San José State University |
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applet-magic.comThayer WatkinsSilicon Valley & Tornado Alley USA |
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The Rate of Revolution of anElectron about an Atomic Nucleus |
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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 essentialy 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 Interpretatiom 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.

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

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 = 4.122×10

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 appearace 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

Relative to the speed of light this is

The relativistic correction factor is then

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.

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

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.,

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.

For the electron the quantum of angular momentum is

L = 1.0559×10

The value the reduced Planck's constant is 1.0545718×10^{−34} J-sec.
This is amazingly close. However what theory requires is ½~~h~~. 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.

For the proton the quantum of angular momentum is

L = 2.9455×10

The ratio of this angular momentum to ½~~h~~ is 5.586. If a Bohr-Mottelson √I(I+1) rule is involved, that ratio
can be accounted for by I=5 since √5*6=5.4772. The whole problem of the spin of a proton is covered by an article in
the July 21, 2014 issue of *Scientific American*.

(To be continued.)

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