Simply \(\hbar\)

Consider,

\(E_{\Delta h}=mc^2(\cfrac{a_{\psi\,f}}{a_{\psi\,i}}-1)\)

In the case of a photon, \(a_\psi\) is the radius of its helical path.  This shows that a change in radius is accompanied by an exchange of energy.

and

\(2\pi a_{\psi\,n}= \lambda_{n}\)

So,

\(E_{ \Delta h }=mc^{ 2 }(\cfrac { \lambda _{ f } }{ \lambda _{ i } } -1)\)

\( E_{ \Delta h }=mc^{ 2 }(\cfrac { f_{ i } }{ f_{ f } } -1)\)

Which suggests a photon can change frequency, given an appropriate energy input, \(E_{ \Delta h }\).  If \( E_{ \Delta h }\) resulted in an emission of a photon,

\(h_ff_p=-E_{ \Delta h }=mc^{ 2 }(1-\cfrac { f_{ i } }{ f_{ f } })\)

\(\lambda_fmcf_p=\cfrac{1}{f_f}mc^2f_p=mc^{ 2 }(1-\cfrac { f_{ i } }{ f_{ f } })\)

\(f_p=f_f(1-\cfrac { f_{ i } }{ f_{ f } })=f_f-f_i\)

Surprise! Surprise!  This suggests that an emission of a photon occurs only if the lower energy states have been folded up, where a lower energy state actually have higher frequency.  This is illustrated on the plot below,

Frequency Intersections

The curve marked \(n=1\) in red, is folded up and has a higher intersection for frequency with the line \(y=x\).  Transition from all other states with lower frequencies will result in an emission of a photon. Transition from states \(n=10\) however, will not result in a photon emission because the associated frequency is higher than that for state \(n=1\).

And,

\(\cfrac{1}{\lambda_p}=\cfrac{1}{\lambda_f}-\cfrac{1}{\lambda_i}\)

for photon emission.

Since,

\(h_if_i=mc^2\)

\(h_ff_f=mc^2\)

\( E_{ \Delta h }=mc^{ 2 }(\cfrac { h_{ f } }{ h_{ i } } -1)\)

which relates the change in energy as a result of a change in \(h_n\) directly.

We still don't have justifications for \(E_o\)