Sujin | Ether | Proof(1): Classic Physics

The emitted wavelengths of the Photon-Ether based on Bohr's atomic model

The wavelengths emitted by the electron can be determined using the Rydberg formula. Below is a summary of the Rydberg formula and the emitted wavelengths in the Lyman series, where n decreases from $n \geq 2$ to $n = 1$ , and the Paschen series, where the transition changes from $n \geq 4$ to $n = 3$ .

\dfrac{1}{\lambda} = R(\dfrac{1}{m^2} - \dfrac{1}{n^2}) \hspace{10pt} \{ R=1.0973731568539 \times 10^7 m^{-1} \}

Rydberg Formula: how did you figure it out, sir?

Wavelength of the Lyman Series
n	2	3	4	5	6	7	8
nm	121.50	102.51	97.20	94.92	93.73	93.02	92.57

Wavelength of the Paschen Series
n	4	5	6	7	8
nm	1,874.60	1,281.46	1,093.52	1,004.67	954.34

At first glance, it may seem that there is a decreasing pattern within the same series, but it is difficult to determine how the values change between different series. It is challenging to infer the transition from 4 to 3 using only the values in the Lyman series. We desire a consistent value for the spacing between n shells, so wavelengths are not suitable. However, if we express the wavelengths in terms of their reciprocals, called wave numbers, it can be represented as follows:

Wave Number of Lyman Series
n	2	3	4	5	6	7	8
cm^-1	82,302.98	97,544.28	102,878.73	105,347.82	106,689.05	107,497.77	108,022.67

Let's focus on the gaps between the wave numbers. In other words, we will examine the values when transitioning from 2 to 1 and from 4 to 3.

Wave Number Differences of the Lyman Series
n	$3 \to 2$	$4 \to 3$	$5 \to 4$	$6 \to 5$	$7 \to 6$	$8 \to 7$
cm^-1	15,241.29	5,334.45	2,469.08	1,341.23	808.72	524.89

Wave Number Differences of the Paschen Series
n	$5 \to 4$	$6 \to 5$	$7 \to 6$	$8 \to 7$
cm^-1	2,469.08	1,341.23	808.72	524.89

The difference in wave numbers is the same for both series. This supports the hypothesis that a specific energy is conserved in the space between the n shells in the form of Ether. Each corresponding Photon-Ether between the shells has a constant energy, and it can be interpreted as being converted into exactly that amount of Photon-Sparkle and emitted. Now, the absorption and emission of energy have been combined as additions.

Emission Energy of Photon-Ether

If photon-Ether has a constant wavelength, we can determine its energy. The energy of light can be calculated using the formula $E = hc / λ$ , where λ represents the previously calculated wavelength. The value obtained has units of Joules (J). Converting this energy value to electron volts (eV) yields the following formula, and organizing it results in the table below.

E = Rhc(\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2})\cdot6.242\cdot10^{32}

Since the Rydberg constant, Planck's constant, and the speed of light are all constants,

E = Rhc(\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2})\cdot6.242\cdot10^{32}

E = 1.0973731568539 \cdot 10^{-7} \cdot 6.62607015 \cdot 10^{-34} \cdot 299792458 \cdot (\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2}) \cdot 6.242 \cdot 10^{32}

E = 13.60676328 \cdot (\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2})

The formula for calculating energy (in electron volts, eV) using the Rydberg formula

Result Energy from the Formula
n	$2 \to 1$	$3 \to 2$	$4 \to 3$	$5 \to 4$	$6 \to 5$	$7 \to 6$	$8 \to 7$
eV	10.2050	1.8898	0.6614	0.3061	0.1663	0.1002	0.0650

In the Bohr atomic model and the Schrödinger equation for the hydrogen atom, the energy of the nth shell satisfies the following equation:

E_n = -\frac{ℏ^2}{2 \mu a_0 ^ 2} \frac{1}{n^2}

By substituting $\dfrac{1}{n^2}$ with $(\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2})$ and J with eV, we can observe that the obtained expression closely matches the one derived earlier.

\frac{ℏ^2}{2 \mu a_0 ^ 2} (\dfrac{1}{(n - 1)^2} - \dfrac{1}{n^2}) \hspace{10pt} \{ n \geqq 2 \}

n	$2 \to 1$	$3 \to 2$	$4 \to 3$	$5 \to 4$	$6 \to 5$	$7 \to 6$	$8 \to 7$
1st values	10.2050	1.8898	0.6614	0.3061	0.1663	0.1002	0.0650
2nd values	10.2009	1.8890	0.6611	0.3060	0.1662	0.1002	0.0650

Rydberg of Photon-Ether simple plus calculation

As seen in the diagram, we can now easily determine the energy possessed by the ether through addition. We are now able to calculate energy, wavelength, and even mass.

Summary

Everything discussed here is quite obvious. It's basic knowledge that can be found in any classical physics textbook. These concepts were already well-established during Niels Bohr's time, specifically in the context of hydrogen atoms. However, they do not hold true for multi-electron atoms. You have been deceived. Haha!

What I am looking for here is to present the conditions under which the ether-spark hypothesis can be convincing.

Firstly, the energy possessed by the ether should be predictable.
Secondly, the energy possessed by the ether should be obtainable through addition.
Thirdly, these conditions should hold true even for multi-electron atoms.

If these conditions are satisfied, it would be possible to demonstrate that energy is stored in the form of ether. In the next section, we will delve into multi-electron atoms.