Calculations – Photons

Calculations – Photon Wavelengths

Hydrogen Wavelengths (Ionization)
Using the Transverse Wavelength Equation – Hydrogen, the wavelengths of photons absorbed during hydrogen ionization were calculated for the ground state and each of the excited states (orbitals 1, 2, 3, 4, 5, 6).  Calculations are compared to measured results for photons wavelengths of hydrogen.

 

Photon wavelength - hydrogen ionization calculations using energy wave theory

 

 

Hydrogen Wavelengths (Shell Transition)
Using the Transverse Wavelength Equation – Hydrogen, the wavelengths of photons emitted during hydrogen for an electron transitioning from an excited state (orbitals 3, 4, 5, 6, 7, 8, 9) to the second (2) orbital.  Calculations are compared to measured results for photons wavelengths of hydrogen.  For example, as the electron transitions from the 3s orbital to 2s orbital (3->2) it emits a photon calculated to be 6.56E-07 meters.

 

Hydrogen photon wavelength calculation for orbital transitions using energy wave theory

 


 

Calculations – Photon Energy

The energy of a photon that is absorbed or emitted by any atom can be calculated using the Transverse Energy Equation, along with 1) the distance and 2) constructive/destructive wave interference (amplitude factor) on an electron transitioning orbitals in an atom.  These two variables are found in the tables from the summary results in the calculations of atoms and then used with the Transverse Energy Equation.

More than 150 calculations are compared to measured results of photon energies required for ionization.  An explanation is provided in the Atomic Orbitals paper for the variation of calculations versus measured results. Ionization energies were calculated in Mj per mole.

 

Ionization Energy of the First (Outermost) Electron – Neutral Element

Ionization Energy of Neutral Elements

 

Ionization Energy of the 1s Electron – Neutral Element (from Spectroscopy Experiments)

A pattern emerges for the first orbital of the elements from hydrogen to calcium that allow a simplified method reducing two variables to one.  In these calculations, the Amplitude Factor Equation – 1s Orbital Ionization is used to solve one variable without knowing orbital distance.  Instead, the Bohr radius is used as the orbital distance when this equation is used.  This equation only applies to the 1s orbital.  

 

Ionization of 1s Electron of Neutral Elements (Spectroscopy)

 

Ionization Energy of 1s1 Electron – Ionized Element

Ionization Energy of 1s Electron - Ionized Atom

 

Ionization Energy of 1s2 Electron – Ionized Element

Ionization Energy of 1s2 Electron - Ionized Atom

 

Ionization Energy of 2s and 2p Electrons – Ionized Element

The charts for the 2s and 2p electron ionization energies are summarized quickly below, but the details can be found in the Atomic Orbitals paper.

Ionization energy of 2nd orbital - Ionized Atoms

 


 

Example Calculations – 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).

n1s

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%).

bohr radius

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 – Hydrogen 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 ionized

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 – Hydrogen.  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 (above) 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.

Transverse Wavelength Equation

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

 


 

Example Calculations – Photon  Energies

In photon energy calculations (above), 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.  

Boton Photon 1 Eq 1

Boron Photon 1 Eq 2

 

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.

Boron Photon 1 Eq 3

Boron Photon 1 Eq 4

 

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.

Boron 2 - Eq 1

Boron 2 - Eq 2

 

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.

Boron 2 - Eq 3

Boron 2 - Eq 4

 

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.

Boron 3 - Eq 1 (Amplitude Factor Equation)

 

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

Boron 3 - Eq 2

Boron 3 - Eq 3

 

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.

 


 

Example Calculation – Conservation of Energy

Examples for proving the transfer of longitudinal wave energy to transverse wave energy and vice versa.

 

Example #1) Photon Creation – Atoms

When an electron is captured into an atom, or moves to a lower orbital energy level, a photon is released. This is a transverse wave as a result of particle vibration as it settles into place. Each particle (proton and electron) has longitudinal, standing wave energy.

 

Conservation of Energy Orbital Transition 1

Before an Electron is Captured by an Atom

 

After the electron is captured, there is a change in longitudinal wave energy as these particles have destructive wave interference. Energy is always conserved, so this means that it is transferred to two, transverse waves that will travel in opposite directions. These transverse waves are photons.  The electron’s vibration is perpendicular to the direction to the nucleus, so one photon will travel towards the nucleus and be absorbed and cause the atom to recoil. The other photon will leave the atom.

Photon Orbital After

After Electron Captured by Atom

 

Transverse Energy – Photon

A single electron and proton is hydrogen. At the first orbital, known as the Bohr radius (a0), the Transverse Energy Equation was used to derive the Rydberg Constant (in joules). This is the energy of a photon that is released for an electron at the Bohr radius in hydrogen:

Orbital Transition Eq 1

Transverse Energy Equation (at Bohr Radius) – Rydberg Constant

 

Longitudinal Energy – Destructive Waves 

The energy for the photon is a result of a difference in longitudinal wave energy as two opposite phase particles destructively interfere, which is the case with a proton and electron. The energy difference in longitudinal wave form is:

Orbital Transition Eq 2

 

The values Ee and re are the short form of the energy equation that is used for force. They are the energy of the electron and classical electron radius, respectively. Q is the particle count that is also used in the Force Equation. This energy equation is the basis of the Force Equation and its derivation. These values can be replaced by their wave constants to form the Longitudinal Energy Equation – Constructive Wave Interference Difference, where it is a function of two groups of particles with count Q, changing their distance (r). The equation above is simplified and then replaced by their wave constants.

Orbital Transition Eq 3

Longitudinal Energy Equation – Constructive Wave Interference Difference

 

To solve for hydrogen and the Bohr radius, the same distances (r) are used in the equation above. However, two photons are created, so only half of this energy value is used to match one photon.

Orbital Transition Eq 4

 

E= 1/2 El. The result from the change in longitudinal energy is -2.1799E-18 joules. This is the same energy that is created in a new, transverse wave energy of the photon which is also -2.1799E-18 joules – the Rydberg constant. Since there are two photons created, the transverse energy is 1/2 of the longitudinal energy. Energy is completed conserved, but it changes wave forms from longitudinal to transverse, or vice versa when a photon is absorbed (transverse to longitudinal).

 

Example #2) Photon Creation – Annihilation

When an electron interacts with a positron, it is known to annihilate and create two photons that match the rest energy of the electron. The section on annihilation describes the process and reason for this interaction, including why these two particles can also appear out of nowhere with pair production. Here, the process is described mathematically as the transfer of energy from wave forms.

Annihilation 1

Before Annihilation – An Electron and Positron

 

The positron is the anti-particle to the electron. It is identical to the electron in mass and charge. In wave theory, the only difference with this particle is that it is anti-phase on the longitudinal wave (180 degree phase shift). This causes destructive wave interference between these two particles.

After being attracted by destructive waves, the particles vibrate until they reach a resting position. This vibration causes two transverse waves – two photons. The particles (defined by their wave centers) are still there but without any standing waves to be measured as mass. They can be separated again later with a photon that is equal to or greater than the sum of their masses (pair production).

Annihilation After

After Annihilation – An Electron and Positron

 

Longitudinal Energy – Destructive Waves 

The conservation of energy applies again to this example. Two photons (Et) are created from two electron masses (El). 2E= 2El, or E= El. The longitudinal wave energy of the electron was calculated in the electron section and also shown as one of the fundamental physical constants, using the Longitudinal Energy Equation and wave constants.

Longitudinal Energy - Mass of Electron

 

Transverse Energy – Photon

The two photons that are created from annihilation are calculated with the Transverse Energy Equation. This equation requires an initial distance (r0) and a final distance (r). Infinity is used as the initial distance as the two particles initially start very far away from each other and can be modeled with infinity. The final distance for annihilation is called r where the “C” is Compton, since this is related to the Compton wavelength. Since annihilation is a phase-shift, which is 180 degrees (or half of the 360 degree rotation), the final resting position is half of the classical electron radius (re). This is the point where waves are completely destructive and a resting position for particles at minimal wave amplitude.

Transverse Energy Annihilation Eq 1

 

These distances are used in the Transverse Energy Equation.

Transverse Energy Annihilation Eq 2

 

The classical electron radius (re) is also solved for in terms of wave constants. These values are inserted and the equation can be solved.

Transverse Energy Annihilation Eq 3

 

E= El. The photon energy calculated with the Transverse Energy Equation is 8.1871E-14 joules, which is identical to the mass of the electron, which is calculated with the Longitudinal Energy Equation at 8.1871E-14 joules. It shows that annihilation is a complete transfer of energy from standing waves of longitudinal wave energy to transverse energy.  Conservation of energy.

Source Data: All graphs shown here, and the calculations for all of the photon wavelengths and ionization energies can be found in the downloadable spreadsheet.  Further information on the derivation of the equations and how to replicate them are in the Particle Energy and Interaction and Atomic Orbitals papers.