Taking the 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 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 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 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 , 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.
Consider a cubic 3D infinite well.
(a) How many different wave functions have the same energy as the one for which ?
(b) Into how many different energy levels would this level split if the length of one side were increased by ?
(c) Make a scale diagram, similar to Figure 3, illustrating the energy splitting of the previously degenerate wave functions.
(d) Is there any degeneracy left? If so, how might it be “destroyed”?
In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave, , where is the length of its home. Given that , we would have , and the kinetic energy, , would thus be . These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: . Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain
for the quantized energies?
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