Like Wave, Like Particle, Not Attracted to Electrons II

From the previous post "Like Wave, Like Particle, Not Attracted to Electrons", the energy in required in moving from $r$ at $\infty$ to ${r}_{eo}$ is

$PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r}}{{r_{eo} }})}+C({ r }_{ eo}-{ r })\}}$

Energy required in moving from $\infty$ to ${r}_{ef}$ is

$PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r}}{{r_{ef} }})}+C({ r }_{ ef}-{ r })\}}$

So, the energy required in moving the electron from from ${r}_{eo}$ to ${r}_{ef}$ is

$E_s =PE_e({r}_{ef})-PE_e({r}_{eo})$

$E_s={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r }}{{r}_{ef}})}-ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ ef}-{ r })-C({ r }_{ eo}-{ r })\}}$

${E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef }-{ r }_{ e o})\}$

which is the same expression as ${E}_{s}$ as before.

From previous calculation of atomic radius we find that, the centripetal force is of the form,

$\cfrac{{m}_{e}c^2}{{r}_{eo}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{eo}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}$

and

$\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}(3-\cfrac{\sqrt{3}}{3})$

In general,

$\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}.A$

where $A$ is a numerical constant dependent on the configuration of the electrons around the positive charge.  Using this expression to bring a charge from infinity to its final configuration position along the line joining it and the positive center.

$PE=-A\int^{{r}_{eo}}_{r\rightarrow\infty}{ \cfrac{-q^2}{4\pi\varepsilon_o{r}^2}d r}$

$PE= -\cfrac{q^2A}{4\pi\varepsilon_o}\{ \cfrac{1}{r}\}|^{{r}_{eo}}_{r\rightarrow\infty}$

$PE= -\cfrac{q^2A}{4\pi\varepsilon_o} \cfrac{1}{{r}_{eo}}$

The negative sign indicates that this potential is lower than that at $\infty$ which is zero.  Numerically, the energy stored when the electron is pushed to a lower orbit, (photon and electron repel) is equal to this PE when the electron is ejected (ie the atom is ionized) after the photon has passed.

$E_s ={ m }_{ e }c^{ 2 }\{ ln{(  \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef}-{ r }_{ eo})\}=\cfrac{q^2A}{4\pi\varepsilon_o} \frac{1}{{r}_{eo}}$

$ln{ ( \cfrac{{r }_{ e o}} {{r}_{ef}}e^{ C({ r }_{ ef }-{ r }_{ eo }) } })=\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }$

$\cfrac{{r }_{ e o}}{{r}_{ef}} e^{ C({ r }_{ ef }-{ r }_{ eo }) } =e^{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }}$

${ r }_{ ef }e^{ -C.{ r }_{ ef } }={ r }_{ eo }e^{ -C.{ r }_{ eo } }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }  }$

because of the $e^{ -\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }  }\lt 1$    factor,  and ${x}{e^{-Cx}}$ is monotonous increasing for $x \lt 1$ for all values of $C$.    ${r}_{ef}\lt {r}_{eo}$ for ${r}\lt 1$.  Which is an acceptable result.

Given an element the constant $A$ is determined from distribution of electrons around the positive center.  The factor $C$....