\(a_\psi=19.34\,nm\)
\(a_\psi=16.32\,nm\)
\(a_\psi=15.48\, nm\)
\(a_\psi=14.77\,nm\)
If,
\(a_\psi=a_{\psi\,c}=19.34\,nm\)
\(a_\psi=a_{\psi\,c}=16.32\,nm\)
\(a_\psi=a_{\psi\,c}=15.48\, nm\)
\(a_\psi=a_{\psi\,c}=14.77\,nm\)
\(a_\psi=a_{\psi\,c}=16.32\,nm\)
\(a_\psi=a_{\psi\,c}=15.48\, nm\)
\(a_\psi=a_{\psi\,c}=14.77\,nm\)
in which case we have,
\(n.\cfrac{4}{3}\pi \left(a_{\psi\,c}\right)^3=\cfrac{4}{3}\pi \left(a_{\psi\,n}\right)^3\)
where \(1\le n\le 77\) and
\(a_{\psi\,c}=a_{\psi\,n=1}\)
That is to say, \(n\) small particles of radius \(a_{\psi\,c}\) coalesce into a bigger particle of \(a_{\psi\,n}\).
\(a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}\)
where \(n=1,\,2,\,3,..77\)
If each of this particle is responsible for a spectra line then, a spectra series due to one type of particle will line up nicely on a \(y=\sqrt[3]{n}\) plot with a common scaling factor. An the maximum number of stable lines due to stable particle, observable or otherwise is \(77\) or \(78\). Unstable particles that grows beyond the plateau on the \(\psi\) vs \(r\) graph where \(\psi\) pinch off with decreasing force will also result in faint spectra line.
Just speculating.
