Question

Draw a graph of potential energy between a pair of nucleons as a function of their separation.

Hint:

U is zero at large r, has a minimum of about -40 MeV at r0 = 0.8 fm, and rises steeply for r less than r0.

Answer 1

Step 1: Understanding the Concept:
Two nucleons attract each other through the strong nuclear force. The potential energy of the pair depends on their separation \(r\). We draw this curve from what we know about the nuclear force.

Step 2: Features of the graph:
1. For large \(r\) (more than about 3 to 4 fm), the nuclear force is negligible, so \(U \approx 0\).
2. As \(r\) decreases below a few fm, the nuclear force is attractive, so the potential energy becomes negative and falls with a steep slope.
3. At \(r = r_0 \approx 0.8\) fm, the potential energy is the minimum, about \(-40\) MeV. Here the force is zero, and the pair is most stable.
4. For \(r < r_0\), the force is repulsive (the hard core of the nucleons), so the potential energy rises very steeply, cuts the r-axis a little below \(r_0\) and goes to large positive values.

Step 3: How to draw it:
Draw the r-axis (in fm) horizontally and the potential energy \(U(r)\) (in MeV) vertically. Mark \(r_0 \approx 0.8\) fm on the r-axis.
Start from a large positive value on the left of the vertical axis, bring the curve down steeply, make it pass through zero just before \(r_0\), reach the lowest point \(U \approx -40\) MeV at \(r_0\), and let it rise slowly and flatten to meet the r-axis from below at large \(r\).

Final Answer:
The curve is steeply positive for \(r < r_0\), has a deep minimum at \(r_0 \approx 0.8\) fm, and approaches zero from below at large separation.
\[ \boxed{U_{min} \approx -40\ \text{MeV at } r_0 \approx 0.8\ \text{fm}} \]