Particles In Orbits

The difference plot of the expression, $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

however, draws a sense of déjà vu...

The concepts leading to the expression $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ reverses the energy sign of conventional electron energy level transitions; $E_{1\rightarrow 2}$ is negative and the particle loses energy.  $E_{1\rightarrow 2}$ is emitted

This would make the plot for $n_1=1$ in the above graph, an absorption line.

For all values of $n_2$, only when $n_1=1$ is an absorption line.  The plots in black are emission lines against which we see the absorption line without direct illuminations.

Values of the plots in black below $y=0$ is ignored because $n_2\ge n_1$.

Rydberg constant is not murdered, instead a new process is given birth.

$\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ occurs at the same time as $\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$ due to quantized energy levels in Bohr model theory.  The two process has reverse energy signs and hence married with a negative sign.  According to Bohr model $E_{1\rightarrow 2}$ is positive and the particle gains energy and transits to a higher energy level.

We have instead,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

where a change $n_1\rightarrow n_2$ is accompanied by a change in orbital energy level $n_{o1}\rightarrow n_{o2}$.

Only now are the particles in orbits.

Note:  $E_2=E_1+\Delta E_{1\rightarrow 2}$