The big idea: Weigh all the protons and neutrons of a nucleus separately, then weigh the finished nucleus. The nucleus is always a tiny bit lighter.
That missing mass is the mass defect. Through E = mc² it equals the binding energy — the energy you would need to pull the nucleus completely apart.
Loose protons + neutrons
- All the nucleons separate, far apart
- Their masses simply add up — the larger total
- Higher energy: it took work to pull them apart
The bound nucleus
- The nucleons stuck together by the strong force
- Slightly smaller total mass — the mass defect Δm
- Lower energy: building it released the binding energy
New words, plainly: Mass defect (Δm) = (mass of the separate nucleons) − (mass of the bound nucleus).
Binding energy = the energy that mass defect is worth (E = mc²); also the energy needed to split the nucleus back into loose nucleons.
Binding energy per nucleon = binding energy ÷ number of nucleons — the fair way to compare different nuclei.
| Mass added up | Mass / u |
|---|---|
| 2 protons (2 × 1.007276) | 2.014552 |
| 2 neutrons (2 × 1.008665) | 2.017330 |
| Separate nucleons, total | 4.031882 |
| Actual helium-4 nucleus | 4.001506 |
| Mass that went missing (Δm) | 0.030376 |
Why 'per nucleon' matters: A big nucleus has a big total binding energy just because it has more nucleons.
To compare how stable two nuclei are, divide by the number of nucleons. The higher the binding energy per nucleon, the more stable the nucleus.
Three steps: find the mass defect (separate nucleons − nucleus), turn it into energy with E = mc², then divide by the number of nucleons to get the binding energy per nucleon.
- energy equivalent of the mass — here the binding energy (J or MeV)
- mass converted — here the mass defect Δm of the nucleus (kg or u)
- speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
Two units, one idea: In joules: put Δm in kilograms (1 u = 1.661 × 10⁻²⁷ kg) and multiply by c².
In MeV (faster): keep Δm in u and multiply by 931.5 (because 1 u = 931.5 MeV c⁻²).
Both give the same energy — pick whichever the question asks for.
IB-style question — binding energy of helium-4
A helium-4 nucleus is made of 2 protons and 2 neutrons. Proton mass = 1.007276 u, neutron mass = 1.008665 u, and the helium-4 nucleus has mass 4.001506 u. Find the total binding energy of the nucleus. (1 u = 931.5 MeV c⁻².)
Solution
- First add up the masses of the separate nucleons (2 protons + 2 neutrons):
- Find the mass defect Δm = separate nucleons − the actual nucleus:
- Turn that mass into energy with the given . Using 1 u = 931.5 MeV c⁻², just multiply Δm by 931.5:
- Work it out — keep the unit:
Final answer
Δm = 0.030376 u, so the binding energy = 0.030376 × 931.5 ≈ 28.3 MeV.
IB-style question — binding energy per nucleon of helium-4
Helium-4 has a total binding energy of about 28.3 MeV and contains 4 nucleons. Find its binding energy per nucleon.
Solution
- Binding energy per nucleon = total binding energy ÷ number of nucleons. Write it first:
- Put in the numbers (Eb = 28.3 MeV, A = 4):
- Work it out — keep the unit:
Final answer
binding energy per nucleon = 28.3 ÷ 4 ≈ 7.1 MeV per nucleon — a tightly bound, very stable nucleus.
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Plot the binding energy per nucleon against the nucleon number A for every nucleus and you get one of the most important graphs in physics. It explains where nuclear energy comes from.
Reading the curve: Higher up = more tightly bound = more stable. The curve peaks near iron (A ≈ 56) — the most stable nuclei of all.
Any reaction that moves a nucleus toward the peak ends up more tightly bound, so it releases energy.
How this is tested: Binding energy shows up in both papers.
- Paper 1A: a quick MCQ — read off which nucleus is most stable, or compare the energy released per unit mass in fusion vs fission. - Paper 2: a calculation — find a mass defect, the binding energy, and the binding energy per nucleon from given masses.
Classic trap: thinking only fission releases energy. Both fusion (light nuclei) and fission (heavy nuclei) release energy, because both move toward iron.
Fusion (light fuel)
- Joins light nuclei, e.g. hydrogen isotopes
- Big jump UP the steep part of the curve
- Releases several MeV per nucleon → most energy per kilogram
Fission (heavy fuel)
- Splits a heavy nucleus, e.g. uranium
- Smaller climb up the gentle part of the curve
- Releases under 1 MeV per nucleon → less energy per kilogram
IB-style question — energy per unit mass: fusion vs fission
A fusion reaction releases 17.6 MeV from a total of 5 nucleons (deuterium + tritium). A fission reaction releases 200 MeV from a total of 236 nucleons (a uranium nucleus). Find the ratio of the energy released per nucleon in fusion to that in fission.
Solution
- Energy per nucleon is a fair per-unit-mass comparison (every nucleon has almost the same mass). Write it first:
- Fusion — put in 17.6 MeV over 5 nucleons:
- Fission — put in 200 MeV over 236 nucleons:
- Take the ratio fusion : fission:
Final answer
Fusion releases about 3.5 MeV per nucleon and fission about 0.85 MeV per nucleon, so fusion gives roughly 4 times more energy per unit mass of fuel.