Q.The potential energy between two nucleons inside a nucleus is minimum at a distance of about ______.
(A) 0.8 fm
(B) 1.6 fm
(C) 2.0 fm
(D) 2.8 fm

The binding energy curve is one of the most important graphs in nuclear physics, and understanding why it has the shape it does reveals the fundamental nature of the strong nuclear force.

Why binding energy per nucleon plateaus

The binding energy per nucleon, $\frac{B}{A}$, measures how tightly bound each nucleon is on average. For medium-mass nuclei (roughly $A = 30$ to $170$), this quantity hovers around $8$ to $8.8$ MeV per nucleon, staying remarkably constant. This plateau exists because of two competing effects:

The saturation property of nuclear forces. The strong nuclear force is extremely short-range—it acts only over distances of about $1$ to $2$ fm (roughly the size of a nucleon itself). Each nucleon in the interior of a nucleus interacts with only its nearest neighbors, typically $8$ to $12$ nucleons in a close-packed arrangement. Adding more nucleons to the nucleus doesn't increase the number of interactions per nucleon in the bulk; it just adds more nucleons experiencing the same local environment.

Think of it like a crowd of people holding hands: each person can only hold hands with their immediate neighbors. Doubling the crowd size doesn't double the number of hands each person holds—it stays constant.

Volume vs. surface effects. The total binding energy grows roughly as $A$ (volume term in the semi-empirical mass formula), while the number of nucleons is also $A$, so $\frac{B}{A}$ remains approximately constant. Surface nucleons are slightly less bound (they have fewer neighbors), but for large $A$, the fraction of surface nucleons becomes negligible.

For medium-mass nuclei, the volume term $a_V$ dominates, and the correction terms are small, yielding a nearly flat curve.

Evaluating the Assertion and Reason

1. Assertion (A): The binding energy per nucleon is practically constant for $30 < A < 170$.

   This is true. Experimental data shows that $\frac{B}{A}$ varies only slightly in this range, peaking near iron-56 at about $8.8$ MeV/nucleon and remaining within $\pm 5\%$ across the entire interval. This is the stable plateau region of the binding energy curve.

2. Reason (R): Nuclear forces are not short-range for $30 < A < 170$.

   This is false. Nuclear forces are always short-range, regardless of mass number. The range of the strong force is approximately $1$–$2$ fm and does not change with $A$. In fact, it is precisely because nuclear forces are short-range that the binding energy per nucleon saturates and remains constant. …

Why this distance matters

The nuclear force between two nucleons (protons or neutrons) is not like gravity or electromagnetism. It has a very short range — it only acts over distances comparable to the size of a nucleus itself. The key feature is that the force is strongly repulsive when nucleons are too close (to stop them from collapsing into each other) and attractive at slightly larger separations, binding the nucleus together.

The potential energy curve for the nucleon-nucleon interaction therefore has a deep minimum at some equilibrium separation. That minimum is the distance where the attractive and repulsive contributions balance — the "sweet spot" where a pair of nucleons would sit if undisturbed.

Step-by-step reasoning

1. Recall the shape of the nuclear potential.

   The nuclear force is modelled by potentials like the Yukawa potential or the Reid soft-core potential. All such models show a repulsive core below about $0.5\ \text{fm}$ and an attractive well that peaks (minimum potential) near $0.8\ \text{fm}$ to $1.0\ \text{fm}$.

2. Eliminate the larger distances.

   At $1.6\ \text{fm}$, the nuclear force is still attractive but much weaker — the potential is already climbing back toward zero. At $2.0\ \text{fm}$ and $2.8\ \text{fm}$, the interaction is negligible (the force range is about $2\ \text{fm}$). The minimum cannot be there because the potential is nearly zero, not a deep well.

3. Identify the correct value. …

The key to this question lies in understanding the range and nature of the strong nuclear force. Unlike the Coulomb force, which is long-range and can be either attractive or repulsive, the nuclear force has a very specific behaviour: it is strongly attractive at typical nucleon separations (around 1–2 fm), but becomes strongly repulsive when nucleons are pushed too close together (below about 0.5 fm). This repulsive core is what prevents nuclei from collapsing.

Let’s examine each statement carefully.

1. Statement (A): "They are stronger than Coulomb forces."

   This is true. At distances of about 1 fm, the nuclear force between two protons is roughly 100 times stronger than the electrostatic repulsion. That’s why nuclei with many protons can stay bound despite the Coulomb repulsion.

2. Statement (B): "They have about the same magnitude for different pairs of nucleons."

   This is also true. The strong nuclear force is charge-independent — it acts almost identically between two protons, two neutrons, or a proton and a neutron (ignoring the small electromagnetic correction). This is why the deuteron (p–n) and the diproton (p–p) would have similar binding if not for Coulomb repulsion.

3. Statement (C): "They are always attractive."

   This is false. While nuclear forces are attractive at typical internucleon distances, they become repulsive at very short separations (the “hard core”). This repulsion is essential to explain nuclear stability and the fact that nucleons do not coalesce into a point. So the force is not always attractive — it has a repulsive core.

4. Statement (D): "They saturate as the separation between two nucleons increases." …