F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityMass–Energy at Work: Binding Energy, Nuclear Power and Antimatter
F2.3 F1.2 F2.1About 70 minutes
Learning goals- I can calculate mass defect and binding energy using atomic mass units and MeV.
- I can use the binding-energy-per-nucleon curve to explain why fission and fusion release energy.
- I can calculate the energy released in fission and fusion reactions from masses.
- I can apply mass–energy conservation to pair annihilation and pair production, and explain how PET scanners use it.
Words to know
atomic mass unit (u)
1/12 of the mass of a carbon-12 atom: 1 u = 1.6605 × 10⁻²⁷ kg, equivalent to 931.5 MeV.
mass defect
The difference between the total mass of the separate nucleons and the mass of the nucleus (or between reactants and products).
binding energy
The energy needed to separate a nucleus into its individual protons and neutrons; equal to the mass defect × c².
binding energy per nucleon
Binding energy divided by the number of nucleons; a measure of how tightly a nucleus is held together. It peaks near iron-56.
antiparticle
A particle with the same mass as another but opposite charge (and other opposite properties); the positron e⁺ is the electron's antiparticle.
pair annihilation
A particle and its antiparticle meeting and converting their total energy into photons, e.g. e⁺ + e⁻ → 2γ.
pair production
A high-energy photon converting into a particle–antiparticle pair near a nucleus; needs at least 2mc² of energy.
Where the mass goes
Remember from Grade 11: in Work, Energy and Power you found that the products of deuterium–tritium fusion have about 3.1 × 10⁻²⁹ kg less mass than the reactants, released as about 2.8 × 10⁻¹² J, and in Thermal and Nuclear Energy you learned how fission, fusion and CANDU reactors work. In this lesson you will use the tools of nuclear and particle physicists (atomic mass units and MeV) to explain why some reactions release energy and others need it, and see the most complete conversion of mass to energy there is: matter meeting antimatter.
Mass defect and binding energy
A helium-4 nucleus has 2 protons and 2 neutrons. Surprisingly, it has less mass than 2 free protons plus 2 free neutrons. Putting the nucleons together released energy (the strong nuclear force pulled them in), and by mass–energy equivalence, the nucleus lost mass Δm = E/c². To pull the nucleus apart again, you would have to supply that energy back: this is the binding energy.
Nuclear masses are listed in atomic mass units: 1 u = 1.6605 × 10⁻²⁷ kg, and
1 u × c² = 931.5 MeV
So a mass defect in u, multiplied by 931.5, gives the energy in MeV. Tables usually list atomic masses (including electrons), so we use the mass of a hydrogen-1 atom (1.007825 u) for each proton; the electron masses then cancel. The neutron mass is 1.008665 u.
Worked exampleExample 1: Binding energy of helium-4
The atomic mass of helium-4 is 4.002603 u. Find its mass defect, binding energy and binding energy per nucleon.
- Separate parts: 2 × 1.007825 u (two ¹H atoms) + 2 × 1.008665 u (two neutrons) = 2.015650 + 2.017330 = 4.032980 u.
- Mass defect: Δm = 4.032980 − 4.002603 = 0.030377 u.
- Binding energy: 0.030377 × 931.5 MeV = 28.30 MeV.
- Per nucleon: 28.30 ÷ 4 = 7.07 MeV.
Δm = 0.030377 u; binding energy ≈ 28.3 MeV (7.07 MeV per nucleon). Keep all six decimal places until the subtraction; the defect is less than 1% of the total.
The curve that explains nuclear energy
Plot binding energy per nucleon against mass number and you get one of the most important graphs in physics. It rises steeply for light nuclei, peaks near iron-56 (about 8.8 MeV per nucleon) and falls slowly for heavy nuclei.
- Fusion: joining very light nuclei (hydrogen isotopes into helium) moves up the steep left side. The products are more tightly bound, have less mass, and energy is released. This powers the Sun.
- Fission: splitting a very heavy nucleus (uranium-235) into two medium nuclei moves up the gentle right side, from about 7.6 to about 8.5 MeV per nucleon. With 235 nucleons, that gain of roughly 0.9 MeV each releases about 200 MeV per fission.
Either way, energy is released when nuclei move toward iron. Going the other way (splitting iron, or fusing beyond iron) needs energy.
Binding energy per nucleon for some nuclei (points; the dashed line just guides the eye, and ⁴He sits above the smooth trend because it is unusually tightly bound). Fusion of light nuclei and fission of heavy nuclei both move toward the peak near iron and release energy.Worked exampleExample 2: Energy from D–T fusion in MeV
Deuterium (2.014102 u) fuses with tritium (3.016049 u) to give helium-4 (4.002603 u) and a neutron (1.008665 u). Find the energy released.
- Mass before: 2.014102 + 3.016049 = 5.030151 u.
- Mass after: 4.002603 + 1.008665 = 5.011268 u.
- Δm = 5.030151 − 5.011268 = 0.018883 u.
- E = 0.018883 × 931.5 = 17.6 MeV.
17.6 MeV per reaction (= 2.82 × 10⁻¹² J, matching the Grade 11 result).
Worked exampleExample 3: One fission of uranium-235
One possible fission is ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n. Atomic masses: ²³⁵U 235.043928 u, n 1.008665 u, ¹⁴¹Ba 140.914411 u, ⁹²Kr 91.926156 u. Find the energy released directly.
- Mass before: 235.043928 + 1.008665 = 236.052593 u.
- Mass after: 140.914411 + 91.926156 + 3(1.008665) = 235.866562 u.
- Δm = 0.186031 u.
- E = 0.186031 × 931.5 = 173 MeV.
About 173 MeV released immediately, mostly as kinetic energy of the fragments; later radioactive decays of the fragments add more, for a total of about 200 MeV per fission. Burning one molecule of methane releases about 9 eV, so fission releases about 20 million times more energy per event.
Nuclear power: E = mc² on the grid
A CANDU reactor, like those at Ontario's Bruce and Darlington stations, uses natural uranium fuel and heavy water to slow neutrons so they keep the chain reaction going. Each fission's 200 MeV becomes thermal energy, which boils water to drive turbines. Ontario gets roughly half of its electricity from nuclear stations.
How much mass actually 'disappears'? A reactor producing 900 MW of electrical power at about 33% efficiency produces thermal energy at about 2.7 × 10⁹ W. In a year (3.16 × 10⁷ s) that is 8.5 × 10¹⁶ J, which corresponds to Δm = E/c² ≈ 1 kg of mass converted into energy per year, from many tonnes of fuel. The social and environmental side of nuclear power is discussed in the next lesson, Relativity in Technology and Society.
Antimatter: total conversion
Every type of particle has an antiparticle with the same mass and opposite charge. The electron's antiparticle is the positron (e⁺). When an electron and a positron meet, they annihilate: both disappear and their entire rest energy becomes photons:
e⁺ + e⁻ → γ + γ, each photon 0.511 MeV (if the particles were nearly at rest).
Two photons, not one: the pair had (almost) zero total momentum, so the photons must fly off in opposite directions with equal momenta.
The reverse also happens: a photon with at least 2 × 0.511 = 1.022 MeV passing near a nucleus can create an electron–positron pair (pair production). Energy becomes matter. In particle accelerators such as CERN's Large Hadron Collider, collision energy creates new, heavier particles, such as the Higgs boson (about 125 GeV/c²) discovered in 2012. Mass–energy is conserved in all of these processes; mass alone is not.
In a PET (positron emission tomography) scanner, a tracer gives off positrons. Each positron annihilates with an electron, producing two 0.511 MeV gamma rays travelling in opposite directions. Detectors that fire at the same moment mark a line through the source; many lines build a 3-D image.PET scanners
Positron emission tomography (PET) is E = mc² in a hospital. A patient receives a small dose of a tracer, such as a sugar-like molecule tagged with fluorine-18, which emits positrons. Active tissue (for example, a growing tumour) takes up more of the sugar. Each positron travels only a millimetre or so before annihilating. A ring of detectors catches pairs of 0.511 MeV photons arriving at the same time, and a computer reconstructs where the annihilations happened. The tracer isotopes are made in cyclotrons, including hospital cyclotrons and TRIUMF in Vancouver.
Nuclear energy checklist
- Use atomic masses with ¹H for protons; keep 6 decimal places until you subtract.
- Δm (in u) × 931.5 = E (in MeV). Δm (in kg) × c² = E (in J).
- Energy is released when total mass decreases (products are more tightly bound).
- Annihilation: E = 2mc² (plus any kinetic energy). Pair production needs at least 2mc².
Just for funA PET scanner is one of the few places where antimatter is used on purpose every day. Unlike in science fiction, it is measured in harmless, carefully controlled trickles of positrons, and nobody's starship is powered by it.
Hands-on activityA magnet model of binding energy
You need: two strong fridge magnets (or two magnetic toy pieces), a kitchen scale is optional, and paper. Safety: small magnets can be very dangerous if swallowed; keep them away from young children and pets.
- Hold the two magnets apart, then let them snap together. Notice that you did not have to do any work; the magnets did work on each other and released energy (you hear a click: a little sound energy, and a tiny bit of thermal energy).
- Now pull them apart. You must do work. That work is the 'binding energy' of the pair.
- Explain the analogy: which is the 'bound nucleus'? Which is 'the separate nucleons'? Which system has more energy?
- Relativity says the bound pair has slightly less mass than the separated magnets. Estimate how much: if pulling them apart takes about 0.05 J, what is Δm = E/c²?
What to look for: the separated magnets have more energy, so they have more mass, by about 0.05 ÷ 9.0 × 10¹⁶ ≈ 6 × 10⁻¹⁹ kg, roughly the mass of a few million iron atoms, and far too small for any scale. In a nucleus the binding energy is millions of electron volts per nucleon, so the mass defect is large enough to measure (almost 1% for helium-4).
Your turn
1Convert a mass defect of 0.0100 u to MeV and to joules.
0.0100 × 931.5 = 9.32 MeV = 9.32 × 10⁶ × 1.602 × 10⁻¹⁹ = 1.49 × 10⁻¹² J.
2Find the binding energy of deuterium (²H, 2.014102 u).
Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u; BE = 0.002388 × 931.5 = 2.22 MeV (1.11 MeV per nucleon).
3Carbon-12 has an atomic mass of exactly 12.000000 u. Find its binding energy per nucleon.
Δm = 6(1.007825) + 6(1.008665) − 12.000000 = 12.098940 − 12.000000 = 0.098940 u. BE = 92.2 MeV; per nucleon 7.68 MeV.
4Use the binding-energy curve to explain why fusing two iron-56 nuclei would not release energy.
Iron-56 is at the peak of the curve. Heavier nuclei have lower binding energy per nucleon, so the product would be less tightly bound and have more mass: the reaction would need energy, not release it.
5An electron and a positron, both nearly at rest, annihilate. What is the total photon energy, and why are there two photons?
1.022 MeV (2 × 0.511 MeV). The initial momentum is zero, so the photons must have equal and opposite momenta; a single photon would carry momentum.
6What is the minimum photon energy for producing an electron–positron pair?
1.022 MeV (2mc² for the electron).
7Estimate the mass converted to energy per year in a nuclear station producing 2.7 × 10⁹ W of thermal power.
E = 2.7 × 10⁹ × 3.16 × 10⁷ = 8.5 × 10¹⁶ J. Δm = E/c² = 0.95 kg, about 1 kg.
More practice: warm-up
2Near which nucleus does the binding energy per nucleon peak?
Iron-56 (around iron and nickel).
3Which releases energy for light nuclei: fusion or fission?
The antiparticle of the electron: same mass, positive charge.
Positron emission tomography.
More practice: on your own
1Oxygen-16 has atomic mass 15.994915 u. Find its binding energy per nucleon.
Δm = 8(1.007825) + 8(1.008665) − 15.994915 = 16.131920 − 15.994915 = 0.137005 u. BE = 127.6 MeV; per nucleon 7.98 MeV.
2About how much energy per fission is released if uranium-235 (7.6 MeV per nucleon) splits into fragments with an average of 8.5 MeV per nucleon?
(8.5 − 7.6) × 235 ≈ 210 MeV, consistent with about 200 MeV.
3Find the number of fissions per second in a reactor producing 2.7 × 10⁹ W of thermal power (200 MeV per fission).
200 MeV = 3.2 × 10⁻¹¹ J. Fissions/s = 2.7 × 10⁹ ÷ 3.2 × 10⁻¹¹ = 8.4 × 10¹⁹ per second.
4A proton and an antiproton (each mc² = 938 MeV) annihilate at rest. What total energy is released?
2 × 938 = 1876 MeV (about 1.88 GeV).
5Explain why the 'law of conservation of mass' from chemistry is only approximately true.
Energy changes change mass by Δm = ΔE/c². Chemical reactions involve only a few eV per molecule, so the mass changes are about 10⁻¹⁰ of the total, too small to detect. Strictly, it is mass–energy that is conserved.
Challenge
1Lithium-6 (6.015123 u) has a binding energy per nucleon well below helium-4. Calculate it and use the curve to explain which is more stable.
Δm = 3(1.007825) + 3(1.008665) − 6.015123 = 6.049470 − 6.015123 = 0.034347 u. BE = 32.0 MeV, per nucleon 5.33 MeV. Helium-4 (7.07 MeV per nucleon) is much more tightly bound, so it is more stable.
2A 2.0 MeV photon produces an electron–positron pair. If the energy is shared equally, find the kinetic energy and speed of each particle (ignore the nucleus's recoil).
Energy left after creating the pair: 2.0 − 1.022 = 0.978 MeV, so each gets 0.489 MeV. γ = 1 + 0.489/0.511 = 1.957; v/c = √(1 − 1/1.957²) = √(1 − 0.261) = 0.860.
3Ontario's nuclear stations convert only about 1 kg of mass per reactor per year into energy, yet use many tonnes of fuel. Explain, using binding energies, why only about 0.1% of the fuel's mass is converted even in fission.
Fission releases about 200 MeV per uranium nucleus, while the uranium nucleus's rest energy is about 235 × 931.5 ≈ 219 000 MeV. 200/219 000 ≈ 0.09%, so only about 0.1% of the mass of the fissioned uranium is converted. (Natural uranium fuel also contains only 0.7% uranium-235, so far less than 0.1% of the whole fuel bundle's mass is converted.)
Check your understandingQuiz
1. What is the energy equivalent of 1 u of mass?
2. Why do both fission of uranium and fusion of hydrogen release energy?
3. A nucleus has a mass defect of 0.0300 u. What is its binding energy?
4. An electron and a positron at rest annihilate. What is produced?
5. What is the minimum photon energy needed for electron–positron pair production?
6. In D–T fusion, Δm = 0.018883 u. How much energy is released?
For parents and teachersBig idea (F2.3, F1.2): nuclear energy and antimatter annihilation are E = mc² in action. Binding energy explains why both fission of heavy nuclei and fusion of light nuclei release energy, and PET scanners use electron–positron annihilation every day.
How to help: Ask your teen to sketch the binding-energy curve from memory and mark where fission and fusion happen. The magnet activity needs care: strong magnets can pinch fingers and are very dangerous if swallowed, so keep them away from young children and pets.
Common difficulties:
- Rounding masses before subtracting. The difference is tiny; keep every decimal place.
- Mixing atomic and nuclear masses. Using ¹H atoms for protons makes the electrons cancel.
- Thinking more binding energy means more mass. More binding energy means less mass (more energy was released when it formed).
- Forgetting that annihilation produces two photons, to conserve momentum.