Don't Worry, Be Creepy

It does not quite matter, given

$f(n)=\left(\cfrac{\sqrt[3]{n}-{1}}{\sqrt[3]{n}}\right)$

n	f(n)
70	0.7573572497
71	0.758501808
72	0.7596250716
73	0.7607277244
74	0.7618104192
75	0.7628737797
76	0.7639184021
77	0.7649448568
78	0.7659536896
79	0.7669454232

The value of $f(n)$ for the range $70\le n\le79$ average to

$f(n)=0.762$

from which we may estimate $f_{\psi\,c}$ from,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

where $n_1=1$

from the post "Sizing Them Up Again..." dated 4 Oct 2016.

But what is $n$ or $n_{\small{large}}$?

Better yet since the emission spectra line as the result of  $(n_2=2,\,n_1=1)$, (two basic particles, $n=1$ coalesce) that produces a pair of photons, $\Delta E_{1\rightarrow 2}$ has double the intensity and hence readily identifiable,

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-\sqrt[3]{1}}{\sqrt[3]{1}\sqrt[3]{2}}\right)$

where $n_1=1$ and $n_2=2$.

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}}\right)$

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

without worry about what $n_{\small{large}}$ might be.

This is different from Planck's relation $E=h.f$.  A factor of about $\frac{1}{5}$ creeps in.