The shape of the electron orbit of a Hydrogen atom.

Figure 1. The shape of electron orbit of a Hydrogen atom. The red sphere represents the nucleus, the green sphere represents the electron, the blue line shows the shape of the electron orbit, and the red lines show electrodynamically induced magnetic fields.

For more details please read below.

Quantum Physics has made huge strides forward in the theoretical description of the atomic structure and behavior. Unfortunately, the quantum operator acting on components of kinetic and potential energy elements of Hamiltonian in multidimensional Hilbert space represent a quite abstract topic for students, especially in the absence of the visual representations of orbits, forces, momentums, spins or any other specific terms.

In addition, the final result which shows the probability to find an electron, which rotates around the nucleus in Hydrogen atoms is less than 50%. This does not create much enthusiasm, especially among students. If the students are knowledgeable, they will find that the actual probability to find an electron on the orbit of a Hydrogen atom, according to Quantum Mechanics, is close to 36%.

After some critical analysis of the modern state of the science of Atomic orbitals, we are going to present our creative part. We shall start with the calculation of the electron orbit in a Hydrogen atom.

In this Project we shall not use Statistics or Probability. Accordingly, we shall not take into account the Uncertainty Principle.

This means that whether you prefer to use the word orbit or orbital, we are simply interested in the trajectory of electrons in atoms.

When we say that an electron orbit must satisfy the laws of Physics, we do not mean picking some specific laws, which help to explain our experimental results while ignoring other laws. Our result should satisfy all of the laws of Physics applicable to our Project. Their sequence of order does not play any specific role.

Here we shall need to satisfy following laws of Physics.

1. Law of energy conservation.

In the absence of an external source of energy, the Law of Energy Conservation indicates that the shape of the electron orbit can be either spherical or ellipsoid for a conservative system as follows from the solutions of classic Hamilton equations. The spherical orbit shape is confirmed by experimental data.

2. Coulomb force.

We shall utilize the idea of the equilibrium between centripetal force and the Coulomb force. This equilibrium determines the radius of the sphere, which contains the electron orbit. We have to write these equations, taking into account the actual shape and length of the electron orbit, which will be discussed in this project.

3. Maxwell equations.

The movement of a negatively charged electron in the field of a positively charged proton, creates an electromagnetic field. Radio and TV transmitters are examples and experimental proof of that fact. We cannot accept the idea that anybody authorized Bohr, Schrodinger or any other scientist to forbid atoms to emit photons. Electrons and protons in any antenna have exactly the same properties as electrons and protons in any atom. We have to find such an orbit configuration which prevents the emission of a quantum of electromagnetic energy.

4. Quantum Mechanics. The length of a quantum resonator.

The length of the orbit of an electron in the ground state must be equal to one wavelength. In order to prevent photon emission, the energy induced during the movement of an electron for the distance of one wavelength, should be minimized.  This condition is quite different from the condition used by Bohr. Bohr formulated his hypothesis in terms of momentum, but in terms of the length of an electron orbit, his hypothesis contradicts this principle.

The solution, which satisfies all of these equations simultaneously, will produce the shape of the lowest S-orbit of an electron for the Hydrogen atom.

In our method, we shall analyze electric and magnetic fields separately in order to simplify orbit calculations. By minimizing the integral of either the magnetic or electrical field, induced during one period of wave, we guarantee that no photon is emitted from atom in the ground state.

Calculations of the magnetic field allows us to produce a simple model. In order to explain the logic of our calculations we shall start from a simple model and show the path to the final solution.

Figure 1. Electric coil.

Electric current in the coil on Figure 1 induces a magnetic field, marked with a blue arrow and the letter B. If we put another coil above the first one with an electric current in the opposite direction, we can expect that the sum of these two magnetic fields will be equal to zero. A photon will not be generated if there are magnetic components that cancel each other out.

It means that we need to calculate such an electron trajectory that in the length of one wavelength, the sum of the magnetic fields from each part of the trajectory is minimized.

Figure 2a.

Figure 2b.

Figures 2a and 2b are not actual results or trajectories. They are a model, which can help us explain our method of calculations. The sizes of the proton, electron and the orbit radius are not up to scale.

The orbit shape on Figures 2a and 2b are created from four identical hemispheres. Every point of these lines belongs to the surface of the sphere (Figure 2b).  At the same time, every point of this trajectory belongs to one of the surfaces of the cube (Figure 2a).

In our initial model, the circular trajectory of the electron propagation creates an induced magnetic field in the orthogonal direction according to the Faraday formula:

$\oint E \cdot ds = {\  } – \frac{\partial \Phi _{mag}}{\partial t}$       (1).

Continuous travel of an electron along this trajectory induces magnetic fields with opposite directions. With this configuration, four opposite magnetic fields are created in the length of one round of an electron orbits travel. These magnetic fields from opposite half circles adds one to another and cancel out each other’s field.

We take into account that the half quantum of an electromagnetic wave cannot be emitted after one full wavelength of an electron rotation along such a trajectory.

Now let us check the list of the laws of Physics and see if we have contradictions to any of them.

The law of energy conservation suggests a spherical orbit. Our orbit is spherical.

For an electron in spherical orbit, Mechanics and Electrostatics give us an equilibrium between the centripetal and Coulomb forces:

$\frac{e^2}{4{\ } \pi {\ }\epsilon_0 {\ } r^2} = \frac{m {\ } v^2}{r}$       (2).

In this formula $e$ – is an electron charge, $m$ – is an electron mass, $r$ – is the radius of the orbit, $v$ – is the electrons speed and $\epsilon_0$ – is the dielectric constant of the vacuum.

Energy of a moving electron can be expressed from Classic and Wave Mechanics as:

$\frac{m {\ } v^2}{2} = h {\ } f$       (3).

These energies are equal, because we describe the same particle.

The length of four hemi spheres on Figure 2a and Figure 2b is equal to:

$L = 4 {\ }\pi {\ } r =n {\ } \lambda$       (4).

The frequency of the rotation can be found as the speed of the electron divided by the length of orbit:

$f = \frac{v}{L} = \frac{v}{4{\ } \pi {\ } r}$       (5).

Substitution of the frequency from (5) into (3) will result in:

$ \frac{m {\ } v^2}{2} =h \cdot \frac{v}{4{\ } \pi {\ } r} = \frac{\hbar {\ } v}{2 {\ } \pi}$       (6).

We used $\hbar = \frac{h}{2 {\ } \pi}$ in expression (6).

As a result we got the expression:

$m \cdot v \cdot r = n \cdot \hbar$       (7).

The rotational moment is equal to the whole number, multiplied by the Planck constant, divided by $2 \pi$.

This is the correct expression, which supposedly cannot be produced by Physics. Formula (7) was first formulated and hypothesized by Niels Bohr. It was later also postulated in Quantum Mechanics.

As a reminder, in the case of a circular orbit, the expression in formula (4) would become:

$L = 2 \pi r =n \cdot \lambda$       (4c).

In the case of a circular model instead of formula (7), the result would be:

$m \cdot v \cdot r = 2 n \cdot \hbar$       (7c).

Formula (7c) represents the wrong result. It contradicts experimental data.

It means that with our approach we succeeded to get rid of the necessity to postulate the result, which contradicted the main idea of a resonator, of Wave Mechanics and of Quantum Mechanics. There is no and cannot be any operators, which could explain why the main mode TEM₀₀₀ of an electromagnetic wave cannot be obtained in the resonator with the length equal to one wavelength.

With this simple model, we’ve restored common sense to this problem. The length of a resonator is equal to one wavelength. We’ve also satisfied Mechanics, Electrodynamics, Quantum Mechanics and the Energy Conservation Laws.

Now we can try to produce the complete process of obtaining a Hydrogen atoms energy level spectrum, which incorporates straightforward and simple transformations that agree with experimental data.

The last and most important part is to put real numbers into our model and to check if these qualitative ideas can be achieved in a real model.

Though the short answer is – no. It is not that simple.

There is no actual numerical solution with the electric and magnetic fields arranged along or orthogonal to the $x, y, z$ axes, which would satisfy our model. The model is unsustainable in real numbers for the type of shape in which an electron orbit consists of four hemi-circles.

$  \left \{ \begin {array} {lll} E_x^2+E_y^2=E_r^2\\ E_y^2+E_z^2=E_r^2\\ E_y^2+E_z^2=E_r^2    \end {array} \right \} $       (8).

In these equations $r$ is the radius of each hemisphere. The radius, which would satisfy any one of equations (8) and still belong to the surface of the sphere, would result in the same radius for the second equation to be equal to zero.

It means that the correct solution for the Faraday equation

$\oint E \cdot ds = – \frac{\partial \Phi _{mag}}{\partial t}$       (1)

should be found in alternative forms.

Although several solutions, including a sinusoidal type wave, can be presented as valid solutions for the case of a single electron in a Hydrogen atom. We’ve found three main arguments which indicate that an ellipse projected on a spherical surface is the only valid solution.

• Classic Mechanics solution of Hamilton equation for planets, rotating around the Sun is an ellipse or sphere. These are the only closed orbit solutions.
• For a Helium atom with two electrons on the orbit, the solution for sinusoidal type orbit results in the value of function amplitude, which is too big for one quarter of the surface of a sphere.  It means that para- and ortho- trajectories for a Helium atom would become impossible.
• The ellipse projected on the surface of a sphere type orbital produces orbital moments, which laid the foundation for the calculation of orbits of the next elements of the Periodic table, which correspond to the experimental data. 

These restrictions leave us with the only possible solution for the electron orbit shape. We have to find the function 

$E(r) = f(a, b, r)$.       (9).

In this expression $a$ and $b$ are the values for two parameters of an ellipse. There is no need for the third parameter. The third parameter is the radius of the sphere $r$.

Figure 1 shows the result of our calculations. 

The values of these two parameters are equal to:

$a = 0.707 \cdot r = \frac{2.829 \cdot \pi \cdot \epsilon_0 \cdot \hbar^2 \cdot n^2}{m \cdot e^2}$ and $b = 1.252 \cdot r = \frac{5.008 \cdot \pi \cdot \epsilon_0 \cdot \hbar^2 \cdot n^2}{m \cdot e^2}$       (13).

Here we expressed the axes sizes in the units of the radius of the electron orbit. In spite of the presence of only one dipole, consisting of a proton and an electron, our solution includes quadrupole momentum. For a Hydrogen atom, quadrupole moment appears only in the integral over the length of one wavelength. In the differential analysis of the instant fields, only the dipole moment plays a role. This instant imbalance of the dipole makes the Hydrogen atom highly reactive in Chemistry. Every first part of full cycle induced energy is accumulated in an atom and every second part of the cycle it is added with an opposite sign. Not all moment is compensated and the resulting residual represents a chemical bond. Chemical activity of an atom is the result of non-compensated orbital moment. The noble elements of the Periodic Table have orbital moments as well as spins equal to zero.

Analysis.

The orbit of an electron in our model belongs to the surface of a sphere of radius $r$. Both the function, which represents the electron trajectory and the derivative of this function are continuous and do not have any specific points.

The energy of a moving electron can be expressed from Classic and Quantum Mechanics as

$\frac{m \cdot v^2}{2} = h \cdot f$       (14).

These energies are equal, because we describe the same particle.

The length of the electron orbit according to Figure 2.2 and Figure 2.3 is equal to:

$L = \pi \cdot [3(a + b) – \sqrt{(3a+b) \cdot(a+3b)]} =4 \pi r$       (15).

We used the Ramanujan formula for the length of the ellipse in our calculations.

And the frequency of rotation can be found as the speed of the electron divided by the length of the orbit:

$f = \frac{v}{L} = \frac{v}{4 \pi r}$       (16).

Substitution of the frequency from (16) into (14) will result in:

$ \frac {m \cdot v^2}{2} = h \frac {v}{4 \pi r} = \frac{ \hbar \cdot v}{2 \cdot r}$       (17a).

We introduced $\hbar = \frac{h}{2 \pi}$ in equation (17a).

Equation (17a) can be rewritten as:

$m {\ } v {\ } r = \hbar$       (17).

Equation (17) is the expression, which supposedly could not be produced by Classic Mechanics and needed the hypothesis of Bohr.  This expression is also known as the orbital postulate of Quantum Mechanics.

The reason for this problem is based in the belief that the orbit of an electron is circular. Electron orbit is not circular. Our new electron orbit does not require this postulate and hypothesis. This result follows from simple physical calculations.

Now we can calculate the radius of the electron orbit and the energy levels for the Hydrogen atom.

Since electron orbit belongs to the surface of a sphere, there should be equilibrium between the Coulomb force and centripetal force:

$\frac{e^2}{4 \pi {\ } \epsilon_0 {\ } r ^2} = \frac{m {\ } v^2}{r}$       (18).

From equation (18) the speed of an electron can be expressed as 

$v = \sqrt{ \frac {e^2}{4 {\ } \pi {\ } \epsilon_0 {\ } r {\ }m} } $       (19).

Combination of (17) and (19) gives us the radius of the spherical surface of the electron orbit:

$m {\ } r {\ } \sqrt{ \frac {e^2}{4 {\ } \pi {\ } \epsilon_0 {\ } r {\ }m} } = \hbar \cdot n$

$m^2 {\ } r^2 {\ } \frac {e^2}{4 {\ } \pi {\ } \epsilon_0 {\ } r {\ }m} = \hbar^2 \cdot n^2$

$m {\ }r {\ } e^2 = 4 {\ } \pi {\ } \epsilon_0  {\ } \hbar^2 {\ } n^2$

$r = \frac {4 {\ } \pi {\ } \epsilon_0  {\ } \hbar^2 {\ } n^2}{m {\ } e^2}$       (20).

The energy of the moving electron can be calculated as:

$E = \frac{m{\ } v^2}{2}$       (21).

From equation (19)  and (20), the second power of the electron speed can be expressed as:

$ v^2 =  \frac {e^2}{4 {\ } \pi {\ } \epsilon_0 {\ } r {\ }m} = \frac {e^2 \cdot e^2 \cdot m } {4 {\ } \pi {\ } \epsilon_0 {\ } m {\ } 4 {\ } \pi {\ } \epsilon_0 {\ } \hbar^2 {\ } n^2} = \frac {e^4}{4 {\ } {\epsilon_0}^2 {\ } h^2 {\ }  n^2} $      (22). 

The energies of electron states in a Hydrogen atom are equal to:

$ E_n = \frac {m}{2} \cdot \frac {e^4}{4 {\ } {\epsilon_0}^2 {\ } h^2 {\ }  n^2} = \frac {m {\ } e^4}{8 {\ } {\epsilon_0}^2 {\ } h^2 {\ }  n^2} = \frac {13.6}{n^2} eV$       (23).

The expression for the Rydberg constant is equivalent to the energy of the first state, expressed in the units of reciprocal centimeters:

 $R= \frac {1}{\lambda} = \frac {E_0}{h{\ }c} =  \frac {m {\ } e^4}{8 {\ } {\epsilon_0}^2 {\ } h^3 {\ }  c} $       (24).

These values for energy levels is not much of a new achievement. The values for the orbit radius and energy spectra are well known. Still it is nice to get rid of postulates, which cannot be explained and understood. 

The important result is the new shape of the electron orbit in a Hydrogen atom.

Our next part will cover the Helium atom.