Question

Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. With the help of the plot, explain why the nuclear force is attractive for $r > r_0$ and repulsive for $r < r_0$.

Hint:

Plot a potential well with its minimum at \(r_0\). Use \(F = -dU/dr\) and read the sign of the slope on each side of the minimum.

Answer 1

Step 1: Understanding the Concept:
The nuclear force acts between nucleons. Its potential energy \(U\) depends on the separation \(r\). The force follows from the slope of the curve: \(F = -\frac{dU}{dr}\).

Step 2: Draw the plot (described in words):
Draw \(r\) on the horizontal axis and \(U(r)\) on the vertical axis.
For large \(r\) (beyond about 3 to 4 fm) the curve is almost on the axis, at \(U \approx 0\). As \(r\) decreases, the curve dips below zero and reaches a minimum at \(r = r_0\), where \(r_0 \approx 0.8\ \text{fm}\). Below this point the curve rises very steeply and becomes positive (a strong repulsive core). Mark \(r_0\) on the axis and mark the minimum value of \(U\) of about \(-40\ \text{MeV}\).

Step 3: Explain the force for \(r > r_0\):
Here the slope \(\frac{dU}{dr}\) is positive, because \(U\) rises from the minimum towards zero as \(r\) grows.
\[ F = -\frac{dU}{dr} < 0 \]
A negative force acts towards smaller \(r\). It pulls the nucleons closer, so the force is attractive. Also, the potential energy is lower when they are closer, so they tend to come together.

Step 4: Explain the force for \(r < r_0\):
Here the slope \(\frac{dU}{dr}\) is negative, because \(U\) falls steeply as \(r\) increases towards \(r_0\).
\[ F = -\frac{dU}{dr} > 0 \]
A positive force acts towards larger \(r\). It pushes the nucleons apart, so the force is repulsive. At \(r = r_0\) the slope is zero, so the force is zero and the nucleons are in stable equilibrium.

Final Answer:
The potential energy has a minimum at \(r_0\). The force is attractive for \(r > r_0\) (positive slope) and repulsive for \(r < r_0\) (negative slope). \[ \boxed{F = -\frac{dU}{dr}} \]