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Upper and Lower Limits of Nuclear Stability |
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Previous work established, contrary to the conventional theory, that like nucleons are repelled from each other and it is only unlike nucleons that are attracted to each other and help hold a nucleus together. Thus a stable nucleus needs a proper mix of neutrons and protons. The stable isotopes of the heavier element have fifty percent more neutrons than protons. If the nucleonic composition of a nucleus deviates too much from the proper ratio the nucleus is unstable.
For the values of proton numbers the maximum and minimum numbers of neutrons in the known nuclides are as shown.
This display can be construed as being elliptical; or rather, being the upper edge of one ellipse and the lower edge of another.
Previous work resulted the following regression estimates.
The equation for an elipse of semimajor axes a and b centered on (0, 0) is
If the ellipse is centered on (h, k) the equation is of the form
If the ellipse is tilted with respect to the horizontal there is an xy term. For a tilted ellipse centered on (0, 0) the equation is
For one centered on (h, k) it is
The expanded version of this is
Stability here is defined as "stable enough to have its mass measured."
Previous work found close fitting equations for the maximum and minimum numbers of neutrons as quadratic functions of the proton number. These did not allow for any quadratic dependence on the neutron number.
To achieve this let the dependent variable be 1 for all cases. The regression equation is then
There of course can be no constant term for the regresion equation. That would just give 1=1.
From the regression coefficients the parameters of an ellipse are found as follows. First the semimajor axes are found by
With these values of a and b the following two equations must be solved.
And finally
The regression equation found for the maximums is
This would correspond to
For the regression equation to correspond to an ellipse the regression coefficients for p² and n² must both be positive. That is not the case for the above regression equation. However solutions for n as a function of p might be found as solutions to the quadratic equation
That is to say
Consider Neon with p=10. The maximum number of neutrons is 22. For this case
Then e² − 4df = −0.583554433 and therefore there is no real solution for n. Something is seriously wrong. When the actual values of p and n are substituted into the regression equation the computed values in no way approximate the dependent variable of 1. What is wrong is the computed regression equation is wrong. For some unknown reason EXCEL cannot handle the case in which all values of the dependent variable are equal to 1.
(To be continued,)
For the minimum limits the equation is
Again the equation cannot be for an ellipse.
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