Nuclear stability — binding energy curve, magic numbers, N/Z ratio

Question

The binding energy per nucleon for $^{56}_{26}\text{Fe}$ is approximately 8.8 MeV and for $^{235}_{92}\text{U}$ is approximately 7.6 MeV. Explain why iron is the most stable nucleus, and predict whether uranium would undergo fission or fusion to become more stable.

Solution — Step by Step

Binding energy per nucleon ( $BE/A$ ) tells us how tightly each nucleon is held inside the nucleus. Higher $BE/A$ means more energy needed to pull a nucleon out — so higher $BE/A$ means more stable.

Iron-56 has $BE/A \approx 8.8$ MeV — the highest of any nucleus. This is the peak of the binding energy curve.

Nuclei want to reach higher $BE/A$ (more stable). There are two paths:

Light nuclei (below Fe): fuse together to increase $BE/A$ → fusion
Heavy nuclei (above Fe): split apart to increase $BE/A$ → fission

Uranium ( $A = 235$ ) is far heavier than iron. Its $BE/A$ of 7.6 MeV is lower than iron’s 8.8 MeV.

Uranium will undergo fission to move toward the peak of the $BE/A$ curve. When U-235 splits into two medium-mass fragments (typically around $A = 90$ to $A = 140$ ), the products have higher $BE/A$ . The difference in total binding energy is released as kinetic energy — this is what powers nuclear reactors and atomic bombs.

Energy released per fission $\approx (8.5 - 7.6) \times 235 \approx \mathbf{200 \text{ MeV}}$

Why This Works

Nuclear stability depends on the competition between the strong nuclear force (attractive, short-range) and the Coulomb repulsion (repulsive, long-range between protons).

graph TD
    A["Nuclear Stability Factors"] --> B["Strong Nuclear Force"]
    A --> C["Coulomb Repulsion"]
    A --> D["N/Z Ratio"]
    B --> E["Attracts all nucleons, range ~1 fm"]
    C --> F["Repels protons only, long range"]
    D --> G["Light nuclei: N ≈ Z stable"]
    D --> H["Heavy nuclei: N > Z needed for stability"]
    A --> I["Magic Numbers: 2, 8, 20, 28, 50, 82, 126"]
    I --> J["Extra stability when N or Z is magic"]

For light nuclei ( $A < 20$ ), $N \approx Z$ gives stability. As nuclei get heavier, more neutrons are needed to dilute the Coulomb repulsion — the stability line curves away from $N = Z$ . Beyond bismuth-209 ( $Z = 83$ ), no nucleus is truly stable.

Magic numbers (2, 8, 20, 28, 50, 82, 126) correspond to complete nuclear shells — analogous to noble gas electron configurations. Nuclei with magic N or Z (or both — “doubly magic” like $^{4}\text{He}$ , $^{16}\text{O}$ , $^{208}\text{Pb}$ ) are exceptionally stable.

Alternative Method

For quick stability prediction without the $BE/A$ curve, use the N/Z ratio test:

If $N/Z \approx 1$ for light nuclei → stable
If $N/Z \approx 1.5$ for heavy nuclei → stable
If $N/Z$ is too high → beta-minus decay (neutron → proton)
If $N/Z$ is too low → beta-plus decay or electron capture (proton → neutron)
If $Z > 83$ → alpha decay to reduce both N and Z

This decision tree covers most nuclear decay questions in CBSE and NEET.

Common Mistake

Confusing binding energy with energy stored. Students often think high binding energy means the nucleus has more energy and is less stable — the exact opposite is true. High binding energy means the nucleons have given up more energy to bind together, so they sit in a deeper potential well. You need to supply that much energy to break them apart. Think of it like a deep hole — the deeper the hole, the harder it is to climb out, and the more stable the system.

BE = [Zm_p + Nm_n - M_{nucleus}]c^2

\text{Mass defect: } \Delta m = Zm_p + Nm_n - M_{nucleus}

1 \text{ u} = 931.5 \text{ MeV}/c^2

Peak of $BE/A$ curve: $^{56}\text{Fe}$ at ~8.8 MeV/nucleon

Magic numbers: 2, 8, 20, 28, 50, 82, 126

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next