Animated Solution for Physics - Atoms and Nuclei: Assume that the nuclear binding energy per nucleon $(B/A)$ versus mass number $(A)$ is as shown in the figure. Use this plot to choose the correct choice(s) given below.

Binding Energy

Before we dive into the graph, let's establish the golden rule of nuclear reactions: Energy is released only if the products are more tightly bound than the reactants.

Think of binding energy as a measure of stability. The higher the binding energy per nucleon ($B/A$), the more stable the nucleus. Nature always prefers stability. So, if a nuclear process (like fusion or fission) results in a shift from a lower $B/A$ to a higher $B/A$, the excess energy is radiated away. Mathematically, the $Q$-value of the reaction must be positive:

Remember, the graph gives us the binding energy per nucleon. To find the total binding energy, we must multiply this value by the total number of nucleons, which is the mass number $A$.

The graph provided is a simplified, piecewise model of the actual binding energy curve. It has three distinct regions: 1. For $A<100$, the $B/A$ is a constant $2\text{ MeV}$. 2. For $100<A<200$, the $B/A$ jumps to a constant $8\text{ MeV}$. 3. For $A>200$, the $B/A$ drops to a constant $6\text{ MeV}$.

Let's use this model to evaluate each option.

The Futile Fusion

Imagine fusing two identical nuclei with mass numbers in the range $1<A<50$. Let's call the mass of each reactant $A_1$. The fused product will have a mass of $2A_1$.

Since $A_1<50$, the product mass $2A_1$ will be strictly less than 100. Looking at our graph, both the reactants and the product fall into the first region where $B/A=2\text{ MeV}$.

Let's calculate the $Q$-value:

Because the binding energy per nucleon didn't change, the total binding energy is perfectly conserved. No extra energy is released. Thus, option (a) is incorrect.

The Power of Stars

Now, let's fuse two nuclei from the range $51<A<100$. Again, let the reactant mass be $A_1$. The product mass is $2A_1$.

This time, $102<2A_1<200$. The reactants are in the $2\text{ MeV}$ region, but the product has crossed the threshold into the highly stable $8\text{ MeV}$ region!

Let's calculate the $Q$-value:

Since $A_1$ is positive, $Q>0$. A massive amount of energy is released because the nucleons have settled into a much more stable configuration. Option (b) is correct.

The Failed Fission

What if we take a nucleus from the highly stable middle region ($100<A<200$) and split it into two equal fragments? Let the parent mass be $A_1$. The fragments will each have a mass of $A_1/2$.

Since $100<A_1<200$, the fragments will fall into the range $50<A_1/2<100$. The parent nucleus had a $B/A$ of $8\text{ MeV}$, but the fragments drop down to the $2\text{ MeV}$ region.

Let's calculate the $Q$-value:

Here, $Q<0$. The reaction is endothermic. You would have to supply energy to break this stable nucleus apart. It will not release energy spontaneously. Option (c) is incorrect.

The Reactor's Core

Finally, let's look at heavy nuclei in the range $200<A<260$. If one of these splits into two equal fragments, the fragments will have masses between 100 and 130.

The parent nucleus is in the $6\text{ MeV}$ region, but the fragments land perfectly in the most stable $8\text{ MeV}$ region.

Let's calculate the $Q$-value:

Since $Q>0$, energy is released. The heavy, slightly unstable nucleus splits into two highly stable medium-mass nuclei, radiating the excess binding energy. This is the fundamental principle behind nuclear fission reactors! Option (d) is correct.

By systematically applying the rule of total binding energy to the provided piecewise graph, we have proven that energy is released during the fusion of medium-light nuclei and the fission of heavy nuclei. The correct choices are (b) and (d).