A-Level · 26 September 2026 · 6 min read
Nuclear binding energy: mass defect and energy-release questions
The main traps in nuclear-energy calculations are choosing inconsistent masses and confusing total binding energy with binding energy per nucleon. Powers of ten matter, but a correct calculator result cannot rescue a model that counts the wrong particles on one side of the calculation.
These original examples use deliberately simplified numerical data to make the reasoning visible. They are A2 practice for Edexcel IAL Physics, not tabulated measurements for named nuclides. Use the masses and constants supplied in an actual question, with the particle conventions stated there.
Define the mass difference before substituting
For a nucleus, mass defect is the total mass of its separated protons and neutrons minus the mass of the bound nucleus. The bound system has lower rest energy, so energy must be supplied to separate it. The missing mass has not vanished without accounting: the mass difference corresponds to the energy transferred when the system is formed.
Write the proton and neutron counts first. If atomic masses are supplied instead of nuclear masses, electron masses must be treated consistently. In many reaction calculations they cancel when both sides are balanced, but do not mix an atomic mass on one side with a nuclear mass on the other without checking what is included.
Worked example: calculate a mass defect and binding energy
An illustrative four-nucleon model has total separated-nucleon mass 6.700 × 10⁻²⁷ kg and bound-nucleus mass 6.650 × 10⁻²⁷ kg. The mass defect is 0.050 × 10⁻²⁷ = 5.0 × 10⁻²⁹ kg. With c = 3.0 × 10⁸ m s⁻¹, binding energy is 5.0 × 10⁻²⁹ × 9.0 × 10¹⁶ = 4.5 × 10⁻¹² J.
Using 1 MeV = 1.60 × 10⁻¹³ J for this example, the total is approximately 28 MeV. Per nucleon it is about 7.0 MeV. The division by four comes after finding the total binding energy. These approximate model values demonstrate the calculation and should not be quoted as precise data for a real isotope.
Change the question: total or per nucleon?
Suppose nucleus A has total binding energy 60 MeV for 8 nucleons, while nucleus B has 160 MeV for 20 nucleons. Their binding energies per nucleon are 7.5 and 8.0 MeV respectively. Comparing 60 with 160 alone mostly reflects the different sizes; comparing per nucleon gives a more useful measure of how tightly nucleons are bound on average.
Be careful with “more stable”. Binding energy per nucleon is useful for energy trends, but the full question of radioactive stability also involves the permitted transformations and composition. Do not use one rounded comparison to declare that any nucleus with a lower value must immediately decay.
Worked example: estimate an energy release from binding values
In a simplified reaction, a bound system containing 240 nucleons initially has average binding energy 7.6 MeV per nucleon. All 240 nucleons finish in bound products with average 8.4 MeV per nucleon. The increase in total binding energy is 240 × (8.4 − 7.6) = 192 MeV, or about 3.1 × 10⁻¹¹ J released per reaction.
The assumption that all 240 nucleons are bound in the products matters. A real fission equation may include free neutrons and several products, so calculate the actual total for each side. Do not multiply one product’s binding energy per nucleon by the original mass number unless that correctly represents every nucleon in the stated model.
Explain why fission and fusion can both release energy
The binding-energy-per-nucleon curve rises strongly among light nuclei and broadly reaches its highest region among medium-mass nuclei. Combining suitable light nuclei can produce more tightly bound products. Splitting suitable very heavy nuclei can also move towards more tightly bound products. The common explanation is the change in total energy, not that splitting always releases energy while joining always requires it.
Energy release does not imply that a reaction starts easily or proceeds without conditions. A fusion process still faces the interaction between positively charged nuclei before they approach closely enough for the relevant nuclear interaction. Separate the question “Is the final state lower in energy?” from “How can the reaction be initiated?”.
Use units and conservation as a final audit
Check nucleon number and electric charge in the reaction equation, then check whether the masses describe the complete systems being compared. Convert electronvolts or megaelectronvolts using the correct factor: 1 MeV is one million eV. Keep enough intermediate digits because subtracting similar masses can magnify rounding errors.
For practice, solve one mass-defect question, one per-nucleon comparison and one complete reaction-energy question. In each answer, write a sentence explaining the sign: energy is released when the final total rest mass is smaller. That explanation shows whether the arithmetic describes a physical change rather than a memorised subtraction order.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
Why is energy needed to separate a nucleus?
The bound nucleus has lower total rest energy than the same nucleons separated and at rest. Supplying binding energy raises the system to that separated state. The mass defect multiplied by c² gives that energy when the masses are defined consistently.
Why divide binding energy by nucleon number?
It gives the average binding energy per nucleon, allowing a more useful comparison between nuclei of different sizes. Do not divide if the question asks for total binding energy or total reaction energy. Read which quantity is requested before calculating.
Which way round do I subtract masses for binding energy?
For a nucleus’s binding energy, subtract the bound nuclear mass from the total mass of its separated nucleons. For reaction energy, compare the total initial and final system masses. State which calculation you are doing, since the lists of particles can differ.
How can both fission and fusion release energy?
Suitable reactions can produce a final system with greater total binding energy and lower total rest mass. Light-nucleus fusion and very-heavy-nucleus fission can both move towards more tightly bound products. Not every possible fusion or fission reaction is energetically favourable.
Do I include electron masses in nuclear-energy calculations?
Use a consistent mass convention. Atomic masses include electrons, whereas nuclear masses do not. Electron contributions may cancel in a balanced calculation, but you must verify that from the particles and masses supplied. Do not combine atomic and nuclear data without accounting for the difference.