# Assumptions and Examples

## Assumptions – Photons

The Transverse Wavelength Equation and Transverse Energy Equation, used in the next section (Examples), are built and derived from these assumptions. Please refer to the Particle Energy and Interaction paper for the complete derivation of the equations. The orbital distance calculations are found in the tables for orbital distances, with further detail in the Atomic Orbitals paper. The following assumptions were made when understanding particle interaction, including atomic orbitals:

• Particle vibration creates a transverse wave. A particle may vibrate upon annihilation, when transitioning between orbitals in an atom, or when an entire atom vibrates due to kinetic energy.
• Longitudinal amplitude difference creates particle motion as particles seek to minimize amplitude.
• The difference in longitudinal energy is transferred to transverse energy in a wave packet known as the photon.
• Particles and their anti-matter counterparts attract because of destructive waves between the particles; like particles (e.g. electron-electron) repel due to constructive waves, seeking to minimize amplitude.
• Electrons in an atomic orbital are both attracted and repelled by the nucleus. A positron is assumed to be at its core to attract the orbital electron; opposing forces in the nucleus repel the orbital electron.  A potential model of the proton with this structure is explained in Fundamental Physical Constants.

## Example – Photon Wavelengths

Orbital Distances – Hydrogen

The orbital distances for hydrogen are calculated and explained in Atoms.  For the purpose of the next two examples, which use wavelength counts (n) to determine transverse wavelengths, the orbital distance in wavelengths is required.

The fine structure constant (αe ) is derived in terms of wave constants in Fundamental Physical Constants, but for simplicity of this equation, its well-known value is used here.  The equation below is solved for the first orbital (1). The remaining orbitals are integers (replacing 1 with integers: 2, 3, 4, etc).

Orbital=1
αe = Fine Structure Constant = 7.29735257 x 10-3
Calculated Value Wavelengths (n):
187,789

To represent wavelengths in meters, it is converted using the number of wavelengths above (187,789) multiplied by the electron radius (K λ).  Units are in meters.  The calculation for the first orbital matches the CODATA value for the Bohr radius with no difference (0.000%).

n1s =187,789
Calculated Value Meters (m):
5.2918E-11
Difference from CODATA (Bohr Radius): 0.000%

Hydrogen Photon Wavelength (Ionization)

In the hydrogen photon wavelength calculations, the wavelengths of absorbed photons for hydrogen were calculated at differing orbitals when the atom is ionized (electron leaves the atom).

To find photon wavelengths, the Transverse Wavelength Equation is used.  In the case of ionization, this equation can be simplified. The electron is ejected from the atom so the final position (n) can be replaced by infinity in the equation.  In the first orbital, n=187,789 (from the example above) and a wavelength value of 9.113E-8 meters is calculated for the wavelength of hydrogen ionization, a difference of 0.004% from measured results.

n1s =187,789
Calculated Value:
-9.113E-8 meters (m)
Difference from Measured Result: 0.004%

Hydrogen Photon Wavelength (Orbital Transition)

Similar to the calculation above, the wavelengths of photons emitted or absorbed can be calculated using the Transverse Wavelength Equation.  In this example, an electron changes orbitals but does not leave the atom in the case of ionization.  When the electron transitions to a lower shell, it emits a photon.  The calculation in this case yields a positive result, noting that a photon is emitted.  When a photon is absorbed and the electron transitions to a higher shell or is ejected from the atom (ionization), the calculation yields a negative result.

The hydrogen photon wavelength calculations table contains calculations of electron transitions from various orbitals to the second orbital.  An example calculation is provided below, as an example of transitioning from the third orbital (3) to the second orbital (2).  This is represented by: 3->2.

Using the orbital calculations, the third orbital has 1,690,098 wavelengths and the second orbital (N=2) has 751,155 wavelengths.  These are used as the values n0 and n respectively.  The resulting calculation is 6.561E-7 meters, or a difference of 0.028% from the measured result.

3->2
n0 =
1,690,098
n =
751,155
Calculated Value:
6.561E-7 meters (m)
Difference from Measured Result: 0.026%

## Example – Photon Energies

In photon energy calculations, calculations are charted for ionized and neutral elements. Energy levels of photons are calculated using the Transverse Energy Equation. The three variables in the equation are the initial distance (r0), final distance (r) and the amplitude factor (δ).  Three examples are provided to demonstrate the calculations of transverse wave energy:

#1) Boron – 2p

The first example is the photon energy required to ionize an electron in the 2p subshell of boron. The initial amplitude factor for Boron 2p B2p) and the 2p orbital distance (r0) come from distances and amplitude factor tables found in the Atoms section. They are summarized below.  Note that radius (1.41) is converted to meters by multiplying by the Bohr radius (found in wave constant terms) – more details on this in the Atomic Orbitals paper.

The Transverse Energy Equation is used and the values from the tables are inserted into the variables.  The final distance (r) is set to infinity because the electron is ionized (leaves the atom).  The remaining are wave constants.

The result is a value of -1.32 x 10-18 joules.  A negative sign in the result indicates that a photon is absorbed.  To convert from joules to megajoule per mole (to compare to measured results) it is multiplied by 6.02214179 x 1017. After converting from joules to megajoules per mole, the calculated result is -0.80 Mj/mol.  This matches the measured result for Boron which is -.80 Mj/mol.

#2) Boron – 1s

A second example is provided for the ionization of the Boron 1s electron ionization from a neutral atom, comparing against results from spectroscopy experiments.  The same methodology is applied where the orbital distance and amplitude factor is found in the tables from the Atoms section.

The Transverse Energy Equation is used again, but now with the amplitude factor and the orbital distance of the affected electron that will be ionized.

After converting from joules to megajoules per mole, the calculated result is -19.3 Mj/mol.  This matches the measured result for boron 1s spectroscopy which is -19.3 Mj/mol.

#3) Boron – 1s  (using Amplitude Factor Equation)

An alternative method for calculating the 1s electron ionization can be performed using the Amplitude Factor Equation – 1s Orbital Ionization.  The results for neutral atoms are found in the last row in the Amplitude Factor table, but the equation will also be shown again to solve for the amplitude factor.  The Amplitude Factor Equation is shown for the 1s2 electron for boron.  For boron, Z=5 for the number protons, e1=2 and e2=3 for the first and second orbital electrons.

Using this method, the initial orbital distance is set to the Bohr radius (a0) in the Transverse Energy Equation.

Again, the calculated result is -19.3 Mj/mol.  This matches the measured result for Boron 1s spectroscopy which is -19.3 Mj/mol.  This alternative version reduces two unknown variables to one, because only the amplitude factor needs to be solved.  However, it only works for the 1s electrons from hydrogen to calcium.