Continuing from the previous post "Another Take On Discrete",
\(n^{ 2 }ln(x)+Ax+C=0\)
It is possible that, if we define a size limit, \(x\ge a_\psi\) then,
\(n_o^{ 2 }ln(a_\psi)+Aa_\psi+C=0\)
and,
\(C=-n_o^{ 2 }ln(a_\psi)-Aa_\psi\)
Since
\(0\lt a_\psi\lt 1\)
in metric units, \(C\) can be positive or negative.
\(C=n_o^{ 2 }|ln(a_\psi)|-Aa_\psi\)
But what is the value for \(n_o\)? \(n_o=1\)? or \(n_o=2\)?...
Should this \(\psi\) wave be at the lowest energy possible? As such, \(n_o=1\). In which case,
\(C=|ln(a_\psi)|-Aa_\psi\)
We see that for smaller \(a_\psi\), \(C\) is larger. The corresponding value of \(f_{o}\) is folded up.
But what does it mean when \(f\) is folded to a higher value?
We are dealing with a single particle NOT in orbit around a nucleus. \(f\) being folded upwards means that \(\psi\) of a lower frequency are found further from the center than expected. In general, \(\psi\) waves further from the center have higher frequencies.
If all is this valid, it means the line spectra that we see are in part due freed ionized particles not in orbit around a nucleus. Energy transitions of particles in orbit around the nucleus produce a set of spectra lines and the freed particles produce another, superimposed on the first. Two processes are involved.
The freed particles are electrons.
This set of spectra lines due to electrons should be present on all observed spectrum, irrespective of the element being investigated. I think these lines do exist and are being attributed to hydrogen that supposedly cannot be gotten rid of.
Interesting indeed. Similar spectra lines should exist for all particles.
