Nuclear binding energy of Fe-56 — why is it the most stable nucleus

Question

The binding energy per nucleon of $^{56}\text{Fe}$ is approximately $8.79\,\text{MeV}$ . Why is iron-56 considered one of the most stable nuclei? Use the binding energy curve to explain why both fission of heavy nuclei and fusion of light nuclei release energy.

(JEE Main 2021, similar pattern)

Solution — Step by Step

Binding energy per nucleon of $^{56}\text{Fe}$ = $8.79\,\text{MeV}$

Total binding energy = $8.79 \times 56 = \mathbf{492.2\,\text{MeV}}$

This is the energy needed to completely disassemble the nucleus into 26 protons and 30 neutrons.

On the binding energy per nucleon (BE/A) curve:

Light nuclei (H, He, Li): BE/A is low and increases steeply
Medium nuclei (around Fe-56): BE/A reaches a maximum of ~ $8.79\,\text{MeV}$
Heavy nuclei (U, Pu): BE/A gradually decreases to ~ $7.6\,\text{MeV}$

The nucleus with the highest BE/A is the most stable because it requires the most energy per nucleon to break apart. Fe-56 sits at this peak.

Fusion of light nuclei: When light nuclei (like H) fuse to form heavier nuclei (like He), the product has higher BE/A. The increase in binding energy is released as kinetic energy. Example: $4 \text{H} \to \text{He}$ releases about $26.7\,\text{MeV}$ .

Fission of heavy nuclei: When heavy nuclei (like U-235) split into medium nuclei, the products have higher BE/A. Again, the difference is released as energy. Example: U-235 fission releases about $200\,\text{MeV}$ .

Both processes move toward Fe-56 on the curve — toward higher stability.

\boxed{\text{Fe-56 has maximum BE/A} \approx 8.79\,\text{MeV, making it the most stable nucleus.}}

Why This Works

Binding energy represents the “mass defect” — the difference between the mass of individual nucleons and the assembled nucleus. A higher binding energy means a larger mass defect, which means the nucleus has converted more mass into binding energy (via $E = mc^2$ ), making it harder to disassemble.

The BE/A curve has its characteristic shape because of competing effects: the strong nuclear force (short-range, attractive) and the Coulomb repulsion (long-range, between protons). For small nuclei, adding more nucleons increases the number of strong-force bonds faster than Coulomb repulsion grows. Beyond Fe-56, the Coulomb repulsion starts winning, and BE/A decreases.

Alternative Method — Mass defect calculation

To verify: the mass of $^{56}\text{Fe}$ is $55.9349\,\text{u}$ .

Mass of constituents: $26 \times 1.00783 + 30 \times 1.00867 = 26.2036 + 30.2601 = 56.4637\,\text{u}$

Mass defect: $\Delta m = 56.4637 - 55.9349 = 0.5288\,\text{u}$

$BE = 0.5288 \times 931.5 = 492.6\,\text{MeV}$

$BE/A = 492.6/56 = 8.80\,\text{MeV}$ ✓

The BE/A curve is a favourite in both CBSE (3-mark conceptual question) and JEE (MCQ on which process is exothermic). Remember: fusion works for nuclei lighter than Fe, fission works for nuclei heavier than Fe. Both release energy by moving toward Fe on the curve.

Common Mistake

Students often confuse “highest total binding energy” with “highest BE per nucleon.” Heavier nuclei like U-238 have higher total binding energy (about $1800\,\text{MeV}$ ), but lower BE per nucleon (~ $7.6\,\text{MeV}$ ). Stability is determined by BE per nucleon, not total BE. Fe-56 wins on the per-nucleon metric, which is what matters for comparing stability.

Question

Solution — Step by Step

Why This Works

Alternative Method — Mass defect calculation

Common Mistake

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