Comparisons are made with experimental observations^{1}^{,}^{2}^{,}^{3}^{,}^{4}

A conventional statement of the Rydberg formula

$\begin{align} E_{\sf{Bohr}} = hc \mathcal{R} _{\mathbf{H}} \left( \frac{1}{ {\rm{n}} _{f}^{2} } - \frac{1}{ {\rm{n}} _{i}^{2} } \right) \end{align}$

where $\mathcal{R} _{\mathbf{H}}$ is the **Rydberg** number for hydrogen

So the models work by requiring that

$\begin{align} hc \mathcal{R} _{\mathbf{H}} = 2 \left| \, H_{chem}^{\mathcal{A}} \vphantom{{H_{chem}^{\large{\gamma}}}^{9}} \right| \end{align}$

This condition is obtained from the distribution of electrochemical quarks, which in the case of atomic hydrogen, is that there are eight quarks of each type. Other conditions and distributions are used to represent observations of atomic bonds, isotopes and hydrogenic atoms.

${\mathcal{R}}_{\mathrm{H}}$ is the Rydberg number for hydrogen, given^{5} by

$\begin{align} {\mathcal{R}}_{\mathrm{H}} \equiv {\mathcal{R}}_{\infty} \frac{m_{\sf{p}}}{m_{\sf{p}} + m_{\sf{e}}} \end{align}$

where

${\mathcal{R}}_{\infty}$ is a constant

$m_{\sf{p}}$ is the rest mass of a proton

$m_{\sf{e}}$ is the rest mass of a electron

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