So far, the hypothesis has been tested based on Bohr's atomic model. However, Bohr's atomic model is no longer valid. It does not apply to multi-electron atoms, and in reality, electrons do not orbit in circular paths. What quantum physics has revealed is that electrons within an atom manifest as shapes (probability) called orbitals by absorbing energy. Within the orbital, the electron exists probabilistically.

Electron Orbital
Electron Orbital credit: Wikipedia

Hypothesis: The nodes of orbitals are photon-Ether

In the diagram, the black regions are referred to as nodes. The shape of the orbitals is determined by the pattern and number of these nodes. The orbital's shape is influenced by the principal quantum number n, azimuthal quantum number l, and magnetic quantum number ml, as introduced in Bohr's atomic model. An increase in n results in an increase in the number of circular patterns, similar to Bohr's atomic model. An increase in l leads to the disappearance of one circular pattern, giving rise to linear nodes. The magnetic quantum number determines the orientation.

An orbital with circular nodes is referred to as an s orbital. If there are no nodes, it is labeled as 1s; with one circular node, it becomes 2s. An orbital with one linear node is designated as a p orbital. With one linear node is 2s, and one linear and one circular node is 3p, and so forth. Following this pattern, orbitals with specific numbers of linear nodes are assigned letters such as d, f, g, etc. Now, let's reinterpret the orbitals assuming the nodes are photon-Ether.

Viewing Orbitals from the Perspective of Ether
Viewing Orbitals from the Perspective of Ether See the circular and linear ethers

An increase in n signifies an increase in the number of Ether entities, and an increase in l transforms the circular Ether into a linear form. An s orbital contains only circular Ether, while a p orbital can be interpreted as having one linear Ether and one circular Ether. The 1s orbital corresponds to a state without Ether, and the 3p orbital represents a state with one circular and one linear Ether

This is commonly understood as a transformation in the shape of the electron. However, from the Ether perspective, the black nodes represent Ether, and the Ether of the electron maintains a spherical shape, with space separated by photon-Ether. It can be understood as if the Ether of the electron is split, similar to the double-slit experiment. Therefore, the electron-Sparkle probabilistically exists in the separated space, shaped according to the form that manifests as the orbital.

The Shape of Nodes in Orbital
The Shape of Nodes in Orbital What if the nodes are electron-Ether?

In order to better illustrate the perspective of Ether, let's establish a notation using circles and lines. [X] represents 1s, [O] represents 2s, and [OO] represents 3s. [-] stands for 2p, [-O] represents 3p, and [--O] signifies 4d. As the digit increases, for readability, we will also use a notation like [2-5O] to represent [--OOOOO]. When there are multiple electrons, for example, 1s2 can be denoted as [X]2, and 2s2 as [O]2. The data used below is referenced from NIST.

What I want to demonstrate through this is that there is a pattern in the accumulation based on circular and linear shapes. The transition of values in multi-electron atoms follows a regular trend, much like in hydrogen atoms. I aim to show that the emitted energy is the sum of Ether and that I need a reference point, similar to the Rydberg formula for hydrogen-like atoms, to explain this trend.

First, let's observe the change in circular Ether values in hydrogen and helium. It involves variations such as [X], [O], [OO], and so on.

Hydrogen and Helium's Circular Ether Values
Hydrogen and Helium's Circular Ether Values The graph represents the Rydberg formula, and helium is located in the top right corner.

When the graph is shifted vertically and horizontally, it appears as if it might overlap. Although it is repeatedly emphasized not to use the Rydberg formula for multi-electron atoms, the temptation is irresistible. I will attempt to create a reference equation by modifying the Rydberg formula. Spoiler alert: finding a perfect equation without any errors in this process would be a great achievement, but beware! I haven't found that method. However, I aim to find an equation that approximates it as closely as possible.

Shifting a Graph

Helium's s and p Orbitals
Helium's s and p Orbitals

Let's examine the changes in the helium orbitals. Hydrogen is not suitable for this analysis due to its very small errors. Each color represents s and p orbitals with the same term symbols (e.g., 1S0, 3S1, 3P*2). Consider these values as the result of the Rydberg formula being shifted left and right. Then, it might be possible to create graphs for each orbital by modifying the Rydberg formula. Let's use the ionization energy values as convergence points. The ionization energy of hydrogen is 13.60676328. Let's assume the ratio for the graphs is same with hydrogen. The ionization energy of helium is 24.58556828. This can be expressed in the following equation. In the upcoming equations, R will be represented with Arabic numerals separated by a dot (.) for the atomic and ion numbers, while the ether or orbital values will be indicated below it.

R1.1(x)=13.60676328ā‹…(1āˆ’1(x+1)2) R^{1.1}(x) = 13.60676328 \cdot (1 - \dfrac{1}{(x + 1)^2})
Rydberg formula of hydrogen (eV)
R2.1(x)=13.60676328ā‹…(1āˆ’1(x+1)2)+24.58556828āˆ’13.60676328R^{2.1}(x) = 13.60676328 \cdot (1 - \dfrac{1}{(x + 1)^2}) + 24.58556828 - 13.60676328
Rydberg formula of helium (eV)

It is vertically shifted Rydberg formula to match the height of helium

the Rydberg Formula Shifted Vertically to Match Helium
the Rydberg Formula Shifted Vertically to Match Helium

It seems to fit together nicely. Now, let's try horizontal movement. The ratio of the graph, 13.60676328, is denoted as r (radius), and the difference between the vertical shift values, 24.58556828 - 13.60676328, is denoted as s (shift). The horizontal shift value is substituted with k.

w=rā‹…(1āˆ’1(v+1+k)2)+sw = r \cdot (1 - \dfrac{1}{(v + 1 + k)^2}) + s

In the given equation, if we substitute w with the energy of [O]ether and v with the ether's number, 1, then adjusting the graph to the first position will result in horizontal movement. Now, let's formulate the equation for the horizontal shift value k.

w=rā‹…(1āˆ’1(v+1+k)2)+sw = r \cdot (1 - \dfrac{1}{(v + 1 + k)^2}) + s
ā†’wāˆ’sr=1āˆ’1(v+1+k)2\to \dfrac{w - s}{r} = 1 - \dfrac{1}{(v + 1 + k)^2}
ā†’1āˆ’wāˆ’sr=1(v+1+k)2\to 1 - \dfrac{w - s}{r} = \dfrac{1}{(v + 1 + k)^2}
ā†’11āˆ’wāˆ’sr=(v+1+k)2\to \dfrac{1}{1 - \dfrac{w - s}{r}} = (v + 1 + k)^2
ā†’11āˆ’wāˆ’sr=v+1+k\to \dfrac{1}{\sqrt{1 - \dfrac{w - s}{r}}} = v + 1 + k
ā†’k=11āˆ’wāˆ’srāˆ’vāˆ’1\to k = \dfrac{1}{\sqrt{1 - \dfrac{w - s}{r}}} - v - 1

Sure, let's substitute the expression we derived for the horizontal shift value k back into the original equation.

R(x,r,s,v,w)=rā‹…(1āˆ’1(x+1+11āˆ’wāˆ’srāˆ’vāˆ’1)2)+sR(x, r, s, v, w) = r \cdot (1 - \dfrac{1}{(x + 1 + \dfrac{1}{\sqrt{1 - \dfrac{w - s}{r}}} - v - 1)^2}) + s
ā†’R(x,r,s,v,w)=r+sāˆ’r(x+11āˆ’wāˆ’srāˆ’v)2\to R(x, r, s, v, w) = r + s - \dfrac{r}{(x + \dfrac{1}{\sqrt{1 - \dfrac{w - s}{r}}} - v)^2}
ā†’R(x,r,s,v,w)=r+sāˆ’r(xāˆ’v+1rāˆ’w+sr)2\to R(x, r, s, v, w) = r + s - \dfrac{r}{(x - v + \dfrac{1}{\sqrt{\dfrac{r - w + s}{r}}})^2}
ā†’R(x,r,s,v,w)=r+sāˆ’r(xāˆ’v+rrāˆ’w+s)2\to R(x, r, s, v, w) = r + s - \dfrac{r}{(x - v + \sqrt{\dfrac{r}{r - w + s}})^2}

Here, r + s represents the peak of the graph, equivalent to the ionization energy. Let's denote r + s as p (peak)

R(x,r,p,v,w)=pāˆ’r(xāˆ’v+rpāˆ’w)2R(x, r, p, v, w) = p - \dfrac{r}{(x - v + \sqrt{\dfrac{r}{p - w}})^2}
Modified Rydberg formula with vertical and horizontal shifting.

r represents the ratio of the graph and is the peak of the hydrogen-like atom in the same phase. In other words, the r value for helium is the same as that for hydrogen, and the r value for Li II is the same as that for He II. prepresents the peak and convergence value of the graph, which is equivalent to the ionization energy. When values are substituted, the graph of the s and p orbitals of helium looks as follows.

The Graphs of Helium's s and p Orbitals
The Graphs of Helium's s and p Orbitals Aren't thy beautiful?

In another example, when expressing the equation for lithium's 2S1/2, shifted to 1s23s as [X]2[OO] notation, it can be formulated as follows. Substituting hydrogen's r value (13.60676328), lithium's ionization energy s value (5.39114472), and the emission energy at the 2nd position (3.373129):

R2S.1/23.1(x)=R(x,13.60676328,5.39114472,2,3.373129)R^{3.1}_{2S.1/2}(x) = R(x, 13.60676328, 5.39114472, 2, 3.373129)
The Graphs of Lithium's s Orbital
The Graphs of Lithium's s Orbital

It's beautiful too

How to Shift Horizontally

Now, by simply plugging in the values into the equation, the graph will shift vertically and horizontally. However, it raises the question of what values to use as a reference and how much to move them for a meaningful comparison. Since the reference equation must remain consistent, continuously shifting the graph may not be practical. Let's examine the values for the helium orbitals.

1S0[X][O][X][OO][X][OOO][X][OOOO][X][5O][X][6O]
20.61622.92023.67424.01124.19124.298
1P*1[X][-][X][-O][X][-OO][X][-OOO][X][1-4O][X][1-5O]
21.21823.08723.74224.04624.21124.311
1D2[X][--][X][--O][X][--OO][X][2-3O][X][2-4O]
23.08723.74224.04624.21124.311

What should be placed in an empty space? Before [X][--], let's position [X][-], which represents a value where the spherical orbital doesn't exist. In this case, we can shift the s orbital symmetrically to the reference graph, and for the rest, we can move them by the same amount as the horizontal shift of the p orbitals.

Horizontal Shifting of Helium Orbital Values
Horizontal Shifting of Helium Orbital Values

If done this way, all points can be aligned towards the reference graph. Some may wonder why we are shifting the values left or right. However, this is exactly the same concept as shifting an equation horizontally. After shifting the equation, both the equation and the points must be moved simultaneously to create a unified reference equation. This allows for a comparison of all values against a single reference. Let's substitute the s orbital into the shifting equation and calculate the shift.

k=11āˆ’wāˆ’srāˆ’vāˆ’1k = \dfrac{1}{\sqrt{1 - \dfrac{w - s}{r}}} - v - 1
ā†’ks=11āˆ’20.6157751334āˆ’(24.58556828āˆ’13.60676328)13.60676328āˆ’1āˆ’1\to k_{s} = \dfrac{1}{\sqrt{1 - \dfrac{20.6157751334 - (24.58556828 - 13.60676328)}{13.60676328}}} - 1 - 1
ā†’ks=āˆ’0.148628918071\to k_{s} = āˆ’0.148628918071
If we move all points in the x-axis direction by this amount, they will align with the reference Rydberg formula.
s, p, and d Orbitals
s, p, and d Orbitals

The s orbital has a lower value (shifting graph to the right), while orbitals like p and d containing linear ether are clustered to the left. Let's move the orbitals excluding s towards the next state, represented by [-].

The kp value for moving the p orbital in the [X][-]direction is 0.00897479319406. Now, let's substitute these values and redraw the table. This time, I will represent them in coordinates instead of values.

1S0[X][O][X][OO][X][OOO][X][OOOO][X][5O][X][6O]
(1+ks, 20.616)(2+ks, 22.920)(3+ks, 23.674)(4+ks, 24.011)(5+ks, 24.191)(5+ks, 24.298)
1P*1[X][-][X][-O][X][-OO][X][-OOO][X][1-4O][X][1-5O]
(1+kp, 21.218)(2+kp, 23.087)(3+kp, 23.742)(4+kp, 24.046)(5+kp, 24.211)(6+kp, 24.311)
1D2[X][--][X][--O][X][--OO][X][2-3O][X][2-4O]
(1+kp, 23.087)(1+kp, 23.742)(1+kp, 24.046)(1+kp, 24.211)(1+kp, 24.311)

The result graph is like below:

The Horizontal Shifting Graph of Helium Orbitals
The Horizontal Shifting Graph of Helium Orbitals

The purpose of this document is to demonstrate the accumulation of ether. Therefore, instead of looking at the total energy, we should focus on the energy changes at each stage. The increase in energy from [X][O] to [X][OO] should be compared by examining the difference between the two values and R(2+ks)āˆ’R(1+ks)R(2+k_{s}) - R(1+k_{s}).

a=22.920317682āˆ’20.6157751334=2.3045425486a = 22.920317682 - 20.6157751334 = 2.3045425486
b=R(2+ks)āˆ’R(1+ks)=2.29784533694b = R(2+k_{s}) - R(1+k_{s}) = 2.29784533694
aāˆ’b=0.00669721166409a - b = 0.00669721166409
ā†’D(x)=v2āˆ’v1āˆ’(R(x+1+k2)āˆ’R(x+k1))\to D(x) = v_2 - v_1 - (R(x+1+k_{2}) - R(x+k_{1}))
The difference D(x) when energy is emitted from energy v2 to v1in the modified Rydberg formula.

The table below is shifting result for all values:

1S0[OO] - [O][OOO] - [OO][OOOO] - [OOO][5O] - [OOOO][6O] - [5O]
0.00669721166409āˆ’0.00322123849668āˆ’0.00157668275873āˆ’0.00076388060241āˆ’0.000401501193807
1P*1[-O] - [-][-OO] - [-O][-OOO] - [-OO][1-4O] - [-OOO][1-5O] - [1-4O]
0.00163108806923āˆ’0.000784728358325-0.00038084816397āˆ’0.000180879400757āˆ’0.0000924805841381
1D2[--O] - [--][--OO] - [--O][2-3O] - [--OO][2-4O] - [2-3O][2-5O] - [2-4O]
0.006423971091680.002357191006030.001063231962240.0005496052858620.000312953223304

Let's move them to a graph:

Changes in the Helium Orbital Values
Changes in the Helium Orbital Values

This way, we can compare the energy changes, which represent the differences in all measured values, on the same standardized basis. The values also seem quite consistent, making them very suitable for representation on a single graph. The error is also very small. It looks delightful at every glance. In the next section, through actual comparisons and interpretations, I aim to explore whether we can consider the emission energy of an atom as the accumulation of ether