Nuclear Binding Energy Calculations — Mass Defect Step by Step
A nucleus weighs less than the particles it is made of. That missing mass, converted by E = mc², is the energy holding the nucleus together — the binding energy. The calculation is short, but it fails for two predictable reasons: students use the wrong mass for the proton, and they divide by the wrong number at the end. This guide fixes both, with four worked examples whose answers you can check against any standard data table.
The formulas
What each symbol means, and the one trap in it
- Z — atomic number, the number of protons. N — number of neutrons. A = Z + N — mass number.
- m(¹H) = 1.007825 u — the mass of a hydrogen atom, which includes one electron.
- mn = 1.008665 u — the neutron mass.
- M(atom) — the atomic mass of the nuclide, which also includes Z electrons.
Why the formula uses m(¹H) and not the proton mass. Data tables list atomic masses, not nuclear masses. Using Z hydrogen atoms on the left puts Z electrons on both sides of the subtraction, so they cancel out. If instead you use the proton mass mp = 1.007276 u together with an atomic mass, you have Z stray electrons and the answer is wrong. Pick one convention: atomic masses with m(¹H), or nuclear masses with mp. Never mix them.
The conversion factor 1 u = 931.494 MeV/c² comes straight from E = mc² with 1 u = 1.66054 × 10⁻²⁷ kg. Many textbooks round it to 931.5 or even 931; that changes the third significant figure, so match whichever value your syllabus prints.
Worked example 1 — helium-4
Atomic mass of ⁴He = 4.002602 u, Z = 2, N = 2.
Step 1 — add up the separated particles.
2 × 1.007825 = 2.015650
2 × 1.008665 = 2.017330
Total = 4.032980 u
Step 2 — mass defect.
Δm = 4.032980 − 4.002602 = 0.030378 u
Step 3 — binding energy.
Eb = 0.030378 × 931.494 = 28.30 MeV
Step 4 — per nucleon.
28.297 ÷ 4 = 7.07 MeV per nucleon
Helium-4 sits unusually high for such a light nucleus. That stability is exactly why alpha particles are emitted as intact ⁴He units rather than as loose nucleons.
Worked example 2 — iron-56, the peak of the curve
Atomic mass of ⁵⁶Fe = 55.934936 u, Z = 26, N = 30.
26 × 1.007825 = 26.203450
30 × 1.008665 = 30.259950
Total = 56.463400 u
Δm = 56.463400 − 55.934936 = 0.528464 u
Eb = 0.528464 × 931.494 = 492.3 MeV
Per nucleon = 492.261 ÷ 56 = 8.79 MeV
Around 8.8 MeV per nucleon is the maximum of the binding-energy curve. Everything below iron can release energy by fusing; everything above it can release energy by splitting. That one sentence explains both fusion in stars and fission in reactors.
Worked example 3 — uranium-238
Atomic mass of ²³⁸U = 238.050788 u, Z = 92, N = 146.
92 × 1.007825 = 92.719900
146 × 1.008665 = 147.265090
Total = 239.984990 u
Δm = 239.984990 − 238.050788 = 1.934202 u
Eb = 1.934202 × 931.494 = 1801.7 MeV
Per nucleon = 1801.70 ÷ 238 = 7.57 MeV
Uranium has by far the largest total binding energy of the three, and the smallest of the three per nucleon. Total binding energy tells you how tightly the whole nucleus is held; only the per nucleon figure tells you how stable it is compared with other nuclides. Read the question carefully to see which is being asked.
Worked example 4 — energy released by a nuclear reaction
The deuterium–tritium fusion reaction: ²H + ³H → ⁴He + n. Masses in u: ²H = 2.014102, ³H = 3.016049, ⁴He = 4.002602, n = 1.008665.
Reactants: 2.014102 + 3.016049 = 5.030151 u
Products: 4.002602 + 1.008665 = 5.011267 u
Δm = 5.030151 − 5.011267 = 0.018884 u
E = 0.018884 × 931.494 = 17.59 MeV released
The electrons balance here without any correction: one electron on each side of the arrow (the deuterium and tritium atoms each carry one, helium carries two), so atomic masses can be used directly.
Compare that with a chemical reaction, where a typical bond energy is a few hundred kJ/mol — roughly 1 to 5 electronvolts per molecule. Nuclear energies are about a million times larger per event, which is why 17.59 MeV from one tiny fusion is worth chasing.
The binding-energy curve, as numbers
Each row below is calculated by the same method used above, so you can reproduce any of them as practice.
| Nuclide | Atomic mass (u) | Δm (u) | Eb (MeV) | Eb/A (MeV) |
|---|---|---|---|---|
| ²H (deuterium) | 2.014102 | 0.002388 | 2.224 | 1.11 |
| ⁴He | 4.002602 | 0.030378 | 28.30 | 7.07 |
| ⁵⁶Fe | 55.934936 | 0.528464 | 492.3 | 8.79 |
| ²³⁸U | 238.050788 | 1.934202 | 1801.7 | 7.57 |
Read down the last column: it rises steeply from deuterium, peaks near iron, then falls away slowly. Deuterium is the least tightly bound of the four — a single proton and a single neutron held by only 2.22 MeV in total.
Common mistakes that cost marks
- Using mp = 1.007276 u with an atomic mass. This is the number-one error. With atomic masses, use m(¹H) = 1.007825 u.
- Dividing by Z instead of A. Binding energy per nucleon divides by the total number of protons and neutrons.
- Getting the subtraction backwards. Separated particles minus the actual nucleus. A negative Δm means you have reversed it.
- Leaving the answer in u. The question asks for energy — multiply by 931.494 MeV/u, and state the unit.
- Rounding the masses. Δm is a difference of numbers that agree to three or four decimal places. Round to two decimals early and the mass defect vanishes entirely. Carry all six decimals until the last line.
- Confusing total binding energy with binding energy per nucleon when comparing stability. Uranium beats iron on the first, loses badly on the second.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 12 | Mass defect and binding energy numericals; nuclear physics chapter |
| JEE/NEET | Energy released in fission or fusion; Q-value calculations |
| IIT-JAM / CUET-PG | Binding-energy curve reasoning; stability comparisons |
| GATE / CSIR-NET | Nuclear and radiochemistry: Q-values, decay energetics, semi-empirical mass formula |
The arithmetic is easy; the constants are where slips happen. The suite has no dedicated binding-energy tool, but its constants reference gives you c, the atomic mass unit and the electronvolt to full precision, and the scientific calculator handles the six-decimal subtraction without rounding on you.
Open the ABC Chemistry Calculator Suite →Nuclear chemistry sits at the boundary of the physics and chemistry papers, and is often taught thinly. ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.