Draw the binding energy curve

Problem

Sketch qualitatively the binding energy per nucleon curve $E_l/A$ as a function of the mass number $A$. Indicate: (i) the fusion region, (ii) the maximum (iron/nickel), (iii) the fission region. Explain why nuclei at the two extremes are less stable.

Solution

The curve.

Asymmetric bell-shaped trend: a steep rise for light nuclei, a peak at iron, a slow decline for heavy nuclei.

Why that shape.

• Initial rise (light nuclei). Nuclei with few nucleons have a high fraction of “surface” nucleons, only partially bound because they are surrounded by fewer neighbours. Adding nucleons rapidly increases the average binding energy: the curve rises steeply.
• Maximum (iron/nickel, $A \approx 56$). Around iron the trade-off between short-range nuclear attraction and Coulomb repulsion is optimal: $E_l/A \approx 8{,}8\ \text{MeV/nucleon}$. These are the most stable nuclei.
• Slow decline (heavy nuclei). As $A$ increases, so does the number of protons, and with it the long-range Coulomb repulsion between all pairs of protons. This progressively erodes the binding: the curve declines gently.

At the two extremes, then, nuclei are less stable for opposite reasons: light ones because of excess surface, heavy ones because of excess electric repulsion. It is precisely this shape that makes it possible to release energy both by fusing the former and by splitting the latter.

$\ev{\text{peak} \approx 8{,}8\ \text{MeV/nucleon near } A\approx 56}$

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Topics: Nuclear physics Concepts: Nuclear binding energy Skills: Reading graphs