G. C. Johnson

### aka Electron Coupling Constant

Alpha, (approximately 1/137), is the ratio of the distance between any two point charges, and the photon wavelength with the same energy as the absolute potential between the charges.  In 1916 Arnold Sommerfeld identified this ratio as the fraction of light-speed for the hydrogen atom's ground state electron. Beyond hydrogen, for all single electron ions, the ground state electron velocity/light-speed ratio appears to increase as a multiple of the atomic number, Z.

### de Broglie Wavelength

In 1924 Louis de Broglie discovered that all moving particles have a wavelike property, with a wavelength given by Planck's  constant divided by the particle's momentum.

$\lambda _{d}=\frac{h}{p}=\lambda _{C}\frac{c}{v}$

It was also proposed that electron shells are standing waves which must contain an integer number of de Broglie wavelengths.  This additional constraint on the electron shell gave a new interpretation of Bohr's model in which electron orbits have an angular momentum that is an integer multiple of $\inline \hbar$.

This new paradigm bought atomic physics out of the realm of orbital mechanics and into the modern domain of wave mechanics.

#### The Guess

The guess to be tested is that the ground state of a single-electron ion is defined by two conditions:

Condition I: The electron velocity-to-light speed ratio is the product of alpha and the atomic number, Z.

$\frac{v}{c}=\alpha Z\approx \frac{Z}{137}$

Condition II: The shell circumference must be equal to the electron's de Broglie wavelength.

#### The Consequences

Condition I defines the relativistic Lorentz factor, gamma.

$\gamma =\frac{1}{\sqrt{1-\alpha ^{2}Z^{2}}}$

This allows us to write the relativistic kinetic energy of the electron as:

It can be demonstrated that condition II defines the potential energy as:

$E_{p}=-(\gamma -\frac{1}{\gamma })m_{e}c^{2}$

Taking the absolute value of the sum of kinetic and potential energy yields the ionization energy of single-electron ions.

The above ionization energy is the positive energy which must be added to completely remove the ground-state electron from a nucleus of infinite mass.  The experimental ionization energy is slightly less, since the kinetic energy of the nucleus is positive. (Making the total energy less negative.)

Nucleus kinetic energy can be computed from conservation of momentum. Since the net momentum of the atom cannot be changed by the internal interaction between the nucleus and the electron, their momenta must at all times be equal and opposite.

Setting the classical momentum of the nucleus equal to the electron's relativistic momentum,

$p^{2}=\left ( \gamma ^{2}-1 \right )m_{e}^{2}c^{2}$

the nucleus kinetic energy becomes:

$E_{k(nucleus)}=\frac{p^{2}}{2m_{n}}=0.5\frac{m_{e}}{m_{n}}(\gamma ^{2}-1)m_{e}c^{2}$

#### The Comparison

For hydrogen:

The above model gives a hypothetical ionization energy for an infinite mass nucleus of,

$E_{\infty}=13.60587eV$,

and a nucleus kinetic energy of:

$E_{k(nucleus)}=0.00740eV$

Since,

$E_{expected}=E_{\infty }-E_{k(nucleus)}$,

$E_{expected}=13.59847eV$

From the CRC Handbook of Chemistry and Physics:

$E_{observed}=13.59844eV$

The following graph plots single-electron (+Z-1 ions) observed ionization potentials with the relativistic model.

For low atomic numbers the electron's velocity is a small fraction of the speed of light and the above model corresponds to the Rydberg equation.

For larger atoms, however, the Rydberg equation progressively under-estimates the experimental values.

Cu+28, for example, has a published ionization energy of 11,567.6eV, while the Rydberg model predicts 11,437.6eV;  130eV low, for a relative error of 1.1%.  (Copper is the largest atom for which single-electron ionization energy is published.)

The relativistic model anticipates 11,573.3eV, a 5.7eV overshoot. (A relative error of 0.05%.)

### Classical bonding energies of excited states

While the ground state electron velocity/light speed ratio is the product of alpha and Z, excited states take on integer fractions of that value.  N1 will represent the integer divisor which defines v/c for the general state.

$\frac{v}{c}=\frac{\alpha Z}{N_{1}}$

Similarly, while the ground state shell circumference is one de Broglie wavelength, excited states may contain an integer multiple of that length.  N2 will represent that multiple.  It will be assumed that N1 and N2 are independent quantum numbers.

The following graph shows the classical total energy for the hydrogen atom for the nine intersections of values 1, 2, and 3 for each of the two quantum numbers.

Hydrogen electron velocity Vs. shell radius

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The vertical scale is the hydrogen electron velocity in units of c/137.  (More precisely, alpha times c.)  For other elements, multiply the hydrogen velocity by Z.  (e.g., the ground state velocity of the hydrogen electron is c/137, for He+1 it is 2c/137.)

The horizontal scale is the hydrogen shell radius in units of the Bohr radius.  For other elements, divide the hydrogen radius by Z.  (e.g., the ground state radius of the hydrogen electron is one Bohr radius, for He+1 it is one-half Bohr radius.)

The net bonding energy for hydrogen is shown below and to the left of each intersection.  For other single-electron ions, multiply the hydrogen bonding energy by Z2.  Intersections labeled '>0' indicate that the electron does not bond to the nucleus at that combination of quantum numbers, since the kinetic energy would be equal or greater than the absolute value of the potential energy.

While this model corresponds to the Rydberg equation for cases where N1 equals N2, it offers a much richer array of possible energy transitions.

It should be noted that the transition from the 3,3 state to the 2,2 state has exactly the same (non-relativistic) energy drop as a transition from 2,3 to 3,2.  This energy difference of 1.89 eV represents a photon wavelength of approximately 656 nm, the same as the red doublet emitted by hydrogen.

The hydrogen atom, according to the CRC handbook, also emits a closely spaced ultraviolet doublet at a wavelength of 121.6 nm.  This corresponds to the 10.2eV drop from either a 2,2 to 1,1 transition, or a capture of a free electron directly to the 2,1 state.

To be continued...