So far photon was treated as a particle, not wave. Photons interaction with electrons however is electrostatic, at a distance. This might seems that photon acts like wave. What is more important is that photons are accelerated in the direction of a reverse E-field; attracted to a positive charge. It should be repelled by electrons.
Does a electron really gets knocked out? If it does what actually happens?
Consider a electron in circular motion
\({F}_{c}=-{m}_{e}\cfrac{c^2}{{r}_{e}}\)
Light speed is the maximum, if this is an phenomenon of nature then, if \({r}_{e}\) is made smaller without changing \(c\), a force should develop to resist such a change in \({r}_{e}\). That is
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=-\cfrac{\partial}{\partial {r}_{e}}\left({m}_{e}\cfrac{c^2}{{r}_{e}}\right)\)
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}} \) this force is in the same direction as \({r}_{e}\), when \({r}_{e}\) is compressed a force develops in the opposite direction, so
\({F}_{s}=\cfrac{\partial {F}_{c}}{\partial {r}_{e}}.\Delta {r}_{e}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\Delta {r}_{e}\)
where \(\Delta {r}_{e}\) is the small change in \({r}_{e}\) at \(r={r}_{e}\), in the negative \({r}_{e}\).
As the photon approaches, the electron from a position \(r={r}_{eo}\), is compressed to, \(r\rightarrow {r}_{ef}\), the electron behave just like a spring. After the photon passes, the stored energy is released and the electron is propelled upwards. If the store energy is greater than its potential energy, the electron leaves the pull of the nucleus. The atom is ionized.
Energy stored = \(\int^{{r}_{ef}}_{{r}_{eo}}{{F}_{s}}\partial {r}_{e}=\int^{{r}_{ef}}_{{r}_{eo}}{\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}(\partial {r}_{e})^2}\)
Energy stored, \(E_s ={ m }_{ e }{ c^{ 2 } }\int _{ {r}_{eo} }^{ { r }_{ ef } }{ (-\cfrac { 1 }{ { r }_{ e } } +C) \partial { r }_{ e } } \)
\(E_s ={ m }_{ e }{ c^{ 2 } }|\begin{matrix} { r }_{ ef } \\ {r}_{eo} \end{matrix}-ln({ r }_{ e })+C.{ r }_{ e }\)
\(E_s ={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\)
And ionization occurs when,
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\ge PE_e\) ----(*)
But what is potential energy of the electron orbiting at \({r}_{eo}\)? Potential Energy of the system is the energy stored when the electron is first placed at \(\infty\) and then moved to \({r}_{eo}\). ie. \({r}_{eo}\rightarrow \infty\) and \({r}_{ef}\rightarrow {r}_{eo}\)
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ eo}-{ r })\}}\) ----(**)
The idea is simply this, compress the spring to give it enough energy such that when it is release the electron will orbit at at a new orbit \(r \rightarrow\infty\). Consider (**) minus (*),
\({ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}) }+C(-{ r }_{ ef }+2{ r }_{ eo }-r)\} }\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}e^{C(-{ r }_{ ef }+2{ r }_{ eo }-r)}) }\} }\)
\(\because \lim _{ r\rightarrow \infty }{r{ e }^{ -Cr } }\rightarrow 0\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r }_{ ef } }{ r^{ 2 }_{ eo } } e^{ C(2{ r }_{ eo }-{ r }_{ ef }) }.r{ e }^{ -Cr } }\} } \rightarrow -\infty\)
ie. \(PE_s-E_s \lt 0\)
That is to say it is possible to ionize this way regardless of \({r}_{eo}\). That it is mathematically plausible. Given \({r}_{eo}\), there is an \({r}_{ef}\) to cause ionization.
