\(\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}\)
and so we have,
\(a_{\psi\,new}=0.2063a_{\psi}\)
thus assuming that the first spectra line is of double intensity, ie from the pair \((n_2=2,n_1=1)\),
\(a_\psi=0.2063*19.34=3.99\,nm\)
\(a_\psi=0.2063*16.32=3.37\,nm\)
\(a_\psi=0.2063*15.48=3.19\, nm\)
\(a_\psi=0.2063*14.77=3.05\,nm\)
If, the first spectra line is the result of \((n_{\small{large}},1)\),
\(a_\psi=0.762*19.34=14.74\,nm\)
\(a_\psi=0.762*16.32=12.44\,nm\)
\(a_\psi=0.762*15.48=11.80\, nm\)
\(a_\psi=0.762*14.77=11.25\,nm\)
Which is just a bunch of numbers. However, both cases point to the fact that the calculated \(f_{\psi}\) from Planck's relation, \(\Delta E=h.f_{\psi}\), is not the frequency \(f_{\psi\,c}\) of \(\psi\), to resonate the particle with. \(f_{\psi\,c}\) is defined as,
\(f_{\psi\,c}=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}\)
where there is one wavelength \(m=1\) around the circular path of \(\psi\) with radius \(a_{\psi\,c}\) .
Depending on which spectra line is used to derive \(f_{\psi}\), a factor of \(0.2063\) or \(0.762\) applies. In the case where both the double intensity emission spectra line and the first absorption spectra line are used, with the appropriate factor, there can be only one value for \(f_{\psi\,c}\).
Where do all these lead us?
\(\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\right)\)
The murder of Rydberg constant?
\(\cfrac{1}{\lambda_{2\rightarrow 1}}=R_{\small{H}}\left(\cfrac{1}{n_1^2}-\cfrac{1}{n_2^2}\right)\)
Comparing the two expressions,
\(R_{\small{H}}=\cfrac{1}{2\pi a_{\psi\,c}}\)
which is a constant given \(a_{\psi\,c}\). Maybe...
