Animated Solution for Physics - Atoms and Nuclei: The below is a plot of binding energy per nucleon $E_b$, against the nuclear mass $M$; $A,B,C,D,E,F$ correspond to different nuclei. Consider four reactions (i) $A+B\to C+\epsilon$ (ii) $C\to A+B+\epsilon$ (iii) $D+E\to F+\epsilon$ and (iv) $F\to D+E+\epsilon$ where, $\epsilon$ is the energy released. In which reactions $\epsilon$ is positive?

Decoding the Binding Energy Curve

Imagine you have a map that tells you exactly which atomic nuclei are content and which ones are restless. In nuclear physics, this map is the Binding Energy per Nucleon ($E_b$) versus Mass Number ($M$) curve. It is one of the most profound graphs in all of science, dictating why stars shine and why nuclear reactors generate power.

The fundamental principle to remember is simple: Nature always seeks the lowest energy state, which corresponds to the highest stability. On our graph, the vertical axis represents the binding energy per nucleon, $E_b$. The higher a nucleus sits on this curve, the more tightly bound its nucleons are, and the more stable it is.

When a nuclear reaction occurs, energy (denoted by $\epsilon$) is released if and only if the products of the reaction are more stable than the reactants. Mathematically, this means the products must have a higher $E_b$ than the reactants. If the products are lower on the curve, the reaction is endothermic and requires an input of energy.

Let's look at the left side of the curve, where light nuclei like $A$ and $B$ reside. These nuclei have relatively low binding energies per nucleon.

If we force them together in a process called Nuclear Fusion:

We see that the resulting nucleus $C$ sits much higher on the curve. Because $E_b(C)>E_b(A)$ and $E_b(C)>E_b(B)$, the product is significantly more stable. The excess energy that was holding the less stable nuclei together is released into the surroundings. Therefore, for this reaction, $\epsilon$ is positive.

Now, let's shift our focus to the far right of the curve, where heavy nuclei like $F$ are found. These massive nuclei are somewhat unstable because the repulsive electrostatic force between their many protons begins to counteract the attractive strong nuclear force.

If a heavy nucleus $F$ splits into two lighter, intermediate-mass nuclei $D$ and $E$ in a process called Nuclear Fission:

We observe that the fragments $D$ and $E$ are located higher up on the curve, closer to the peak of maximum stability (near Iron-56). Since $E_b(D)$ and $E_b(E)$ are greater than $E_b(F)$, the products are more stable than the original heavy nucleus. Once again, this transition to a more stable state releases a tremendous amount of energy, making $\epsilon$ positive.

By simply reading the topography of the $E_b$ curve, we can predict the energetic nature of nuclear reactions. Reactions that climb up the curve—whether by fusing light nuclei (Reaction i) or splitting heavy ones (Reaction iv)—will always release energy. Thus, $\epsilon$ is positive for reactions (i) and (iv).