\({E}_{s}={m_pc^2}\left\{ln{( \cfrac{{r_{po} }}{{r}_{pf}}) +\cfrac { 1-{ r }_{ pf } }{ { r }_{ pf }{ r_{ po } } }(r_{pf}-r_{po})}\right\}=-\cfrac{q^2}{4\varepsilon_o}\left\{\cfrac{1}{r_{ef}}-\cfrac{1}{r_{eo}}\right\}=\Delta PE_{re}\)
where \(r_{pf}\) and \(r_{pf}\) are the initial and final radius of the photon in its helical path, \(r_{eo}\) and \(r_{ef}\) are the initial and final radius of the electron orbit around the nucleus. The expression on the left was derived in the post "Light Dispersion?" from considering the work done along the centripetal force in moving the photon to a different orbital radius. The expression on the right is the potential energy change in a field of magnetic repulsive force \(F_L=-i\pi qE\), also the Lorentz's force between the electron and the nucleus. The negative sign indicates that a gain in energy in one equals the lost of energy in the other.
Up till now, the only postulation as to why a photon and an orbiting electron repel each other is thermal gravity between two hot bodies. If we examine the likely magnetic interaction between an orbiting electron and a passing photon,
We find that when the electron orbits are also in the same anti-clockwise sense looking onward to the direction of the approaching photon, the particles should be attracted to each other based on the their interacting magnetic fields alone. The electrons are pulled further from their orbits and the photons pulled down to a smaller radii. The electron on reaching high orbit is not going to to be ionized. The photon is still above the electron, the charge cannot escape the hold of the atom. Given the \(\cfrac{1}{r_e}\) relationship for \(PE_{re}\), the energy change from moving a distance \(\Delta r_e\) is always higher downwards than upwards. (When the photon has a smaller radius, the electrons are pulled down to smaller orbital radii. This can cause ionization.)
The orbits can of course be re-oriented to an clockwise sense, in which case the particles are like opposite current in parallel wires and be repelled from one another. This was the scenario proposed previously where the particles repel each other resulting in diffraction, ionization, glare, etc. (In the case of a smaller photon radial path the electrons are pushed upwards to higher orbit. This will not cause ionization.)
Note: From the above expression by setting \(r_{ef}\rightarrow\infty\) for ionization and \(r_{po}\lt r_{re}\), it is not possible to prove conclusively for all cases that the atom cannot be ionized by pushing the electron to higher orbit.
Since, both orientations are equally likely and all orientations in between, any photon at and beyond the threshold frequency (at the appropriate interaction radius \(r\)) on approach to an atom has only half the chance of ejecting a electron. As a whole, only half the photons would have a photoelectric effect. This conclusion is the same as we consider the relative size of the orbital radius and helical path radius.
If the interaction is purely electrostatic, we would expect the photon to eject the electron in both orientations and that all photons at the threshold frequency would cause a photoelectric effect.
We have assumed that the electron orbits re-orientate themselves slowly, not at the light speed that the photon is approaching. We have also ignored the \(B\) field due to the spinning positive charge which is always perpendicular to the \(B\) field due to the negative charge as they both travel down the helical path.
It is possible that the photon-electron interaction is limited to photon radius larger than the electron orbit, ie \(r_{ph}\gt r_e\). The following graph shows why pushing the electron down is likely to cause ionization.
Given a \(\cfrac{1}{r_e}\) profile for \(PE_{re}\), an equal distance upwards and downwards, the electron pushed downwards gain more energy.
\(\Delta PE_{ \Delta r- }>\Delta PE_{ \Delta r+ }\)
If this energy gain is greater than ionization energy, the electron will break the hold of the nucleus.
\(\Delta PE_{ \Delta r- }>\Delta PE_{ ion }\implies ionization\)
Pushing the electron out is not going to cause ionization unless \(\Delta r_e\rightarrow\infty\).
