Orbital Distances

Orbital Distance (r)


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 with the examples.


Mathcad – Orbital Equations Derivation

The following is the derivation of the orbital equations for use with Mathcad or an equivalent program to simultaneously solve a series of 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 F2 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 F3 but this force scales similarly for any additional electrons, so the same steps are repeated for aF4,F5, etc.

Sum of forces equation


2) F1 is the attractive Coulomb force based on the number of protons (Z) and one electron.   F2 is the axial, repelling force where Q1 and Q2 are the number of same-spin protons in alignment.  To simplify the equation, they are set to equal and now become Q2.  F3 is an electron (ey) that will affect the electron considered for its orbital, from the distance r3.  F3 can be repeated for other electrons with forces at other radii.

electron orbitals derivation


3) Simplify above and expand r3 from distance rule.

Mathcad electron orbitals equation


4) Temporarily remove Bohr radius (re/⍺2) from above equation.  This makes the solution easier, but the result is now a ratio of the Bohr radius.  This value (re/⍺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.

Mathcad electron orbitals equation without Bohr radius


Note: Eq. 2.3.4 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.


Mathcad equation for atomic orbitals graphical format

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



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.



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 for orbital distances

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.



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 for orbital distances

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 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 (θp) needs to be considered.  Again, annotations below explain the changes at boron to construct the equations that yield the orbital distances.

boron mathcad solution

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.



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 orbital equations


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.