San José State University |
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applet-magic.comThayer WatkinsSilicon Valley & Tornado Alley USA |
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A Statistical Investigation of theAngular Momenta of Nuclei |
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The magnetic moment of a nucleus is due to the spinning of its charges. One part comes from the net sum of the intrinsic spins of its nucleons. The other part is due to the rotation of the positively charged protons in the nuclear structure.

However nucleons form spin pairs with other nucleons of the same type but opposite spin. Therefore the net magnetic moment due to the net intrinsic spin of any singleton proton and/or singleton neutron.

The net magnetic dipole moments of a proton and a neutron, measured in magneton units, are 2.79285 and −1.913. respectively. Therefore the magnetic moment for a nuclide due to the intrinsic spins of its nucleons should be zero for even-even nuclides, 2.79285 magnetons for the nuclides with odd p and even n, −1.913 magnetons for nuclides with an even p and odd n and +0.87985 magnetons for odd-odd nuclides.

The magnetic moment of a nucleus μ due to the rotation of its charges is proportional to ωr²Q, where ω is the rotation rate of the nucleus, Q is its total charge and r is an average radius of the charges' orbits. The angular momentum L of a nucleus is equal to ωr²M, where M is the total mass of the nucleus. The average radii could be different but they would be correlated. Thus the magnetic moment of a nucleus due solely to its rotation could be computed by dividing its angular momentum by its mass and multiplying by it charge; i.e.,

where α is a constant of proportionality depending upon the units of measurement. Angular momentum may be quantized. This would make μ directly proportional to Q and inversely proportional to M. But Q and M can be expected to be approximately proportional to each other. That means that if L is quantized then μ is quantized.

More precisly Q is proportional the proton number p. The mass of a nucleus is proportional to p+γn, where γ is the ratio of the mass of a neutron to that of a proton. Thus (Q/M)≅p/(p+γn) and (M/Q)≅(1+γn/p).

The total magnetic moment μ_{T} of a nucleus is sum of that due to the net intrinsic spin of its nucleons, σ,
and that due to its rotation, μ; i.e.,

and hence

L = (1/α)(μ

Therefore for a nucleus with p protons and n neutrons

The quantities (μ_{T}−σ)(1+γn/p)
were computed for the four categories of nuclides and frequency distributions tabulated.
Those frequency distributions are:

The results for the even-even nuclides clearly show a quantization.
But note that the minimum quantum number for a structure is not necessarily unity. That minimum depends upon
the number of separate modes of vibration and rotation. This is also called *the degrees of freedom* of the system.
In the case of a ring structure there are four modes: 1. As a vortex ring, 2. Spinning like a wheel, 3. Flipping like a coin
about a horizontal axis, 4. Spinning like a coin about a vertical axis.

The results for the even p, odd n also show a quantization, but a ± profile shifted due to the effect of the singleton neutron.

When there is a singleton proton the results are much more complex. There is a quantization but with secondary peaks muc more significant than in the first two cases.

Since the magnetic moment of an even-even nuclide is given by

and (Q/M) as a function of p has the form

the plot of magnetic moment versus proton number would be expected to look like this

Here is the actual plot of the data

(To be continued.)

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