From the post "de Broglie Per Person"
\(2\pi a_{ \psi }=n\lambda\)
\(h=2\pi a_{ \psi }mc=mc.n\lambda\)
with \(n=1\).
\(2\pi a_{ \psi }=\lambda_1\)
\(h=2\pi a_{ \psi }mc=mc.\lambda_{1}\) --- (*)
This is the Planck's constant in the expression \(E=hf\).
In general however, from the post "Other Time Dimensions Counts"
\(h=x.mv=x.mc\)
where \(x\) is the length of periodic path of the particle, \(m\) is the mass of the particle and \(v=c\) is the particle's velocity at light speed.
From the post "Quantized x, Discretely"
\(x=e^{ -\cfrac { Ax+C }{ n^{ 2 } } }\)
Consider other solutions to \(a_\psi\) subjected to the constraint above,
\(a_{\psi\,n}=e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } } }\)
then
\(2\pi a_{\psi\,n}=2\pi e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } } }= \lambda_{n}\)
specifically,
\(2\pi a_{\psi,1}=2\pi e^{ -\left(Aa_{\psi,1}+C\right) }= \lambda_{1}\)
\(\cfrac{\lambda_1}{\lambda_{n}}= e^{ -\left(Aa_{\psi,1}+C\right)+\cfrac { Aa_{ \psi \, n }+C }{ n^{ 2 } } }\) --- (*)
since we know from the post "H Bar And No Bar", that all harmonics around \(2\pi a_{ \psi\,n }\) have the same energy, we will only consider the fundamental frequency at \(2\pi a_{\psi\,n}=\lambda_n\).
In which case, when we formulate,
\(h=2\pi a_{\psi\,n}.mc=\lambda_n.mc\)
and
\(E=hf_n=\lambda_n.mc.f_n\)
as \(c=f_{n}\lambda_{n}\)
\(E=mc^2\)
All photons has the same energy but their \(a_\psi\) is different. From a early post in May 2014 "Wave Front and Wave Back", \(r=a_\psi\), photons with different \(a_\psi\)s will be refracted to different angles.
But if we have mistaken and instead formulate,
\(h_o=2\pi a_{ \psi }mc\)
where \(h_o\) is the Planck constant as defined.
\(E=h_of_n=2\pi a_{ \psi }.mc.f_n=\lambda_{1}mc.\cfrac{c}{\lambda_n}\)
from (*),
\(E=mc^2 e^{ -\left(Aa_{\psi,1}+C\right)+ \cfrac { Aa_{ \psi \, n }+C }{ n^{ 2 } } } \)
Since,
\(e^{ x }=1+x+\frac { 1 }{ 2 } x^{ 2 }+\frac { 1 }{ 6 } x^{ 3 }+O(x^{ 4 })\)
\(E\approx mc^2.\left\{1-\left( Aa_{ \psi ,1 }+C \right) +\cfrac { Aa_{ \psi \, n }+C }{ n^{ 2 } }\right\}\)
As such the energy difference between two \(\psi\) energy levels is
\(\Delta E=E_f-E_i\)
\(\Delta E\approx mc^{ 2 }\left\{ \cfrac { Aa_{ \psi \, f }+C }{ n^{ 2 }_{ f } } -\cfrac { Aa_{ \psi \, i }+C }{ n^{ 2 }_{ i } } \right\} \)
where \(n_f\) is the final energy state and \(n_i\) is the initial energy state.
\(\cfrac { Aa_{ \psi \, n }+C }{ n^{ 2 }_{ n } } =ln(\cfrac { 2\pi }{ \lambda _{ n } } )=ln(\cfrac { 2\pi f _{ n } }{ c } )\)
\( \Delta E\approx mc^{ 2 }\left\{ ln(\cfrac { 2\pi f _{ f } }{ c } )-ln(\cfrac { 2\pi f _{ n } }{ c } ) \right\} \)
\( \Delta E\approx mc^{ 2 }\left\{ ln(f _{ f } )-ln(f _{ i }) \right\} \)
However this is still the energy state of one particle and not a particle in orbit around any nucleus.
If we mistaken that \(\Delta E\) results in a photon of wavelength \(\lambda_p\),
\(-\Delta E=h_of_p\)
as the photon represent loss from the system.
\(h_o=2\pi a_{ \psi }mc=\lambda_1mc\)
\( \Delta E\approx mc^{ 2 }\left\{ ln(f_{ i } )-ln(f_{ f }) \right\}=\cfrac { \lambda_1mc^2 }{ \lambda _{ p } } \)
So,
\(\cfrac { 1 }{ \lambda _{ p } }\approx \cfrac{1}{\lambda_1}\left\{ ln(f_{ i } )-ln(f_{ f }) \right\}\)
which is wrong in the first place, as all solutions of \(a_{\psi\,n}\) of a particle not orbiting around any nucleus, has the same energy.
In this picture, all photons have the same energy but their \(a_\psi\) are different. Photons with different \(a_\psi\) are refracted to different locations after a prism and are detected as separate spectra lines.
No particles here are orbiting about an attractive nucleus. Not an ugly one either.
Note:
If \(A\gt\gt a_\psi\) then we can approximate,
\(Aa_{ \psi \, f }+C =Aa_{ \psi \, i }+C \) in which case,
\(\Delta E\approx mc^{ 2 }(Aa_{ \psi \, f }+C)\left\{ \cfrac { 1 }{ n^{ 2 }_{ f } } -\cfrac { 1}{ n^{ 2 }_{ i } } \right\} \)
\(\Delta E\approx mc^{ 2 }n^2_flog(\cfrac{2\pi}{\lambda_f})\left\{ \cfrac { 1 }{ n^{ 2 }_{ f } } -\cfrac { 1}{ n^{ 2 }_{ i } } \right\} \)
when \(n_f=1\)
\(\Delta E\approx mc^{ 2 }log(\cfrac{2\pi}{\lambda_1})\left\{ 1 -\cfrac { 1}{ n^{ 2 }_{ i } } \right\} \)
