Consider F_ pp , F_ pn , and F_ nn as the magnitudes of the net...

At a separation of $1 \text{ fm}$, the strong nuclear force between nucleons is attractive.

The strong nuclear force is charge-independent, meaning the magnitude of the nuclear force between any pair of nucleons is the same. Let this force be $F_N$.

For a proton-neutron pair and a neutron-neutron pair, the electrostatic force is zero. Thus, the net force is entirely due to the strong nuclear force:

$F_{pn} = F_{nn} = F_N$

For a proton-proton pair, in addition to the attractive strong nuclear force, there is a repulsive electrostatic force $F_E$ between the two positively charged protons.

Since the nuclear force is much stronger than the electrostatic force at this distance, the net force is attractive but its magnitude is reduced by the electrostatic repulsion:

$F_{pp} = F_N - F_E$

Therefore, we can conclude that $F_{pp} < F_{pn} = F_{nn}$.