Hydrogenic Atoms

Comparisons are made with experimental observations1,2,3,4,5,6

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, given7 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

Related WikiMechanics articles.

page revision: 4, last edited: 09 Nov 2019 21:07