Even in the stability of solids (i.e.) why things don't pass through each) relies on the Pauli exclusion principle (and not just Coulomb repulsion as is commonly thought).
> In some ways it does not provide the best description of the electron distribution, since the region around r= 0, where the wavefunction has its largest values, is a relatively small fraction of the volume accessible to the electron. Larger radii represent larger physical regions since.
If you were to do any kinds of measurements, you are most likely to find it at the Bohr radius, not closer to the nucleus.
What phenomenon would make them lose energy?
This gif is for a 1d oscillation rather than an orbit, but it does a good job of conveying the idea.
https://thumbs.gfycat.com/ExhaustedAlarmedDeinonychus-size_r...
Interestingly enough, if the electron were a continuous ring (like Saturn's rings) rather than a point charge, it could orbit without losing energy, generating only a static magnetic field. I don't know why this model rarely comes up in explorations of classical physics models of the atom.
I'd say the intuition here is misleading. S orbitals (including the ground state) really are spherically symmetric. There are no waves "around" the orbital: they only vary in the radial direction.
But "must smoothly connect" is important. As you walk outwards from the nucleus, the electron wavefunction oscillates until it reaches the classical turning point, when it switches to exponential decay. Only a very few functions can smoothly join these two modes, which is one explanation for why bound state energies are quantized.
In that case the drawing was 2-dimensional, but the wave function was defined as a third dimension (it was orthogonal to the plane in which the electron was orbiting the nucleus).
It was reasonably clear to me that either the size of the wave was incredibly small compared to the size of the orbital or it was oscillating in some other dimension.