Hartree energy is defined as
\(E_H=2R_Hhc\)
From the post "Looking for Murder" dated 13 Oct 2018 it was proposed that,
\(R_H=\cfrac{1}{2\pi a_{\psi\,c}}\)
where \(R_H\) is the Rydberg constant
So,
\(E_H=2*\cfrac{hc}{2\pi a_{\psi\,c}}\)
but
\(f=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}\)
as
\(\lambda_{\psi\,c}=2\pi a_{\psi\,c}\)
we have,
\(E_H=2*hf_{\psi\,c}\) --- (*)
which only confirms that the prposed definition for \(R_H\) in the post "Looking for Murder" dated 13 Oct 2018 is dimensionally correct.
In the definition for Harttree Energy, this energy is approximately the electric potential energy of the hydrogen atom in its ground state and, by the virial theorem, approximately twice its ionization energy; the relationships are not exact because of the finite mass of the nucleus of the hydrogen atom and relativistic corrections.
\(a_{\psi}=14.77\,nm\) is a particle defined by \(\psi\) in circular motion; Hartree Energy sees an electron in orbit around a hydrogen nucleus.
This contention is not new and has been resolved by thinking that the electron is a wave going in orbit.
The posts presented here however, takes the view that the electron is made up of a \(\psi\) wave warped around a sphere. A \(\psi\) wave in circular motion. Enegry changes of this wave result in the spectral lines. The electron need not be in orbit around the hydrogen nucleus.
The size of a hydrogen atom is in the range of \(\times 10^{-15}\) meters, \(fm\),
but a consistent \(a_{\psi\,c}\), according to (*) is in the range of tenth of \(\times 10^{-9}\).
An electron at \(a_{\psi\,c}\) from the hydrogen nucleus cannot be at the ground state.
A particle, however, can have a force field around it up to \(\times10^{-9}\) meters.
Good night...
