Unfortunately, there’s no simple formula to calculate the strong force or the potential energy of two nucleons interacting via the strong force. FIGURE 42.8 is an experimentally determined potential-energy diagram for two interacting nucleons, with $r$ the distance between their centers. The potential-energy minimum at $r\approx 1\text{fm}$ is a point of stable equilibrium.

FIGURE 42.8 The potential-energy diagram for two nucleons interacting via the strong force.

Recall that the force is the negative of the slope of a potential-energy diagram. The steeply rising potential for $r<1\text{fm}$ represents a strongly repulsive force. That is, the nucleon “cores” strongly repel each other if they get too close together. The force is attractive for $r>1\text{fm,}$ where the slope is positive, and it is strongest where the slope is steepest, at $r\approx 1.5\text{fm}$. The strength of the force quickly decreases for $r>1.5\text{fm}$ and is zero for $r>3\text{fm}$. That is, the strong force represented by this potential energy is effective only over a very short range of distances.

Notice how small the electrostatic energy of two protons is in comparison to the potential energy of the strong force. At $r\approx 1.0\text{fm,}$ the point of stable equilibrium, the magnitude of the nuclear potential energy is $\approx 100$ times larger than the electrostatic potential energy.

We earlier asked what role neutrons play. Why does a nucleus need neutrons? The answer is related to the short range of the strong force. All protons in the nucleus exert repulsive electrostatic forces on each other, but, because of the short range of the strong force, a proton feels an attractive force only from the very few other nucleons with which it is in close contact. Even though the strong force at its maximum is much larger than the electrostatic force, there wouldn’t be enough attractive nuclear bonds for an all-proton nucleus to be stable. Because neutrons participate in the strong force but exert no repulsive forces, the neutrons provide the extra “glue” that holds the nucleus together. In small nuclei, where most nucleons are in contact, one neutron per proton is sufficient for stability. Hence small nuclei have $N\approx Z$. But as the nucleus grows, the repulsive force increases faster than the binding energy. More neutrons are needed for stability, causing heavy nuclei to have $N>Z$.