\(h=2\pi a_{ \psi\,e }m_ec\)
\(a_{ \psi\,e }=\cfrac{h}{2\pi}\cfrac{1}{m_ec}\)
\(a_{ \psi\,e }=\hbar\cfrac{1}{m_ec}=(1.054571726e-34)/(9.10938291e-31*299792458)\)
\(a_{ \psi\,e }=3.862e-13\) m
And \(a_{ \psi\,e }\) can be obtained numerically. This is a small value for one electron. This does not explain the repulsion of two negatively charged conductors.
Nor does this value match any of the four values obtained previously of basic particles. The assumption of \(n=1\) could be wrong in those calculations.
More importantly, beyond \(a_\psi\) an electron exert a small positive repulsive force.
In the case when \(x_z=a_\psi\) then the extend of \(\psi\) around the electron mass is,
\(r_{\psi\,e}=2a_\psi=7.724e-13\) m
where \(r_{\psi\,e}\) is the radius of a sphere around \(m_e\).
But \(m_e\) itself is a flat disc. Should \(\psi\) also be a flat disc?? If \(m_e\) is at the center of \(\psi\) then,
\(r_{\psi\,e}=a_\psi=3.862e-13\) m
And so, \(r_{\psi\,e}\) is between \(3.862e-13\) to \(7.724e-13 \) m.
