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The image of the spatial structure of the nucleus that is emerging is that of coupled concentric shells which might be rings, spheres or cyclinders. The capacities of the shells are determined by sets of the allowable values of quadruplets of quantum numbers. The nucleons (protons and neutrons), being fermions, cannot have wave functions which overlap with the same set of quantum numbers. The shells' structures and positions are maintained by rotation and/or Pauli Principle exclusion. The nucleons within a shell form pairs of nucleons, but these pairs of like nucleons are repelled from each other and thus are dispersed away from each other in the shell as much as possible. There are separate shells for the neutrons and the protons. The proton shells are located just beyond the corresponding neutron shells because, among other things, the protons experience the electrostatics repulsions in addition to that due to the strong force.
The protons in an outer shell are attracted to the neutrons in the inner shells and repelled by the protons in the innershells. There is also a partial repulsion of a proton due to protons in the same shell.
One quantity that gives some information on the relative scale of the shells is the relative size of the incremental binding energy of the last nucleon in a shell compared to the incremental binding energy of the first nucleon in the next shell. The first shell has a capacity of 2; the second a capacity of 4 according to a modified version of the nuclear magic numbers.
The nuclide with 6 protons and 6 neutrons is Carbon 12, which has a binding energy of 92.161728 million electron volts (MeV). The binding energy of the nuclide with 6 protons and 5 neutrons is 73.4399 MeV. The incremental binding energy of the 6th neutron (the 4th in the second shell) is the difference of these two numbers, 18.721828 MeV. The binding energy of the nuclide with 6 protons and 7 neutrons is 97.108063 MeV, so the incremental binding energy of the 1st neutron in the third shell is only 4.946335 MeV. It is lower because that neutron is located farther away from the center of the nucleus and hence farther away from the protons and neutrons in the inner shells. In addition the other neutrons on the second shell had only a partial effect on the last neutron in the second shell. For the neutron in the third shell the full effect of the repulsion of the neutrons in the second shell is operative. Thus the incremental binding energy for the first neutron in the third shell is only 4.946335 MeV instead of something on the order of 18.721828 MeV. The ratio of these two incremental binding energies is 3.785.
The third shells for protons and neutrons both have a capacity of eight. If third proton shell is filled then the last neutron in the third neutron shell has an incremental binding energy of 17.17969 MeV. The first neutron in the fourth neutron shell has an incremental binding energy of 8.47355 MeV. The ratio of these two numbers is 2.02745.
To see how these ratios of incremental binding energies of the last nucleons in a shell to the first nucleon in the next shell provide information about the relative scale of the shells it is necessary to resort to some drastic simplifications. If the nuclear strong force were an inverse distance squared force like the electrostatic force the potential energy would be of the form
where K is a force constant, Z is the net attraction of the inner shell particles and r is the distance to the center of the nucleus. Binding energy can be considered the negative of the potential energy. The net attractive charge Z is the sum of the number of attractive nucleons in the inner shell less the number of repelling nucleons in the the inner shells and also less the shielding by nucleons in the same shell. To keep matters simple it is assumed that half of the charge of the nucleons in the same shell shields the nucleon.
For the nuclide with 14 protons and 14 neutrons, 6 in inner shells and 8 in the same shell, the value of Z for the last neutron in the third shell would be 14−6−½7=4.5. For the first neutron in the fourth shell the value of Z would be 14−6−8=0. There would be no binding energy for that 15th neutron. The strong force is not the same as the electrostatic force. The shielding of the neutrons in the inner shells is less than full.
For the sake of illustration suppose the shield of the inner shell neutrons is 0.7. The value of Z for the last neutron in the third shell would be 14−0.7(6+½7)=7.35. That of the first neutron in the fourth shell would be 14−0.7(6+8)=4.2. Then according to the formula for V.
Thus the relative radii of the shells would be determined. Of course, this computation was based upon arbitrary assumptions for purposes of illustration.
The incremental binding energy for the last neutron in the fourth shell when the fourth shell of protons is completely filled is 16.641 MeV, whereas the incremental binding energy for the first neutron in the fifth shell is 10.247 MeV. The ratio of these two numbers is 1.624.
In the conventional version of the nuclear magic numbers the second shells both have a capacity of 6. If the above procedure is applied to the nuclide with 8 protons and 8 neutrons the incremental binding energies are 15.663736 MeV and 4.143324 MeV and the ratio is 3.7805. This is notable close to the value of 3.785 found using 4 as the capacity of the second shells.
The capacity of the third shells for the conventional version of the nuclear magic numbers is 12. The procedure applied to the nuclide with 20 protons and 20 neutrons gives incremental binding energies of 15.6412 MeV and 8.3627 MeV with a ratio of 1.87035.
As stated previously, the nuclide with 6 protons and 6 neutrons, Carbon 12, has a binding energy of 92.161728 MeV. The binding energy of the nuclide with 5 protons and 6 neutrons is 76.2048 MeV. The incremental binding energy of the 6th proton (the 4th in the second shell) is the difference of these two numbers, 15.956928 MeV. The binding energy of the nuclide with 7 protons and 6 neutrons is 94.1053 MeV, so the incremental binding energy of the 1st proton in the third shell is only 1.943572 MeV. The ratio of these two incremental binding energies is 8.2101.
The third shells for protons and neutrons both have a capacity of eight. If third neutron shell is filled then the last proton in the third neutron shell has an incremental binding energy of 11.58494 MeV. The first proton in the fourth neutron shell has an incremental binding energy of 2.74811 MeV. The ratio of these two numbers is 4.2156.
First a functional form for the nuclear strong force as a function of separation distance would be needed; i.e.,
where H is a constant. Such a force formula would generate a corresponding potential energy function V(r). In general the potential energy associated with adding a unit nucleon to a nucleus having an attractive nucleonic charge Z would be of the form
where H is the force constant and W(r) = −∫_{r}^{∞}f(s)ds.
Let B(r) be the incremental binding energy of a nucleon at a distance r from the center of the nucleus and B(r)=−V(r). Thus
Consider the last nucleon in the nth shell and denote the radius of the nth shell as r_{n}. Then
Likewise the binding energy of the first nucleon in the (n+1) shell would be
Let Z_{0} be the net attractive charge to the nucleons in the nth shell and m_{n} the number of nucleons in the mth shell. Then
where γ is the share that nucleons in the same shell shield each other from the net attractive force of the other nucleons in the nucleus. Then
It is this relation that must be solved to obtain r_{n+1}/r_{n}.
Consider as an exercise the case in which the potential function is the Yukawa potential and W(r)=exp(−r/r_{0})/r. The parameter r_{0} is known from the Yukawa relation to be about 1.522 fermi. Let (B(r_{n})/B(r_{n+1})) be denoted as β and ((Z_{0} − m_{n}/(Z_{0} − γ(m_{n}−1)) as ζ. The equation to be solved is then
Let the function ρ*exp(−ρ) be denoted as g(ρ). The solution to the equations can be represented as solving the pair of equations
(To be continued.)
A strong candidate for the functional form of the nuclear strong force is
The above force formula means that the potential energy V(r) is given by
(To be continued.)
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