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Q13CQ

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Found in: Page 278

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

Taking the${\mathbit{n}}{\mathbf{=}}{\mathbf{3}}$ 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.

See the step by step solution

Step 1:  Nodes

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.

Step 2: Explanation

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.