Nuclear Binding Energy
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Direct answer
Iron-56 sits at the summit of the binding energy per nucleon curve at about 8.8 MeV — the reason nuclei heavier than iron release energy by splitting (fission) and lighter ones by merging (fusion). The arithmetic behind every question: a nucleus weighs less than its separated parts, and the missing mass converts through binding energy, BE = Δm × 931.5 MeV, where 1 u = 931.5 MeV/c². Compute the mass defect as Δm = Z m_H + (A − Z) m_n − M_atom, using the hydrogen atom's mass so the electrons cancel on both sides. The BE/nucleon curve rises steeply to iron then slopes gently down — that single shape is the whole logic of nuclear reactors and stellar furnaces.
What you must remember
- Mass defect convention: Δm = Z m_H + (A − Z) m_n − M_nucleus; with atomic masses the electron books balance automatically — NCERT's stated method.
- Conversion constants: 1 u = 1.66 × 10^-27 kg = 931.5 MeV/c²; m_n = 1.00866 u, m_H = 1.00783 u — the numbers every numerical needs.
- Curve milestones: deuteron about 1.1 MeV per nucleon; helium-4 about 7.07 MeV per nucleon (28.3 MeV total); iron-56 near the top at about 8.8 MeV; uranium-235 down at about 7.6 MeV.
- Fission yield: U-235 plus a neutron splits into fragments plus 2-3 neutrons, releasing about 200 MeV per fission — the reactor number NCERT quotes.
- Fusion yield: light nuclei climbing the steep left edge (hydrogen to helium in the Sun) release several MeV per nucleon, more per kilogram than fission, but need thermonuclear temperatures to beat the Coulomb barrier.
- Stability reading: higher BE per nucleon means tighter binding; N ≈ Z for light nuclei, N > Z for heavy ones; the heaviest stable nuclei sit near lead-208 and bismuth-209.
- One-line energy test: a nuclear reaction releases energy whenever the products have higher total binding energy than the reactants — fission and fusion are two directions up the same hill.
Computing binding energy the exam way
Take helium-4: two protons, two neutrons. Δm = 2 × 1.00783 + 2 × 1.00866 − 4.00260 = 4.03298 − 4.00260 = 0.03038 u. Multiply: BE = 0.03038 × 931.5 ≈ 28.3 MeV, or 7.07 MeV per nucleon. That is the full technique — add the parts, subtract the whole, multiply by 931.5. The curve then does the physics: fission fragments of uranium land near 8.4-8.5 MeV per nucleon, so each of the 235 nucleons gains roughly 0.9 MeV, and 235 × 0.9 ≈ 200 MeV — the textbook yield predicted from the curve alone, before anyone names the fragments. The same logic in reverse explains the Sun: four protons fuse into helium-4, each nucleon gaining about 6 MeV on the steep left flank, which is why a star's fuel lasts billions of years.
How NEET frames it
The recurring distinction is total binding energy versus binding energy per nucleon: uranium has a larger total BE than iron yet is less stable, because stability is decided per nucleon — a classic assertion-reason pair. Numerical traps: writing 931 instead of 931.5 is forgivable, but forgetting that Δm uses the hydrogen mass (not the bare proton) leaves a 0.00055 u × Z error that shifts answers a full MeV. Conservation questions: mass number A is conserved in reactions while mass itself is not — the missing mass is the energy released. Conceptual one-liners worth securing: the Sun's energy source is fusion of hydrogen into helium; energy release requires products more tightly bound than reactants; and 1 u = 931.5 MeV of energy, the chapter's most quoted conversion.
Frequently asked questions
What is the binding energy of the helium-4 nucleus?
About 28.3 MeV total, or 7.07 MeV per nucleon, from a mass defect of 0.03038 u converted at 931.5 MeV per u.
Which nucleus has the maximum binding energy per nucleon?
Iron-56, at about 8.8 MeV per nucleon — the most tightly bound nucleus, which is why both fission and fusion release energy by moving toward it.
Why does uranium fission release energy?
The fragments bind their nucleons more tightly (about 8.5 versus 7.6 MeV per nucleon), and the difference — roughly 200 MeV per fission — is released.
Why does fusion need extremely high temperatures?
Positive nuclei must be driven close enough to overcome their Coulomb repulsion, so collisions must be violent — hence thermonuclear temperatures, as in stellar cores.
Which mass convention avoids electron errors in mass defect calculations?
Use the atomic mass of hydrogen (1.00783 u) plus the neutron mass minus the atom's mass: the electrons on both sides cancel exactly.