Nuclear Binding Energy
Specification: OCR A H556 | Section: 6.4.4 | Teaching time: ~3 hours
- Use Einstein's mass–energy equation ΔE = c²Δm to calculate energy changes in nuclear reactions
- Define mass defect and binding energy, and explain their relationship
- Calculate binding energy per nucleon and interpret the binding energy curve
- Explain why fission of heavy nuclei and fusion of light nuclei both release energy
- Work confidently with atomic mass units (u) and convert between u, kg, MeV and J
Mass and Energy Equivalence
One of the most profound results of Einstein's theory of special relativity is that mass and energy are equivalent — they are different forms of the same thing. Mass can be converted to energy, and energy can be converted to mass.
Where c = 3.00 × 10⁸ m s⁻¹ is the speed of light in a vacuum. This equation tells us that:
- A small amount of mass corresponds to an enormous amount of energy (because c² is huge)
- In nuclear reactions, measurable mass changes occur, releasing significant energy
- Energy has mass — even a hot object is slightly more massive than a cold one!
Units for Nuclear Physics
In nuclear physics, we typically work with much smaller masses and energies than in macroscopic physics. Two convenient units are:
| Unit | Definition | Value |
|---|---|---|
| Atomic mass unit (u) | 1/12 of the mass of a carbon-12 atom | 1 u = 1.661 × 10⁻²⁷ kg |
| Electronvolt (eV) | Energy gained by an electron crossing 1 V | 1 eV = 1.60 × 10⁻¹⁹ J |
Using E = mc² with the atomic mass unit:
1 u → E = (1.661 × 10⁻²⁷) × (3.00 × 10⁸)² = 1.49 × 10⁻¹⁰ J = 931.5 MeV
This is given in your data booklet. Memorise: 1 u = 931.5 MeV
Particle–Antiparticle Pairs
Energy can be converted into mass in the form of particle–antiparticle pairs. For example, a gamma photon with sufficient energy can spontaneously create an electron–positron pair:
The reverse process, annihilation, occurs when a particle meets its antiparticle — all the mass is converted to energy as gamma photons:
The minimum photon energy for pair production equals the rest energy of the particles created. For an electron–positron pair:
Knowledge Check
- E = mc² = (1.673 × 10⁻²⁷) × (3.00 × 10⁸)² ✓
- E = 1.51 × 10⁻¹⁰ J ✓
- In MeV: E = 1.51 × 10⁻¹⁰ / 1.60 × 10⁻¹³ = 941 MeV ✓
- (Or: mass = 1.0073 u, so E = 1.0073 × 931.5 = 938 MeV)
- In joules: E = 3.02 × 10⁶ × 1.60 × 10⁻¹⁹ = 4.83 × 10⁻¹³ J ✓
- m = E/c² = 4.83 × 10⁻¹³ / (3.00 × 10⁸)² = 5.37 × 10⁻³⁰ kg ✓
- Mass of proton = 1.0073 u, so rest energy = 1.0073 × 931.5 = 938 MeV ✓
- Need to create proton AND antiproton: E_min = 2 × 938 = 1876 MeV (or 1.88 GeV) ✓
Mass Defect and Binding Energy
The mass of a nucleus is always less than the sum of the masses of its constituent nucleons (protons and neutrons). This difference is called the mass defect.
Where Z = proton number, N = neutron number, and the masses are measured when the particles are separated.
Why Does Mass Defect Exist?
When nucleons come together to form a nucleus, energy is released (the nucleus is more stable). Since mass and energy are equivalent, this released energy corresponds to a loss of mass.
The energy released when the nucleus forms (or equivalently, the energy required to separate the nucleus into its individual nucleons) is called the binding energy:
Binding energy is the energy that would be required to break apart the nucleus, not the energy holding it together. A higher binding energy means the nucleus is more tightly bound and more stable. The mass defect is the mass "missing" from the nucleus — it's the mass that was converted to energy when the nucleus formed.
Data: m_p = 1.0073 u, m_n = 1.0087 u
Binding Energy per Nucleon
To compare the stability of different nuclei, we use binding energy per nucleon:
Where A is the mass number (total nucleon count). Higher BE per nucleon means greater stability.
Binding Energy Curve
Hover over the curve to see individual nuclei. Notice that iron-56 (A=56) has the highest binding energy per nucleon. Nuclei to the left can undergo fusion; nuclei to the right can undergo fission — both release energy by moving towards the peak.
Fission and Fusion
The binding energy curve explains why both fission (splitting heavy nuclei) and fusion (combining light nuclei) can release energy.
Fission
Heavy nuclei (A > 56) have lower BE per nucleon than medium-mass nuclei. When a heavy nucleus splits into two smaller nuclei:
- The products have higher BE per nucleon (more tightly bound)
- The total binding energy increases
- Energy is released (equal to the increase in total binding energy)
Fusion
Light nuclei (A < 56) have lower BE per nucleon than medium-mass nuclei. When two light nuclei combine:
- The product has higher BE per nucleon
- Energy is released
⚛️ Fission
- Splits heavy nuclei (U-235, Pu-239)
- Releases ~200 MeV per fission
- Used in nuclear power stations
- Chain reaction sustained by neutrons
- Produces radioactive waste
🌟 Fusion
- Combines light nuclei (H isotopes)
- Releases ~17 MeV per reaction
- Powers the Sun and stars
- Requires extreme temperatures (~10⁸ K)
- Cleaner — helium as main product
Both fission and fusion move nuclei towards the peak of the binding energy curve (around iron). Any process that increases the average binding energy per nucleon releases energy. This is why fusion of very light elements releases more energy per unit mass than fission of heavy elements.
Calculating Energy Released
To find the energy released in any nuclear reaction:
Reaction: ²H + ²H → ³He + ¹n
Data: m(²H) = 2.0141 u, m(³He) = 3.0160 u, m(n) = 1.0087 u
Exam-Style Questions
(b) The mass of a lithium-7 nucleus is 7.0144 u. Calculate the binding energy of lithium-7 in MeV.
Data: m_p = 1.0073 u, m_n = 1.0087 u [3 marks]
(c) Calculate the binding energy per nucleon for lithium-7. [1 mark]
(a) [1]:
- Mass defect is the difference between the mass of separated nucleons and the actual mass of the nucleus ✓
- Or: Δm = Zm_p + Nm_n − m_nucleus ✓
(b) [3]:
- Lithium-7: Z = 3, N = 4, so separated mass = 3(1.0073) + 4(1.0087) = 7.0567 u ✓
- Δm = 7.0567 − 7.0144 = 0.0423 u ✓
- BE = 0.0423 × 931.5 = 39.4 MeV ✓
(c) [1]:
- BE per nucleon = 39.4 / 7 = 5.63 MeV/nucleon ✓
(a) Explain why fusion reactions only release energy for light nuclei (A < 56). [3 marks]
(b) Explain why fission of heavy nuclei releases energy. [2 marks]
(c) Suggest why iron-56 is effectively the "ash" of stellar nucleosynthesis. [2 marks]
(a) [3]:
- Fusion combines light nuclei into a heavier nucleus ✓
- For A < 56, the product has higher BE per nucleon than the reactants ✓
- Energy is released because total binding energy increases ✓
(b) [2]:
- Fission splits heavy nuclei into medium-mass nuclei with higher BE per nucleon ✓
- Energy released equals the increase in total binding energy ✓
(c) [2]:
- Iron-56 has the maximum BE per nucleon — it's the most stable nucleus ✓
- No nuclear process (fusion or fission) can release energy from iron — it's the end point of stellar fusion ✓
Data: m(²³⁵U) = 235.0439 u, m(¹n) = 1.0087 u, m(¹⁴¹Ba) = 140.9144 u, m(⁹²Kr) = 91.9262 u
(a) Calculate the mass defect in this reaction. [2 marks]
(b) Calculate the energy released in MeV. [1 mark]
(c) How does this energy compare to typical chemical reaction energies (~few eV)? [1 mark]
(a) [2]:
- Reactants: 235.0439 + 1.0087 = 236.0526 u ✓
- Products: 140.9144 + 91.9262 + 3(1.0087) = 235.8667 u ✓
- Δm = 236.0526 − 235.8667 = 0.1859 u ✓
(b) [1]:
- E = 0.1859 × 931.5 = 173 MeV ✓
(c) [1]:
- Nuclear energies are ~10⁸ times larger than chemical energies (MeV vs eV) ✓
Mass of electron = 9.11 × 10⁻³¹ kg. [3 marks]
(b) In pair production, a single gamma photon creates an electron-positron pair. Explain why this cannot happen in a vacuum. [2 marks]
(a) [3]:
- Rest energy of electron: E = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² ✓
- E = 8.20 × 10⁻¹⁴ J = 0.511 MeV ✓
- Total energy (electron + positron) = 2 × 0.511 = 1.02 MeV (or 1.64 × 10⁻¹³ J) ✓
(b) [2]:
- Pair production cannot conserve both energy AND momentum in vacuum ✓
- A nearby nucleus is needed to absorb excess momentum while taking negligible energy ✓
4¹H → ⁴He + 2e⁺ + 2ν_e + energy
(a) Show that the mass defect in this reaction is about 0.026 u.
Data: m(¹H) = 1.0078 u, m(⁴He) = 4.0026 u, m(e⁺) = 0.00055 u [2 marks]
(b) Calculate the energy released per fusion of 4 protons. [1 mark]
(c) The Sun fuses about 6 × 10¹¹ kg of hydrogen per second. Estimate the power output of the Sun. [3 marks]
(a) [2]:
- Reactants: 4 × 1.0078 = 4.0312 u ✓
- Products: 4.0026 + 2(0.00055) = 4.0037 u (positrons have mass, neutrinos ≈ 0) ✓
- Δm = 4.0312 − 4.0037 = 0.0275 u ≈ 0.026 u ✓
(b) [1]:
- E = 0.026 × 931.5 = 24.2 MeV ✓
(c) [3]:
- 4 protons (mass 4 × 1.0078 × 1.661 × 10⁻²⁷ ≈ 6.69 × 10⁻²⁷ kg) release 24.2 MeV ✓
- Energy per kg = (6 × 10¹¹ × 24.2 × 1.6 × 10⁻¹³) / (6.69 × 10⁻²⁷) ✓
- Or: Number of reactions per second = (6 × 10¹¹) / (4 × 1.67 × 10⁻²⁷) ≈ 9 × 10³⁷
- Power ≈ 9 × 10³⁷ × 24.2 × 1.6 × 10⁻¹³ ≈ 3.5 × 10²⁶ W (actual value ~3.8 × 10²⁶ W) ✓
Topic Summary
Mass–Energy
ΔE = c²Δm. Mass can be converted to energy and vice versa. 1 u = 931.5 MeV is the essential conversion.
Mass Defect
Nuclei have less mass than their separated nucleons. This "missing" mass was converted to binding energy when the nucleus formed.
Binding Energy
Energy required to separate a nucleus into individual nucleons. BE = Δm × c². Higher BE/nucleon = more stable.
Fission & Fusion
Both release energy by increasing BE/nucleon. Fission splits heavy nuclei; fusion combines light nuclei. Peak stability at iron-56.