Question: The potential energy between two nucleons inside a nucleus is minimum at a distance of about:

Step 1 (Governing Law & Concept): [0.25 Mark]
The nuclear force between two nucleons is strongly attractive at distances larger than $r_0$ and strongly repulsive at distances smaller than $r_0$. The potential energy $U(r)$ of the two-nucleon system is defined by the nuclear potential curve, where the minimum potential energy corresponds to the equilibrium separation distance $r_0$.

Step 2 (Step-by-Step Analysis): [0.5 Mark]
According to the properties of nuclear forces as described in the NCERT Physics textbook (Chapter 13: Nuclei):

• The nuclear force is short-ranged and becomes negligible for distances $r > 10 \text{ fm}$.
• The potential energy $U(r)$ is negative (attractive) and reaches its minimum value at a specific separation distance.
• Experimental data and the nuclear potential energy graph indicate that this minimum occurs at approximately $r \approx 0.8 \text{ fm}$.

Step 3 (Final Concluded Answer): [0.25 Mark]
Comparing the theoretical value with the given options, the correct choice is (A) 0.8 fm.

💡 EL10 Concept Insight: Think of the nuclear force like a spring between two nucleons; they want to stay at a "sweet spot" distance of about 0.8 fm where they are most stable and have the lowest energy, pushing away if they get too close and pulling back if they drift too far apart.