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Binding energy problems are mostly careful subtraction, which is exactly why they go wrong: the mass defect is a small difference between two large numbers, so one rounded digit or one forgotten electron changes the answer. This post works three problems in full with real atomic masses and shows the bookkeeping that keeps them right.
What is on the syllabus?
The JEE (Main) 2026 syllabus lists "atomic masses, mass-energy relation, mass defect, binding energy per nucleon and its variation with mass number, nuclear fission and fusion". The NEET (UG) 2026 syllabus and CBSE 2025-26 curriculum list the same ideas, and JEE Advanced 2026 adds "binding energy and its calculation" and energy calculations in fission and fusion. The CBSE 2025-26 sample paper includes a three-mark question asking for the energy released in a fission reaction from given atomic masses. Check your year's bulletin.
What exactly is binding energy?
The NCERT chapter on nuclei defines it this way: a nucleus has less mass than its separate protons and neutrons. The difference is the mass defect ΔM, and the binding energy is E_b = ΔMc², the energy you would need to pull the nucleus apart into nucleons. Dividing by the mass number A gives the binding energy per nucleon.
NCERT shows that 1 u is equivalent to 931.5 MeV. The current NIST value is 931.494 MeV, so 931.5 is fine for problems.
Why use atomic masses?
Tables give atomic masses, which include electrons. If you use the mass of a hydrogen atom (one proton plus one electron) for each proton, the electrons cancel against the electrons included in the atomic mass of the product. So:
ΔM = Z × m(¹H) + (A − Z) × mₙ − m(atom).
Worked example 1: helium-4
Masses from the NIST atomic mass tables: m(¹H) = 1.007825 u, mₙ = 1.008665 u, m(⁴He) = 4.002603 u.
Separate nucleons: 2 × 1.007825 + 2 × 1.008665 = 2.015650 + 2.017330 = 4.032980 u.
ΔM = 4.032980 − 4.002603 = 0.030377 u.
E_b = 0.030377 × 931.5 ≈ 28.30 MeV.
Per nucleon: 28.30/4 ≈ 7.07 MeV.
The common slip: using the bare proton mass, 1.007276 u, with the helium atomic mass. That leaves two electron masses unbalanced and gives about 27.27 MeV, roughly 1 MeV too low.
Worked example 2: iron-56
m(⁵⁶Fe) = 55.934936 u (NIST), with Z = 26 and 30 neutrons.
Separate nucleons: 26 × 1.007825 + 30 × 1.008665 = 26.203450 + 30.259950 = 56.463400 u.
ΔM = 56.463400 − 55.934936 = 0.528464 u.
E_b = 0.528464 × 931.5 ≈ 492.3 MeV.
Per nucleon: 492.3/56 ≈ 8.79 MeV.
NCERT's binding-energy curve peaks at about 8.75 MeV near A = 56 and falls to about 7.6 MeV at A = 238. That shape explains both kinds of nuclear energy: splitting a heavy nucleus or fusing light ones moves the products towards the peak, and the gain in binding energy is released.
Worked example 3: a fusion Q-value
For deuterium–tritium fusion, ²H + ³H → ⁴He + n, the energy released is Q = (mass before − mass after) × 931.5 MeV. NIST gives m(²H) = 2.014102 u and m(³H) = 3.016049 u.
Before: 2.014102 + 3.016049 = 5.030151 u.
After: 4.002603 + 1.008665 = 5.011268 u.
Δm = 0.018883 u, so Q = 0.018883 × 931.5 ≈ 17.6 MeV.
You can check this with binding energies: Q equals the binding energy of the products minus that of the reactants. Helium-4's 28.30 MeV minus deuterium's 2.22 MeV and tritium's 8.48 MeV gives 17.6 MeV again.
For fission, NCERT estimates the energy by the same idea: a nucleus with A = 240 at about 7.6 MeV per nucleon splits into two fragments at about 8.5 MeV per nucleon, a gain of about 0.9 MeV per nucleon, or roughly 216 MeV in total, "of the order of 200 MeV". The evaluated masses behind tables such as NIST's come from the AME2020 atomic mass evaluation.
Common mistakes and how to avoid them
Mixing nuclear and atomic masses. Use ¹H atoms with atomic masses, or bare protons with nuclear masses, never one of each.
Rounding masses early. Two decimal places can wipe out the whole defect.
Calling binding energy "energy stored in the nucleus". It is the energy needed to separate it, or released when it forms.
Confusing total and per-nucleon values. Uranium has a larger total binding energy than iron but a smaller value per nucleon.
Dropping neutrons from a fission equation. Balance A and Z on both sides before calculating Q.
For the decay side of the chapter, read our guide to radioactive decay and half-life, and our post on Lise Meitner and the discovery of fission for where the energy argument began.
When a long subtraction keeps giving you a different answer from the key, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can compare your working line by line.
A practice routine for this chapter
Day 1: binding energy and per-nucleon values for ²H, ⁴He and ¹⁶O from atomic masses.
Day 2: redo one problem deliberately with bare proton masses to see the electron error.
Day 3: four Q-value problems: two fusion, two fission, each with a balanced equation.
Day 4: sketch the binding-energy curve and mark where fission and fusion release energy.
Day 5: the fission question in the latest CBSE sample paper, timed.
Today, calculate the binding energy of oxygen-16 and check that it comes out close to the 127.5 MeV in NCERT Example 13.3.
