Ionization Redefined

If the force between an orbiting electron and the nucleus is repulsive what then is ionization energy, $E_i$?

${E}_{i}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }(r_{ef}-r_{eo})}\right\}-\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}+G\cfrac { m_{ a }m_{ e } }{ r_{ e } }+\int^{\infty}_{r_e}{\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }} d\,r$

The terms are,

${m_ec^2}\left\{ln{( \cfrac{{r_{po} }}{{r}_{pf}}) +\cfrac { 1-{ r }_{ pf } }{ { r }_{ pf }{ r_{ po } } }(r_{pf}-r_{po})}\right\}$  

work done against the centripetal force, while the circular velocity remains constant.

$\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}$

work recovered from the repulsive force of the nucleus $B$ orbit.

$G\cfrac { m_{ a }m_{ e } }{ r_{ e } }$

work done against gravitational attraction.

and

$\int^{\infty}_{r_e}{\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }} d\,r$

$=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{\sqrt{4a^2_e}}=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{2a_e}$

work done against the attraction of neighboring electron $B$ orbits.

The work against centripetal force explodes immediately.  The centripetal force is provided by drag at terminal velocity.  Work done against the centripetal force is essentially work against drag.  The electron gains velocity initially as a result of electrostatic attraction.  Upon achieving light speed, drag curls its path into a helix. The electron perform circular motion in the plane perpendicular to its direction of travel.

To resolve this issue, we see that this work is zero for incremental change in $r_{ef}$, ie. when

$r_{ef}=r_{eo}+\Delta r$    as  $\Delta r\rightarrow0$

this work is zero.  In other words, move slowly lo.

And we are left with,

${E}_{i}=-\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}+G\cfrac { m_{ a }m_{ e } }{ r_{ e } }+\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{2a_e}$

$E_i=G\cfrac { m_{ a }m_{ e } }{ r_{ e } } +\cfrac { q^{ 2 } }{ 4\varepsilon _{ o } } \left\{ \cfrac{n_e}{2a_e} -\cfrac { 1 }{ r_{ e } }  \right\}$

which boldly is the new expression for ionization energy.  $n_e$ is the number of electrons in the stack minus one,  $a_e$ is the radius of an electron and  $m_a$ is the mass of the nucleus.