The Hydrogen Spectrum
The Balmer series of hydrogen. The violet line on the left is Balmer-delta at 410 (nm). The red line on the right is Balmer-alpha at 656 (nm).
The Balmer series of hydrogen. The violet line on the left is Balmer-delta at 410 (nm). The red line on the right is Balmer-alpha at 656 (nm).

Photons that are absorbed or emitted by atomic hydrogen are collectively known as the spectrum of hydrogen. This spectrum includes a wide range of ultraviolet, visible and infrared photons. They are involved in the atomic and molecular interactions of everyday experience and benchtop chemistry. Energies are typically measured in (eV) rather than (MeV).

Gross Structure

Models of photons in the hydrogen spectrum are built exclusively from rotating and electrochemical quarks1. As required by definition, they all have a total angular momentum quantum number of $𝘑 =1$. Comparisons are made with experimental observations2,3,4,5 which are exceptionally precise! Some discern a few parts in $10^{12}$. The following models can accurately reproduce the hydrogen spectrum to only about one part in a million; that is, just the gross structure. We can determine the mechanical energy, $E$, of a hydrogen photon, $\large{\gamma}$, by first assessing its momentum. The hydrogen spectrum does not include gamma-rays, so the overall energy-scale is set by the enthalpy of any chemical quarks, $H_{chem} \,$. Then the coefficients of down quarks, $N^{\mathsf{D}}$ and ${\Delta}n^{\mathsf{D}}$, determine individual energies as

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

The Rydberg Formula

All of the photons in the spectrum of atomic hydrogen, $\mathbf{H}$, are linked to changes in the excited states of $\mathbf{H}$. So if $\mathbf{H}$ goes from some inital state $i$, to some final state $f$, by emitting a photon $\gamma$, then we write

$\begin{align} {\mathbf{H}} _{i} \to {\mathbf{H}} _{f} + \gamma \end{align}$

These states are identified by their principal quantum number, $\rm{n}$. And $\rm{n}$ is always conserved because quarks are indestructible. So for any absorption or emission within the spectrum, the conservation of down-quarks guarantees that

$\begin{align} {\rm{n}} \left( {\mathbf{H}} _{i} \right) = {\rm{n}} \left( {\mathbf{H}} _{f} \right) + {\rm{n}} \left( \gamma \right) \end{align}$

Let us write ${\rm{n}}_{i} \equiv {\rm{n}} \left( {\mathbf{H}}_{i} \right)$ and ${\rm{n}}_{f} \equiv {\rm{n}} \left( {\mathbf{H}}_{f} \right)$. Then this conservation rule can be rearranged as

$\begin{align} {\rm{n}}_{i} - {\rm{n}}_{f} &= {\rm{n}} \left( \gamma \right) \equiv \frac{ n^{\sf{d}}\left( \gamma \right) }{4} = \left[ \frac{n^{\sf{d}}\left( \gamma \right) }{8} + \frac{n^{\sf{d}}\left( \gamma \right) }{8} \right] \\ \\ &= \left( \frac{n^{\sf{d}}\left( \gamma \right) }{8} + \frac{n^{\sf{\overline{d}}}\left( \gamma \right) }{8} \right) - \left( \frac{n^{\sf{\overline{d}}}\left( \gamma \right) }{8} - \frac{n^{\sf{d}}\left( \gamma \right) }{8} \right) \\ \\ &= \frac{N^{\sf{D}}\left( \gamma \right) }{8} - \frac{\Delta n^{\sf{D}}\left( \gamma \right) }{8} \end{align}$

The conservation law will be automatically satisfied if the quantum numbers that describe the excited-states of hydrogen, are related to the quark coefficients of hydrogen-spectrum photons, by

$\begin{align} {\rm{n}} _{i} = \frac{N^{\sf{D}}\left( \gamma \right) }{8} \end{align}$ and $\begin{align} {\rm{n}} _{f} = \frac{\Delta n^{\sf{D}}\left( \gamma \right) }{8} \end{align}$

Substituting these expressions into the equation for photon-energies, as derived above, gives

$\begin{align} E(\gamma) = 2 \left| \, H_{chem}^{\mathcal{A}} \vphantom{{H_{chem}^{\large{\gamma}}}^{9}} \right| \cdot \left( \frac{1}{ {\rm{n}} _{f}^{2} } - \frac{1}{ {\rm{n}} _{i}^{2} } \right) \end{align}$

Notice: this page is under construction
Notice: this page is under construction

a conventional statement of the Rydberg formula

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

How the Quark Models Work

the Rydberg constant for hydrogen from the electrochemical-quarks that are common to all photons in the hydrogen spectrum.

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


$\begin{align} H_{chem} \equiv \sum_{\zeta =11}^{16} \Delta n^{\zeta} U^{\zeta} \end{align}$

$\begin{align} \mathcal{R} _{\mathbf{H}} = 2 \left| \, \sum_{\zeta =11}^{16} \Delta n^{\zeta} U^{\zeta} \, \right| \end{align}$


$\begin{align} \mathcal{R} _{\mathbf{H}} = 2 \left| \, \sum_{\zeta =11}^{16} \Delta n^{\zeta} U^{\zeta} \, \right| \end{align}$

where the sum is over the quarks in ${\mathcal{A}}$

Photons are also described by their wavelength, $\lambda _{\sf{o}}$. Here is a list of photons named after the scientists that first observed them, along with their component quarks. Note that they all share the same electrochemical-quarks because they are all associated with hydrogen.


These models also depend on how quarks are distributed between phase components. For that level of detail, please see the spreadsheet Bonds and Photons in the wiki-files stored here. An X indicates that a calculated result is outside of nominal uncertainty.6

Related WikiMechanics articles.

Fine Structure

Ultrafine Structure

See Footnote 1 in this article about field quanta.

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