Looking for Murder

From the previous post "" dated 13 Oct 2016,

$\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}$

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...