3.8.1.6
Calculating energy released: binding energy after minus binding energy before
Nuclear Energy & Binding Energy — AQA A-Level Physics
- In any nuclear reaction, the energyThe capacity to do work. Measured in joules (J). released equals the difference in total 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 energy. multiplied by c². between products and reactants.
- Energy $released = (total binding energyThe energy required to completely separate a nucleus into its individual protons and neutrons. Equal to the mass defectThe difference between the total mass of the individual nucleons and the actual mass of the nucleus. This mass is converted to binding energy. multiplied by c².$ of products) - (total 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. of reactants).
- Alternatively: find the mass defect of the reaction and use $E = \Delta mc^{2}$.
- Both methods give the same answer. The binding energy method avoids needing individual nuclear masses.
Worked Example
U-235 undergoes fission releasing 218 MeV per event. Calculate the energy released when 1.0 kg of uranium (3% U-235 by mass) undergoes fission.
Show Solution
1
Mass of U-235 in the sample
$$m = 0.03 \times 1000 = 30 \text{ g of U-235}$$
2
Number of atoms
$$N = \frac{30}{235} \times 6.02 \times 10^{23} = 7.68 \times 10^{22}$$
3
Total energy in MeV
$$E = 7.68 \times 10^{22} \times 218 = 1.67 \times 10^{25} \text{ MeV}$$
4
Convert to joules
$$E = 1.67 \times 10^{25} \times 1.6 \times 10^{-13} = 2.67 \times 10^{12} \text{ J}$$
Answer
$E \approx 2.7 \times 10^{12}$ J from 1 kg of 3% enriched uranium