Question

Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation $(r)$. Using this plot, show that the nuclear force is (i) attractive for $r > r_0$, and (ii) repulsive for $r < r_0$.

Hint:

Force is minus the slope of the U against r curve. The slope changes sign at the minimum at r0.

Answer 1

Step 1: Draw the plot (described in words):
Take the separation \(r\) along the horizontal axis and the potential energy \(U(r)\) of the pair of nucleons along the vertical axis. The curve has these features:
1. At a very small separation \(r\), the potential energy is large and positive. It falls steeply as \(r\) increases.
2. It becomes negative and reaches a minimum value (about \(-40\ \text{MeV}\)) at \(r = r_0\), where \(r_0 \approx 0.8\ \text{fm}\).
3. For \(r > r_0\), it rises gradually and approaches zero from below as \(r\) becomes large (beyond a few fm).
Mark \(r_0\) on the horizontal axis at the point of the minimum.

Step 2: Relate force to the slope:
The force between the nucleons is the negative slope of the potential energy curve:
\[ F = -\dfrac{dU}{dr} \]
A positive \(F\) means repulsion (the nucleons are pushed apart). A negative \(F\) means attraction (the nucleons are pulled together).

Step 3: Case (i) - r greater than r0:
To the right of the minimum, \(U\) increases with \(r\). So \(\dfrac{dU}{dr} > 0\) and \(F < 0\). The force is directed towards decreasing \(r\), so the nuclear force is attractive.

Step 4: Case (ii) - r less than r0:
To the left of the minimum, \(U\) decreases as \(r\) increases. So \(\dfrac{dU}{dr} < 0\) and \(F > 0\). The force is directed towards increasing \(r\), so the nuclear force is repulsive.
At \(r = r_0\), the slope is zero, so the force is zero and the pair is in stable equilibrium.

Final Answer:
The slope of the \(U\)-\(r\) curve is positive for \(r > r_0\) (attractive force) and negative for \(r < r_0\) (repulsive force). \[ \boxed{F = -\dfrac{dU}{dr}} \]