Mass defect and nuclear binding energy
Section: Nuclear Physics | Syllabus: Cambridge A Level Physics (9702)
Mass Defect and Nuclear Binding Energy
Mass and Energy Equivalence
Einstein showed that mass and energy are interchangeable. The mass of a body is a measure of its energy content.
E = mc^2where c is the speed of light (3.00 × 10^8 m s^-1). Conservation of Mass-Energy
In nuclear reactions, we must apply the principle of conservation of mass-energy. A decrease in mass (Δ m) results in a release of energy (Δ E).
Unified Atomic Mass Unit (u)
For atomic masses, the unified atomic mass unit is used.
Unified Atomic Mass Unit (u) One-twelfth of the mass of a carbon-12 atom. 1 u ≈ 1.66 × 10^-27 kg. Energy equivalent of 1 u:1 u rightarrow 931 MeV Energy Equivalent of Mass
For 1 kg of mass: E = mc^2 = 1 × (3.00 × 10^8)^2 = 9.0 × 10^16 J
For 1 u of mass: E = 1.66 × 10^-27 × (3.00 × 10^8)^2 = 1.49 × 10^-10 J = 931 MeV
This shows that even a tiny mass corresponds to an enormous amount of energy.
Mass Defect and Binding Energy
The mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons (nucleons).
Mass Defect The difference between the total mass of the separate nucleons and the mass of the nucleus.Note: Mass defect is always positive because separated nucleons have greater total mass than the bound nucleus.
Binding Energy The energy required to separate all the nucleons in a nucleus to infinity, where they exert no forces on each other.Equivalently: the energy released when separate nucleons come together to form the nucleus.
Nuclear Equations
Nuclear reactions must be represented by balanced equations showing conservation of:
- Nucleon number (A) – total nucleons before = total nucleons after
- Proton number (Z) – total charge before = total charge after
Binding Energy per Nucleon
To compare stability of different nuclei, we calculate the binding energy per nucleon.
BE per nucleon = (Total Binding Energy/Nucleon Number (A)) Figure 23.3: Binding Energy per Nucleon vs Nucleon NumberKey features to sketch:
- Y-axis: Binding energy per nucleon (0 to ~9 MeV)
- X-axis: Nucleon number A (0 to ~250)
- Curve rises steeply for light nuclei (A < 20)
- Peak region at Fe-56/Ni-62 (~8.8 MeV per nucleon)
- Gradual decrease for heavy nuclei (A > 62)
Label key nuclides: ^2H, ^4He, ^12C, ^56Fe, ^235U
- Higher BE per nucleon means the nucleus is more stable (more energy needed to remove each nucleon).
- Iron-56 and Nickel-62 have the highest BE per nucleon (~8.8 MeV), making them the most stable nuclides.
- Nuclei to the left of the peak can release energy by fusion.
- Nuclei to the right of the peak can release energy by fission.
Both processes result in products with higher BE per nucleon than the reactants. The increase in total binding energy equals the energy released in the reaction.
Nuclear Fusion and Fission
| Process | Description | Energy Change |
|---|---|---|
| Fusion | Two light nuclei combine to form a heavier nucleus. | Releases energy because the product has a higher BE/nucleon than reactants. |
| Fission | A heavy nucleus splits into two lighter nuclei. | Releases energy because the products have higher BE/nucleon than the parent nucleus. |
Radioactive decay (α, β, γ) is spontaneous – it occurs naturally without any external input.
Nuclear fission (in reactors) and fusion are induced – they require external conditions:
- Induced fission: requires absorption of a neutron to make the nucleus unstable
- Fusion: requires extremely high temperatures (>10^8 K) to overcome electrostatic repulsion between positive nuclei
Calculating Energy Released Using E = c^2 Δ m
The energy released in a nuclear reaction can be calculated from the mass difference before and after the reaction.
Worked Examples
Worked Example: Binding Energy of HeliumQuestion: Calculate the binding energy of a Helium-4 nucleus (mass = 4.00151 u). (Proton mass = 1.00728 u, Neutron mass = 1.00867 u)
Solution1. Mass of nucleons: 2(1.00728) + 2(1.00867) = 4.03190 u
2. Mass defect: Δ m = 4.03190 - 4.00151 = 0.03039 u
3. Convert to Energy: E = 0.03039 × 931 = 28.3 MeV
Example: Alpha DecayRadium-223 decays by alpha emission to form radon-219:
^223_88Ra arrow ^219_86Rn + ^4_2He + γCheck nucleon numbers: 223 = 219 + 4 ✓
Check proton numbers: 88 = 86 + 2 ✓
Example: Nuclear Reaction (Syllabus Example)Nitrogen bombarded with alpha particles produces oxygen and a proton:
^14_7N + ^4_2He arrow ^17_8O + ^1_1HCheck nucleon numbers: 14 + 4 = 17 + 1 = 18 ✓
Check proton numbers: 7 + 2 = 8 + 1 = 9 ✓
Example: Beta-minus DecayIodine-131 decays by beta-minus emission:
^131_53I arrow ^131_54Xe + ^0_-1e + νCheck nucleon numbers: 131 = 131 + 0 + 0 ✓
Check proton numbers: 53 = 54 + (-1) + 0 ✓
Note: ν represents an antineutrino (zero charge, negligible mass).
Worked Example: Energy Released in a Nuclear ReactionQuestion: In the reaction ^14_7N + ^4_2He arrow ^17_8O + ^1_1H, calculate the energy that must be supplied.
Masses: N-14 = 13.99923 u, He-4 = 4.00151 u, O-17 = 16.99474 u, H-1 = 1.00728 u
Solution1. Total mass before: 13.99923 + 4.00151 = 18.00074 u
2. Total mass after: 16.99474 + 1.00728 = 18.00202 u
3. Mass difference: Δ m = 18.00202 - 18.00074 = 0.00128 u
4. Convert to kg: Δ m = 0.00128 × 1.66 × 10^-27 = 2.125 × 10^-30 kg
5. Energy: E = c^2 Δ m = (3.00 × 10^8)^2 × 2.125 × 10^-30
E = 1.91 × 10^-13 J = 1.19 MeV
Since mass increases, this energy must be supplied (endothermic reaction).
Worked Example: Energy Released in FusionQuestion: Calculate the energy released when deuterium and tritium fuse:
^2_1H + ^3_1H arrow ^4_2He + ^1_0n
Masses: ^2H = 2.01355 u, ^3H = 3.01550 u, ^4He = 4.00151 u, n = 1.00867 u
Solution1. Total mass before: 2.01355 + 3.01550 = 5.02905 u
2. Total mass after: 4.00151 + 1.00867 = 5.01018 u
3. Mass difference: Δ m = 5.02905 - 5.01018 = 0.01887 u
4. Energy released: E = 0.01887 × 931 = 17.6 MeV
Mass decreases, so energy is released (exothermic reaction).
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