San José State University |
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The Quantization of the Magnetic Moments and Angular Momenta of Nuclides: Part II |
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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 of any nucleus due to the net intrinsic spin of its nucleons is due entirely to any singleton proton and/or singleton neutron which it might have.

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 could be computed by dividing its angular momentum L by its mass M and multiplying by its charge Q; i.e.,

where α is a constant of proportionality. Angular momentum may be quantized. This would make μ directly proportional to the charge/mass ratio (Q/M).

More precisely 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) is proportional to p/(p+γn).

There could be a slight variation in μ with proton number p and neutron number n because of their effects on the ratio (Q/M).

The above relationship can be read the other way around; i.e., used to compute angular momentum from the magnetic moment, That is to say

Here is the graph of the data for the quantity (μ(1 + γ(n/p)) as a function of the proton number p

The distribution function for the quantity (μ(1 + γ(n/p)) is as follows

The graph indicates a high concentrations at one level but two peaks at other levels symmetrically on eac side. The extreme values at either end of the range are left out of the analysis. The shape of the graph indicates some sort of centrality for the distribution function for angular momentum.

The standard theory for the total magnetic moment for a nuclide ν indicates that the following equation should hold:

where σ is the intrinsic moment of a nuclide due to its unpaired nucleons. The value of σ is given by

Thus the regression equation ν=c_{0} + c_{1}σ + c_{2}[p/(p+γn)/p]
should give a value for c_{1} not statistically different from 1.0.

The results of the regression on the 1003 observations are:

[0.15] [1.4] [15.5]

The t-ratios in the square brackets indicate that the magnetic moment is strongly dependent on the Q/M ratio. But the regression coefficient for σ is not significantly different from zero at the 95 percent level of confidence. Of course since the regression coefficient is closer to 1.0 than it is to 0.0 this means that it is not statistically different from 1.0 at the 95 percent level of confidence. So the standard theory of nuclear magnetic moments is confirmed in this awkward way.

(To be continued.)

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