Atomic Hydrogen |

$\Large{ k }$ | $\large{ \delta _{\hat{m}} }$ | $\large{ \delta _{\hat{e}} }$ | $\large{ \delta _{\theta} }$ | $\large{ \sf{P}_{\it{k}} }$ |

1 | +1 | 0 | +1 | $\sf{d}\bar{\sf{d}} \ \sf{m}\bar{\sf{m}} \ \bar{\sf{t}}\sf{b}$ |

2 | 0 | -1 | +1 | $\bar{\sf{u}} \ \bar{\sf{b}}\sf{t} \ \mathrm{2}\bar{\sf{g}}$ |

3 | -1 | 0 | +1 | $\sf{d}\bar{\sf{d}} \ \sf{a}\bar{\sf{a}} \ \bar{\sf{t}}\sf{b}$ |

4 | 0 | +1 | +1 | $\bar{\sf{u}} \ \bar{\sf{s}}\sf{c} \ \mathrm{2}\sf{e}$ |

5 | +1 | 0 | -1 | $\sf{d}\bar{\sf{d}} \ \sf{m}\bar{\sf{m}} \ \bar{\sf{t}}\sf{b}$ |

6 | 0 | -1 | -1 | $\bar{\sf{u}} \ \bar{\sf{b}}\sf{t} \ \mathrm{2}\bar{\sf{g}}$ |

7 | -1 | 0 | -1 | $\sf{d}\bar{\sf{d}} \ \sf{a}\bar{\sf{a}} \ \bar{\sf{t}}\sf{b}$ |

8 | 0 | +1 | -1 | $\bar{\sf{u}} \ \bar{\sf{s}}\sf{c} \ \mathrm{2}\sf{e}$ |

Definition: **Atomic hydrogen** is formed by the union of a proton $\sf{p}^{+}$, with an electron $\sf{e^{-}}$, bound together by a *magnetic* field $\mathscr{F}$. The letter $\mathbf{H}$ is used to note an atom of hydrogen.

$\mathbf{H} \equiv \left\{ \sf{p}^{+}, \sf{e}^{-}, \mathscr{F}_{\mathbf{H}} \right\}$

The proton is represented by these quarks

$\sf{p^{+}} \leftrightarrow 2\sf{d} + 2\bar{\sf{d}} + 4\sf{b} + 4\bar{ \sf{t} }$

The electron is

$\sf{e^{-}} \leftrightarrow 4\bar{\sf{u}} + 2\bar{\sf{b}} + 2\sf{t} + 2\bar{\sf{s}} + 2\sf{c} + 4\bar{\sf{g}}+ 4\sf{e}$

And the field is described by these $\sf{q}\bar{\sf{q}}$ pairs

$\mathscr{F}_{\mathbf{H}} \leftrightarrow \sf{ 2d\bar{d} + 2m\bar{m} + 2a\bar{a} }$

Then atoms of hydrogen are objectified from space-time events like the one shown in the accompanying table and movie.

## Stability of Atomic Hydrogen

The stability of a particle is described by its mean life. And a quantitative analysis of particle lifetimes is set by a customary assumption that the temperature of a hydrogen atom in its ground state is $\ T=0 \ \sf{\text{(K)}}$. Our current quark models for *all* nuclear particles have been carefully adjusted to respect this convention.

But, despite much effort, the closest to zero we can get in our working calculations and spreadsheets is $\ T=-8.9 \times 10^{-6} \ \sf{\text{(K)}}$ for an atom of hydrogen in its ground state. This number has no physical significance. Rather, it shows the limits of our calculation technique.

Fundamentally, the stability of a nuclear particle is premised on a balance of competing effects. And so temperature calculations depend on a few small differences between some very large numbers. Some rounding errors are inevitable. And if the temperature is near zero, then these errors can add up to being a significant problem.The non-zero result for hydrogen suggests that our temperature calculations are questionable for any result more exact than a few parts in a million.

This seems to be near the limit of what we can obtain from our present computing arrangements^{1} that are ordinary office equipment. This problem is trivial for large computers run by professionals with expertise doing highly precise calculations.

## Gross Structure of Hydrogen Spectrum

Define $\mathit{ň}$ as the **principal** quantum number

$\begin{align} \mathit{ň} \equiv \frac{ n^{\sf{d}} }{4} \end{align}$

Consider an excited hydrogen atom $\mathbf{H}^{\ast}$ formed by the union of some photon $\large{\gamma}$ with a hydrogen atom in its ground state

$\mathbf{H}^{\ast} \equiv \left\{ \mathbf{H}, {\large{\gamma}} \right\}$

Down quarks are indestructible, the prinicpal quantum number is conserved when particles are combined or decomposed. So

$\mathit{ň} \left( \mathbf{H}^{\ast} \right) = \mathit{ň} \left( \mathbf{H} \right) + \mathit{ň} \left( {\large{\gamma}} \right)$

A hydrogen atom in its ground state $\mathbf{H}$ contains four down quarks, so $\begin{align} \mathit{ň} \left( \mathbf{H} \right) = 1 \end{align}$ and

$\mathit{ň} \left( \mathbf{H}^{\ast} \right) = 1 + \mathit{ň} \left( {\large{\gamma}} \right) > 1$

## Photon Energy

When chemical quarks are considered, we can define photons that include no baryonic and no leptonic quarks, just rotating and chemical quarks. These photons are not gamma-rays. They are x-rays, radio-waves and all the photons in between; ultraviolet, visible and infrared. Their energies are typically measured in (eV), rather than (keV). For such low energies, effects from down-quarks are apparent.

From the penultimate equation in the discussion of the mechanical energy of photons

$\begin{align} E \left( {\large{\gamma}} \right) = 2 \left| \, H_{chem}^{\mathcal{A}} \vphantom{{H_{chem}^{\large{\gamma}}}^{9}} \right| \cdot \left[ \frac{64}{\left( \Delta n^{\sf{D}} \right)^{2} } - \frac{64}{ \left( N^{\sf{D}} \right)^{2} } \right] \end{align}$

## The Rydberg Formula

Let

$\begin{align} \mathit{ň} \left( \mathbf{H}_{i} \right)= \frac{ \, N^{\sf{D}} (\gamma) \, }{8} \end{align}$

$\begin{align} \mathit{ň} \left( \mathbf{H}_{f} \right) = \frac{ \Delta n^{\sf{D}} (\gamma) }{8} \end{align}$

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

Then

$\begin{align} E(\gamma) &= \mathcal{R} \left( \frac{1}{ \mathit{ň} _{f}^{2} } - \frac{1}{ \mathit{ň} _{i}^{2} } \right) \end{align}$