission causes the next, so N − 1 are lost or absorbed. Lose more → dies out (subcritical); lose fewer → grows (supercritical).
- Energy per fission from a mass defect in u: use the ×931.5 shortcut. Reactor power ÷ energy per fission gives the number of fissions per second — watch the negative power of ten.
- Reactor parts by their verb: fuel FISSIONS, moderator SLOWS neutrons (water/graphite), control rods ABSORB neutrons (boron/cadmium), heat exchanger REMOVES heat. Lower the rods to slow the reaction, raise them to speed it up.