3.8.1.6
Binding energy is the energy needed to completely separate a nucleus
Nuclear Energy & Binding Energy — AQA A-Level Physics
Key Definition
Binding energy — The amount of energy required to separate a nucleus into its constituent protons and neutrons. Equal to the energy released when the nucleus was formed.
$$E = \Delta mc^2$$
$$E = \Delta m \, c^2$$
- $E$: binding energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). required to completely separate a nucleus into its individual protons and neutrons. Equal to the mass defectThe difference between the total mass of the individual nucleons and the actual mass of the nucleus. This mass is converted to binding energyThe capacity to do work. Measured in joules (J).. multiplied by c². (J)
- $Δm$: mass defectThe difference between the total mass of the individual nucleons and the actual mass of the nucleus. This mass is converted to binding energy. (kg)
- $c$: speed of light (m s⁻¹)
- Nuclear binding energies are of the order MeV. Electron binding energies are only a few eV.
- This is why nuclear reactions release far more energy than chemical reactions.
- Forming a nucleus from isolated nucleons is an exothermic process — energy is released.
- Splitting a nucleus back into isolated nucleons is endothermic — energy must be supplied.
Examiner Tips and Tricks
- Never describe binding energyThe energy required to completely separate a nucleus into its individual protons and neutrons. Equal to the mass defectThe difference between the total mass of the individual nucleons and the actual mass of the nucleus. This mass is converted to binding energy. multiplied by c². as 'the energy stored in the nucleus'.
- It is the energy you must put IN to break the nucleus apart.
- Think of it like the energy needed to pull apart superglued objects.
Worked Example
Calculate the binding energy per nucleonThe binding energyThe energy required to completely separate a nucleus into its individual protons and neutrons. Equal to the mass defect multiplied by c². of a nucleus divided by its nucleon number (mass number). Higher values indicate greater nuclear stability. of K-40 (Z = 19). Nuclear mass = 39.953548 u, proton mass = 1.007276 u, neutron mass = 1.008665 u.
Show Solution
1
Calculate mass defect
$$\Delta m = (19 \times 1.007276) + (21 \times 1.008665) - 39.953548 = 0.36666 \text{ u}$$
2
Convert to kg
$$\Delta m = 0.36666 \times 1.661 \times 10^{-27} = 6.090 \times 10^{-28} \text{ kg}$$
3
Calculate binding energy
$$E = \Delta m \, c^2 = 6.090 \times 10^{-28} \times (3.0 \times 10^{8})^2 = 5.5 \times 10^{-11} \text{ J}$$
4
Find binding energy per nucleonThe binding energy of a nucleus divided by its nucleon number (mass number). Higher values indicate greater nuclear stability. in MeV
$$\frac{E}{A} = \frac{5.5 \times 10^{-11}}{40} = 1.375 \times 10^{-12} \text{ J}$$
$$= \frac{1.375 \times 10^{-12}}{1.6 \times 10^{-13}} = 8.59 \text{ MeV}$$Answer
Binding energy per $nucleon = 8.59 MeV$