Atomic Hydrogen

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Notice: this page is under construction

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.

A tour around a quark model of atomic hydrogen. Note that quarks and anti-quarks of the same type may be mixed-up. See spreadsheets and tables for the most up-to-date models.


Stability of Atomic Hydrogen


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

By definition of the mechanical energy

$E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \vphantom{\sum^{2}} \ }$

But photons are ethereal and $m(\gamma)=0$ so

$E (\gamma) = cp$

Earlier analysis, which ignored chemical quarks, gives

$\begin{align} c p = 2 k_{\sf{F}} \left\| \, \overline{\rho}^{\mathcal{A}} \right\| \end{align}$

where $\mathcal{A}$ is a single phase-component of the photon, and $\left\| \, \overline{\rho} ^{\mathcal{A}} \right\| \,$ is the norm of its radius vector. Recall that

$W^{\mathcal{A}} = k_{\sf{F}} \left\| \, \overline{\rho}^{\mathcal{A}} \right\|$

is the work required to build one phase-component. Then

$\begin{align} E (\gamma) = 2 W^{\mathcal{A}} \end{align}$

Now, 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, and our earlier assumption that $U^{\sf{D}}$ was zero, is now taken as an approximation. So to include chemical quarks, the work is redefined such that

$\begin{align} E(\gamma) &= 256 \frac{ \, n^{\sf{d}} n^{\sf{\overline{d}}} \left( 2W^{\! \mathcal{A}} + U_{\! chem} \right) }{ \left( {n^{\sf{\overline{d}}}}^{2} - {n^{\sf{d}} \vphantom{n^{\sf{\overline{d}}}} }^{2} \right)^{2} } \end{align}$

or

$\begin{align} E(\gamma) &= 256 \frac{ \, n^{\sf{d}} n^{\sf{\overline{d}}} \left( 2 k_{\sf{F}} \left\| \, \overline{\rho}^{\mathcal{A}} \right\| + U_{\! chem} \right) }{ \left( {n^{\sf{\overline{d}}}}^{2} - {n^{\sf{d}} \vphantom{n^{\sf{\overline{d}}}} }^{2} \right)^{2} } \end{align}$

The photon's energy is 'diluted' by the down-quarks, it is reduced by a factor that varies with the inverse square of the coefficients of down-quarks. The foregoing expression can be expressed in terms of the sum and difference of quark coefficients as

$\begin{align} E(\gamma) &\equiv 256 \frac{ \, n^{\sf{d}} n^{\sf{\overline{d}}} \left( 2W^{\! \mathcal{A}} + U_{\! chem} \right) }{ \left( {n^{\sf{\overline{d}}}}^{2} - {n^{\sf{d}} \vphantom{n^{\sf{\overline{d}}}} }^{2} \right)^{2} } \\ \\ &= 256 \frac{\, n^{\sf{d}} n^{\sf{\overline{d}}} \left( 2W^{\! \mathcal{A}} + U_{\! chem} \right) }{ \left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right)^{2} \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right)^{2} } \\ \\ &= 64 \left( 2W^{\! \mathcal{A}} + U_{\! chem} \right) \cdot \frac{ \left[ \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right) +\left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right) \right] \cdot \left[ \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right) -\left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right) \right] } { \left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right)^{2} \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right)^{2} } \\ \\ &= 64 \left( 2W^{\! \mathcal{A}} + U_{\! chem} \right) \cdot \frac{ \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right)^{2} - \left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right)^{2} } { \left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right)^{2} \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right)^{2} } \\ \\ &= 64 \left(2W^{\! \mathcal{A}} + U_{\! chem} \right) \cdot \left[ \frac{1}{\left( n^{\sf{\overline{d}}} - n^{\sf{d}} \right)^{2} } - \frac{1}{ \left( n^{\sf{\overline{d}}} + n^{\sf{d}} \right)^{2} } \right] \end{align}$

But $N^{\sf{D}} \equiv n^{\sf{\overline{d}}} + n^{\sf{d}}$
and $\Delta n^{\sf{D}} \equiv n^{\sf{\overline{d}}} - n^{\sf{d}}$
so

$\begin{align} E(\gamma) &= \left(2W^{\! \mathcal{A}} + U_{\! chem} \right) \cdot \left[ \frac{64}{\left( \Delta n^{\sf{D}} \right)^{2} } - \frac{64}{ \left( N^{\sf{D}} \right)^{2} } \right] \end{align}$

When written this way, it is easy to see that the effect of the down-quarks is neglegible if $\Delta n^{\sf{D}} =8$ and $N^{\sf{D}}$ gets large.


The Rydberg Formula

Let

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

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

$\begin{align} \mathcal{R} \left( \mathbf{H} \right) = 2W^{\! \mathcal{A}} + U_{\! chem} \end{align}$

Then

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

Right.png Next step: displacement of atomic particles.
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