**Background**

The original explanation of an electron’s orbital was proposed by Niels Bohr in 1913 using a model that is very similar to the Earth’s orbit around the Sun – hence the name orbital. The Earth is attracted to the Sun through the gravitational force but it remains in orbit due to its velocity – an equal centripetal force to counter the force of gravity. In the Bohr model, it was assumed that a proton attracts an electron similar to the Sun attracting the Earth. And gravity is replaced by the electrostatic force.

The problem with this model is that it worked well for hydrogen and helium but not for atoms with multiple electrons. It also cannot account for the probabilistic nature of the electron cloud or the electron’s spin. In reality, the electron has a probable distance from the atomic nucleus and the orbital is the average distance it is likely to be.

Another issue is the difference between a proton and a positron that have the same charge. In the case of the positron, the electron is attracted to it and annihilates. In the case of a proton, it finds its orbital. What is the difference between the proton and positron that would cause this to occur?

**Bohr Model – Electron Orbital**

## Explanation

An electron does not circle an atom like a planet circles the Sun. Instead, it is both pushed and pulled at the same time. Bohr originally assumed that the proton was an elementary particle with a single, positive charge. By 1968, the proton was found to be a composite particle consisting of smaller particles (three quarks). In the proton page, the newer discovery of the pentaquark (discovered 2015) is used as the model of the proton, which shows a composite particle of 4 quarks and 1 anti-quark. In this theory the pentaquark is believed to be 4 highly-energetic electrons and a positron at the center.

**Proton**

The composite structure of the proton allows both an attractive force and a repelling force. The orbital electron is attracted to the proton at any angle by the Coulomb force between the positron in the center and the electron. This force is the square of the distance between the particles. The orbital electron is also repelled at the axis between tetrahedral quarks. Within the proton, this is the strong force. Beyond standing waves, it is a repelling force and decreases rapidly at the cube of distance.

**Orbital Distance**

This model allows the position of the electron to be calculated with classical mechanics, which is how orbital distances were calculated. The equation and calculations compared to measured results are provided for this method.

Using the tetrahedral structure of the quarks in the pentaquark model, there are six axes in which the axial force of gluons would repel the electron. As the nucleus continues to spin, this changes the point where the sum of force is zero. The Coulomb force, which is the attractive force, remains constant. But the repelling force is continually changing based on the spin of the proton. The constant attractive force is why the photon ionization energies of electrons are constant and predictable yet the distance of the electron is not. This is modeled below.

This proposed structure of the proton not only leads to accurate distance and photon energy calculations but it explains the curious probability nature of the electron and the shape of atomic orbitals. The electron does not take a straight path to orbit the nucleus like a planet circling the Sun. Instead, it has the probability of being in a particular location around the the atomic nucleus such as below.

Orbital distances for elements with more than one orbital becomes a complex calculation of the constructive wave interference of other electrons in the atom, in addition to the constructive and destructive waves from the nucleus. The distances and angles of the other electrons are required to compute the final orbital distance of the affected electron. Rules for establishing these angles and distances in equations were previously shown in the Equation page.

**Mathcad – Orbital Equations Derivation**

The Equation page showed the steps to calculate an orbital by hand, which is helpful for hydrogen and helium. Beyond helium, a series of equations needs to be solved simultaneously. This section highlights the simplified method to use Mathcad as a software tool to solve the equations. To simplify the complex Mathcad solution, the Force Equation was simplified to only the orbital radii as the unknowns. The Bohr radius was removed from F_{2} but is added back after the solution of radius (r) to get to meters. Thus, the Mathcad solution **provides a ratio of distance to the Bohr radius.**

1) Sum of forces of extra electrons using shorthand notation. Now, another electron is added as F_{3} but this force scales similarly for any additional electrons, so the same steps are repeated for aF4,F5, etc.

2) F_{1} is the attractive Coulomb force based on the number of protons (Z) and one electron. F_{2} is the axial, repelling force where Q_{1} and Q_{2} are the number of same-spin protons in alignment. To simplify the equation, they are set to equal and now become Q^{2}. F_{3} is an electron (e_{y}) that will affect the electron considered for its orbital, from the distance r_{3}. F_{3} can be repeated for other electrons with forces at other radii.

3) Simplify above and expand r_{3} from distance rule.

4) Temporarily remove Bohr radius (r_{e}/⍺^{2}) from above equation. This makes the solution easier but the result is now a ratio of the Bohr radius. This value (r_{e}/⍺^{2}) needs to be re-added to convert to orbital distances from a ratio of the Bohr radius to the actual orbital distance in meters.

Note: The equation expands to the right with more electrons at same distance. *x* is the axis between the affected electron and the nucleus. *y* will expand for each electron and the angle needs to be determined. Electron angle rules were established to estimate these angles.

**Mathcad – Orbital Equation**

Each electron affects the others and so all distances need to be solved simultaneously. Equations were arranged to be solved with Mathcad to generate the orbital distances for each electron in an atom (illustrated below). This solution provides the radius in terms of a **ratio to the Bohr radius.**

**Equation format used in Mathcad to simultaneously solve electron distances in an atom**

## Proof

Proof of the energy wave explanation for the atomic orbital distances is the calculations of:

- 20 orbital distances from hydrogen to calcium for neutral elements.*
- Over 400 orbital distances for ionized elements from hydrogen to calcium.

### Mathcad Solutions for Orbital Distances

The equations in Mathcad are available to download for the first 20 elements, but here, some of the solutions are **annotated** to describe the pattern for the equations and how the solutions are built.

**Helium **

Beyond helium, it is too difficult to manually calculate simultaneous equations because the electrons begin to have two or more distances beginning with lithium (Z=3). Therefore, helium is replicated now using the Mathcad equations to show the same results as above. This matches the manual method to calculate helium for in the example calculations.

**Helium: Mathcad solution of 1s orbital distance (ratio of Bohr radius)**

Using Mathcad, the result is the same as the manual method – 30.2 pm. Note: to arrive at meters, the solution is multiplied by the Bohr radius.* E.g. 0.571 * 5.29177 x 10 ^{–}^{11} = 3.02 x 10 ^{–}^{11 }meters, or 30.2 pm.*

**Lithium**

The equations become more complex beginning with lithium (Z=3) because it begins a new orbital (2s). Therefore, a second equation is required to simultaneously solve the 1s and 2s orbital distances. Each new orbital requires a new equation and appends more repulsive electrons to each equation being solved. These explanations are annotated along with the Mathcad solution below.

**Lithium: Mathcad solution of 1s and 2s orbital distances (ratio of Bohr radius)**

The solution provides the 1s and 2s orbital distances as a ratio of the Bohr radius as 0.397 and 3.272 respectively. In picometers, these distances are **21 pm and 173 pm**.

**Boron **

Boron is the next example, as it now begins the transition to the 2p orbital. Since this is a third distance to calculate, a third equation is added and each equation expands to the right to include the effect of the electron at the 2p orbital distance. Also, this is the first time that the electron angles for the p orbital (θ_{p}) needs to be considered. Again, annotations below explain the changes at boron to construct the equations that yield the orbital distances.

**Boron: Mathcad solution of 1s, 2s and 2p orbital distances (ratio of Bohr radius)**

The solution provides the 1s, 2s and 2p orbital distances as a ratio of the Bohr radius as 0.226, 1.643 and 1.41 respectively. In picometers, these distances are **11.9 pm, 86.9 and 74.6 pm**.

**A****rgon **

The pattern continues for equations and electron angles. Argon, with 18 protons, now includes five orbital distances for 1s, 2s, 2p, 3s and 3p. Mathcad now simultaneously solves 5 unknowns (orbital radii) using 5 equations. The electron angles have been color coded below because there is a distinct pattern.

**Argon: Mathcad solution of 1s, 2s, 2p, 3s and 3p orbital distances (ratio of Bohr radius)**

The complete set of **Mathcad solutions for all neutral elements from hydrogen to calcium are available to download in Mathcad file format.** The ionized elements from hydrogen to calcium were also calculated and put into the tables in the downloadable spreadsheet, although their Mathcad solutions were not provided. Ionized elements simply need to change the Z variable in the Mathcad solution to obtain the ionized element distance. For example, Li1+ uses the configuration and Mathcad solution for He since they both have two electrons. However, the Z value is modified to be Z=3, which is lithium.