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Q13CQ

Expert-verifiedFound in: Page 278

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**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?**

The complexity in radial direction and angular direction are inversely proportional to each other because to balance out the total energy of orbital’s, if radial energy/complexity increases, the angular energy/complexity decreases, and vice – versa.

**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.

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