Understanding Nucleon Force and Potential Energy with Separation

The force between a pair of nucleons (like protons and neutrons) is described by the nuclear force. This force is often visualized by plotting the potential energy \(V\) of the pair as a function of their separation distance \(r\). The relationship between the force \(F(r)\) and the potential energy \(V(r)\) is given by:

\[F(r) = -\frac{dV}{dr}\]

This equation tells us that the force is related to the negative of the slope of the potential energy curve at a given separation \(r\). Let's break down what the slope tells us about the force:

• If the slope \(\frac{dV}{dr}\) is positive (\(V\) increases as \(r\) increases), then \(F(r)\) is negative. A negative force indicates an attractive force.
• If the slope \(\frac{dV}{dr}\) is negative (\(V\) decreases as \(r\) increases), then \(F(r)\) is positive. A positive force indicates a repulsive force.

The typical potential energy curve for a pair of nucleons has a characteristic shape:

• At very small separations (less than about 0.4 fm), the potential energy is very high and positive, indicating a strong repulsive force (steep negative slope \(\frac{dV}{dr}\)). This is the repulsive core.
• At intermediate separations (around 0.8 fm to 2 fm), the potential energy is negative and reaches a minimum, indicating a strong attractive force (positive slope \(\frac{dV}{dr}\) on the outer side of the minimum, and negative slope \(\frac{dV}{dr}\) on the inner side of the minimum, with zero force at the minimum). The force is most strongly attractive near the minimum.
• At larger separations (greater than a few fm), the potential energy quickly approaches zero, indicating that the nuclear force becomes negligible. In the region where the potential energy is negative and increasing towards zero, the slope \(\frac{dV}{dr}\) is positive, meaning the force is attractive.

Now let's analyze the given options based on this understanding of the nucleon-nucleon potential energy curve:

Option 1: The force is attractive when separation between them is greater than 0.8 fm.

Looking at the standard nucleon-nucleon potential curve, at separations greater than approximately 0.8 fm (which is around the region of minimum potential energy), the potential energy \(V(r)\) is negative and increases towards zero as the separation \(r\) increases. This means the slope \(\frac{dV}{dr}\) is positive in this region. Since \(F(r) = -\frac{dV}{dr}\), a positive slope corresponds to a negative force, which is an attractive force. This statement aligns with the behavior of the nuclear force at these separations.

Option 2: The force is repulsive when separation between them is greater than 0.5 fm.

While the force is repulsive at very small distances (less than ~0.4 fm), for separations greater than 0.5 fm, the force is primarily attractive, especially in the region around the potential well minimum (near 0.8 fm) and extending outwards. So, this statement is incorrect.

Option 3: The force is attractive when separation between them is less than 0.8 fm.

For separations less than 0.8 fm, particularly at very small distances (e.g., less than 0.4 fm), the force is strongly repulsive due to the repulsive core. Therefore, the force is not attractive for all separations less than 0.8 fm. It is attractive in a range of distances leading up to the minimum potential (around 0.8 fm), but not for very small distances.

Option 4: The force is independent of their separation.

This is incorrect. The potential energy curve explicitly shows that the interaction energy, and thus the force, varies significantly with the separation distance between the nucleons. The nuclear force is highly dependent on separation.

Based on the analysis of the typical nucleon-nucleon potential energy curve and the relationship between force and potential energy, the statement that the force is attractive when separation between them is greater than 0.8 fm accurately describes the behavior of the nuclear force in that region.

Revision Table: Nucleon Potential Energy and Force

Additional Information: Properties of Nuclear Force

The nuclear force is the force that binds protons and neutrons together in the nucleus. It has several key properties:

• Strongest Fundamental Force: It is the strongest of the four fundamental forces over nuclear distances.
• Short Range: It is a very short-range force, effective only up to distances of a few femtometers (\(1 \text{ fm} = 10^{-15} \text{ m}\)). Beyond this range, it drops off rapidly.
• Attractive and Repulsive Components: As seen from the potential energy curve, it is attractive at intermediate distances but becomes strongly repulsive at very short distances.
• Charge Independent: The nuclear force between two protons, a proton and a neutron, and two neutrons is approximately the same, ignoring electromagnetic effects (Coulomb force) between protons.
• Spin Dependent: The strength of the nuclear force depends on the relative orientation of the spins of the interacting nucleons.
• Non-Central (Tensor) Component: Part of the nuclear force does not act purely along the line connecting the two nucleons.
• Saturation: Each nucleon in a nucleus interacts only with a limited number of its nearest neighbors, rather than with all other nucleons. This property is related to the constant binding energy per nucleon for most nuclei.