along the line joining the spin orbit center and the positive particle center. The photon is being modeled as a dipole with a negative spinning end. The emitted photon need not pass through the proton, as the proton is in orbit and has moved. As both particles are relaxed after the photoelectric photon has passed, both return to their initial states, leave behind an energy shadow of a dipole with spinning negative field at one end and a positive field at the other. This energy shadow of a dipole is self propelled. It moves in direction of the positive end, perpendicular to the plane of the negative particle spin.
\(P_{e^{-}}\) is static calculation. We consider the situation to be the superposition of two parts, an electron orbiting around a proton at light speed, and a displacement of the center of this orbit along a line joining the proton and the center of the other orbit of the orbit pair. As the photoelectric photon passes the electron (orbiting at constant, light speed) is pushed towards the proton. From the post "Like Wave, Like Particle, Not Attracted to Electrons" dated 08 Jun 2014, the energy stored is,
\(E_s ={ m }_{ e }c^{ 2 } \{ln{( \cfrac{r_s}{{r }_{ ph}})}+C({ r}_{ ph }-{ r_s })\}\) --- (*)
The energy stored due to the displacement of the electron orbit center from the proton center is much more difficult because of the second weak field that holds the proton in place. Both weak fields are orientated to cancel each other. The resultant weak field is difficult to quantify.
If we consider the effect of the passing photon is to shield the electron completely from the distance weak field, then the negative particle returned to its original position around the proton, ie zero displacement, \(\small{d_{ph}=0}\). Then the energy stored due to displacement, \(E_d\) is the energy required to move the electron orbit to its normal position when the orbital separation is \(d\).
This should be half the bond formation energy as we are dealing with only one of the two pairs of particles involved. Where bond formation is assumed to be simply, two unpaired orbits moving into parallel position of separation \(d\).
\(E_d=\cfrac{1}{2}E_b\)
where \(E_b\) is the bond formation energy. So,
\(E_P=E_s+\cfrac{1}{2}E_b\)
where \(E_s\) is given by (*).
The resultant weak field has first to be quantified for an algebraic expression for \(E_d\). This is an simplification, because \(r_s\) would have changed as the particle is pushed towards the proton along the line joining the center of the other orbit and the proton. But this discrepancies is captured by \(E_b\), assuming that the electron is returned to its initial position as before the orbits are paired. The effect of the passing photoelectric photon is to unpair the paired orbits locally. In general, to include situation where such unpairing does not happen completely,
\(E_P\le E_s+\cfrac{1}{2}E_b\)
Have a nice day.
Note:
\(E_s ={ m }_{ e }c^{ 2 } \{ln{( \cfrac{r_s}{{r }_{ ph}})}+C({ r}_{ ph }-{ r_s })\}\\={ m }_{ e }c^{ 2 } \{f(r_s)-f({ r}_{ ph })\}\\={ m }_{ e }c^{ 2 } \{C_o-f({ r}_{ ph })\}\)
where, \(f(r)=ln(r)-C.r\) and \(C_o\) a constant for a given \(r_s\).
Illustrative plot of \(f(r)=ln(r)-C.r\) where \(C\lt0\).
