Taking the$\mathit{n}\mathbf{=}\mathbf{3}$states as representative, explain the relationship between the complexity numbers of nodes and antinodes-of hydrogen's standing waves in the radial direction and their complexity in the angular direction at a given value of n. Is it a direct or inverse relationship, and why?

Nodes are the places where the quantum mechanical wave function $\psi$changes its phase. Since, it changes phase from positive to negative or vice-versa, it is equal to 0 at the nodes. Hence, its square${\psi }^2$ is also zero at nodes, which is also called electron density. Hence at nodes, probability of finding electrons is zero.

As given in the Figure 7.15, for$n=3$ there is only one radial antinode in d and the has three radial antinodes. d has multiple angular antinodes, while s has no angular node at all.

It seems, the angular and radial complexities are inversely related.

For a fixed$n=3$ , the energy of the orbitals should be same.

Hence, to balance out the total energy of orbitals, if radial energy/complexity increases, the angular energy/complexity decreases, and vice – versa. That’s how they are inversely proportional.