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 $E_o$