﻿ The Rate of Revolution of an Electron about an Atomic Nucleus
San José State University

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The Rate of Revolution of an

In the Bohr model of an atom an electron revolves about the center of mass of the electron and the nucleus. Even a single proton has 1836 times the mass of an electron so the electron essentially revolves about the center of the nucleus.

That model does a marvelous job of explaining the wavelengths of spectrum of hydrogen. The relative errors in the calculated wavelengths of the spectrum are typically on the order of a few tenths of a percent.

There are some physicists who assert that an electron does not really revolve around a nucleus. But that is just the Copenhagen Interpretation of quantum theory. That interpretation of the solutions to Schrödinger's equations is not believed by many physicists. In fact at a recent conference of quantum physicists a survey was carried out asking each physicist which interpretation of quantum theory he or she believed in. Only 42 percent said they believed in the Copenhagen Interpretation. This was more than any other interpretation, but it was still less than half.

The Copenhagen Interpretation would treat the translucent disk of a rapidly rotating fan as though it was a static particle in which the fan is smeared over a disk-shaped region. In actually of course the translucent disk is the dynamic appearace of the fan which has a perfectly normal physical existence.

## The Rate of Revolution of an Electron in a Hydrogen Atom

The basis for Bohr's model of an atom is that the angular momentum of an electron is an integer multiple of Planck's Constant divided by 2π, h. The angular momentum of a particle of mass m moving in an orbit of radius r at an angular rate of rotation of ω is mr²ω. Thus

#### mr²ω = nh and hence ω = nh/(mr²)

For n=1 the orbit radius is 5.3×10−11 meters. The mass of an electron is 9.11×10−31 kg. Planck's constant divided by 2π is 1.05457×10−34 joule-sec. Thus

#### ω = 1.05457×10−34/[9.11×10−31(5.3×10−11)²] ω = 4.122×1016 radians per second = 6.56×1015 turns per second

This is about 7 quadrillion revolutions per second or equivalently 7 thousand trillion revolutions per second. This is fantastically fast but it is much slower than the rate of rotation of nuclei. At this rate of revolution all that can be observed of the electron in its orbit is its dynamic appearance. Its dynamic appearance is that of a torus.  The question immediately arises as to how high is the tangential velocity of the electron relative to the speed of light. The velocity is given by rω. Thus

#### v = (5.3×10−11)(2.59×1017) = 1.37×107 /sec.

Relative to the speed of light this is

#### v/c = 0.045757

The relativistic correction factor is then

#### 1/(1−(v/c)²)½ = 1.00105

This means corrections for relativistic effects are not significant numerically. Thus the calculated rate of revolution is not inconsistent with the Special Theory of Relativity.

## Magnetic Moments

For a particle with a net charge of Q that is spinning at a rate of ω (radians per second) or ν (turns per second) the effective current is i=Qν=Qω/(2π). The magnetic moment generated by the motion of the electron in its trajectory is the product of the effective current times the area surrounded by the path of the particle. The area of the loop which the current surrounds is πr². Thus

#### μ = iA = Qνπr² or, equivalently μ = iA = Q(ω/(2π))πr² = (Q/2)ωr² = (Q/2)vr

where v is the tangential velocity of the charge. This is analogous to the angular momentum of a particle. The angular momentum involves the mass of the particle rather than the term (Q/2).

It is worthwhile to establish the relationship between the magnetic moment and angular momentum. Let angular momentum be denoted by L. The if L is divided by the mass of the particle and the result multiplied by (Q/2) the result is the magnetic moment μ.; i.e.,

#### μ = QL/(2m)

Angular momentum is quantized so magnetic moment is inversely proportional to mass. Thus the magnetic moment of an electron should be 1836 times the magnetic moment of a proton. The magnetic moment of an electron is 9284.76×10−27 J/sec whereas that of a proton is 14.106×10−27 J/sec. The ratio of these two figures is 658.2135. This is 1836/2.79.

Another way of looking at the problem is that the above relation give the quantum of angular momentum; i.e.

#### L = 2mμ/Q

For the electron the quantum of angular momentum is

#### L = 2(9.10938×10−31)(9284.76×10−27)/(1.602×10−19) L = 1.0559×10−34 J-sec

The value the reduced Planck's constant is 1.0545718×10−34 J-sec. This is amazingly close. However what theory requires is ½h. The problem is resolved by introducing a g-ratio equal to 2 in the analysis. One could equally well account for the result by saying the minimum spin quantum number is 2 instead of 1.

The minimum quantum is not necessarily 1; it is the number of degrees of freedom of the system; i.e., the number of modes of rotation, oscillation and vibration. Albert Einstein derived the implication that the major axes of the planets of the solar system should rotate. The was confirmed for the planet Mercury. If this phenomenon occurs for the planet Mercury with its period of revolution of 88 days consider how much more important it is for an electron revolving about a nucleus 7 quadrillion times per second. So definitely an electron has at least two degrees of freedom.

For the proton the quantum of angular momentum is

#### L = 2(1.6726×10−27)(14.106×10−27)/(1.602×10−19) L = 2.9455×10−34 J-sec.

The ratio of this angular momentum to ½h is 5.586. If a Bohr-Mottelson √I(I+1) rule applies, that ratio can be accounted for by I=5 since √5*6=5.4772. That would mean five degrees of freedom for the motion of a proton. The whole problem of the spin of a proton is covered by an article in the July 21, 2014 issue of Scientific American. For a review of the material of that article see Proton Motion.

There are physicists of the Copenhagen Interpretation persuasion who assert that electrons do not really revolve around nuclei. This is based upon the Uncertainty Principle and the Larmor proposition that accelerated charged particles radiate away energy. A charged particle traveling in a curved orbit will experience acceleration. However the Larmor proposition is for point particles. A charged point particle would have infinite energy in the singularity of its field. Spatially distributed charged particles do not radiate energy.

For an examination of the orbit of an electron and the Uncertainty Principle see Electron and Uncertainty.

(To be continued.)