Particle Spectrum

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 $n_f=1$

$\Delta E\approx mc^{ 2 }log(\cfrac{2\pi}{\lambda_1})\left\{ 1 -\cfrac { 1}{ n^{ 2 }_{ i } }  \right\}$