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Investigations of the Radii of
Shells of Alpha Particles in Nuclei

This is an investigation of shell radii for nuclides that could contain an integral number of alpha particles. The procedure used is to find the ratio of shell radii that would occur for several different potential energy functions.

Structural Binding Energies

The masses of nuclides are less than the sum of the masses of the protons and neutrons which they contain. This mass deficit translated into energy equivalence is called the binding energy of the nuclide. The binding energy of a nuclide could be composed of two components. One component could be the binding energy of substructures which are composed of protons and neutrons and another the binding energy due to the potential energy involved in putting these substructures together into an arrangement. Previous work indicates that the protons and neutrons within a nucleus form alpha particles whenever possible. Any additional protons and neutrons beyond the alpha particles substructures form pairs; neutron-neutron, neutron-proton and proton-proton. These pair formations are not mutually exclusive; i.e., a neutron may form a pair with a proton as well as with another neutron.

A major component of the binding energy of a nuclide would then be that due to the formation of alpha particles and nucleon pairs. The rest of the binding energy would be due to the configuration of those alpha particles and nucleon pairs and any excess protons or neutrons. This binding energy will be referred to as structural binding energy (SBE).

There are accepted values of binding energies for alpha particles (28.3 MeV) and for neutron-proton pairs (2.2 MeV) but no accepted value for neutron pairs.

Alpha Particle Substructures of Nuclides

The structural binding energy for those nuclides that could contain an integral number of alpha particles has an interesting form.

The binding energy of a nuclide which could contain multiple alpha particles is in excess of the binding energies of the alpha particles it might contain. The above graph suggests that there are shell structures of the alpha particles within the nuclei. A shell is a collection of particles with the same quantum number(s) and hence at the same distance from the center of the nucleus. There is no significant increase in binding energy for two alpha particles but for three there is. The additional structural binding energy for the number of alpha particles above two is roughly constant at about 7.3 MeV per additional alpha particle until a level of 14 alpha particles is reached. Thereafter the increase is about 2.7 MeV per additional alpha particle.

The numerical stability of the increments in structural binding energy can be examined by computing the increase in binding energy as the number of potential alpha particles increases. This is the incremental structural binding energy (ISBE) of an alpha particle.

Radii of Filled Shell Nuclides

There is a formula in nuclear physics for the radius of a nuclide; i.e.,

R = 1.25A1/3

where R is the distance in fermi from the center of the nuclide to its edge and A is the number of nucleons in the nuclide. This formula is of uncertain empirical validity but it is worth investigating. The radius r that is being considered in this webpage is the distance from the center to the center of the alpha particles of the outer shell. This means that r is R minus the radius of an alpha particle. Here are the values of R for the filled shell nuclides based upon the above formula.

ShellFilled Shell
R (fermi)
Radius to
Shell Particle
r (fermi)
FirstHelium 41.98430
SecondCarbon 122.861790.87753
ThirdSilicon 283.795741.81148
FourthNickel 564.782332.79808
FifthTin 1005.801993.81773
SixthLead 1646.842135.76219

The best estimate of the root-mean-square (rms) radius of the alpha particle (Helium 4) is 1.68 fermi, but the formula gives the value 1.984 fermi. Since the values of R are based on the formula the value of 1.984 fermi was used to compute the radii to the centers of the alpha particles in the shells.

The computed values of r imply intershell separations of 0.87753, 0.93395, 0.98659, 1.0196, and 1.04014 fermi. These are approximately equal to the rms radii of protons and neutrons. The ratio of r for the sixth shell to that of the fifth shell is about 1.27.

Potential Energy Functions

Strictly Inverse Distance Potential Function

A particle in the attractive force field of an aggregation of Z particles in which the force is inversely proportional to the distance squared has a potential energy function of the form

V(r) = KZ/r

where K is a constant and r is the distance from the center of the force field to the center of the particle. The value of Z is in the nature of a net charge experienced by the particle. This means that the ratio of the radii for two particles is given by

r2/r1 = (Z2/Z1)/(V2/V1)

The first shell has only one alpha particle. The second has two; the third four and the fourth seven. The fifth shell of alpha particles has eleven.

Consider the comparison of the incremental binding energy (7.995326 MeV) due the addition of the last alpha particle to the fourth shell and the (2.708326 MeV) due to the first alpha particle to the fifth shell. There are 7 alpha particles interior to the fourth shell. For the seventh alpha particle in the fourth shell there are then seven alpha particle interior to it and six in the same shell. The six which are in the same shell have only a fractional effect. For the moment let us say that the fraction is 0.5. Thus the Z for the last alpha particle in the fourth shell is Z1=7+0.5(6)=10. For the first particle in the fifth shell all of the particles in the fourth shell are interior so the Z for it is Z2=7+7=14. Therefore (Z2/Z1=1.4. The ratio of potential energies is (V2/V1)=2.708326/7.995326=0.338739. There the ratio of shell radii would be

r2/r1 = 1.4/(0.338739) = 4.133

The fraction of the effectiveness of the particles in the same shell is more like 0.3 rather than 0.5. This would make the Z's 8.8 and 14 and hence the ratio of the Z's would be 1.509 and the ratio of the shell radii 4.6966. Actually the effectiveness of all the interior particles is more like 0.7 rather than 1.0. Thus the Z's are Z2=0.7(7+7)=9.8 and Z=0.7(7+0.5(6))=7.0 and thus the ratio of the Z's is 1.4 and the ratio of the shell radii is 4.133.

The Bohr model of the quantization of angular momentum implies that the radius for a particle with principal quantum number n, rn, subject to a central field of charge Z is given by

r = γn²/Z

where γ is a constant.

Thus the ratio of the radii would be given by

r2/r1 = (n2/n1)²/(Z2/Z1)

For the case above in which n2/n1=5/4 and Z2/Z1=1.4

r2/r1 = (5/4)²/1.4 = 1.116

Since the first shell has only one alpha particle and it must be located at the center of the nucleus. Thus the shell radius would be zero. This indicates that the principal quantum number for the first shell is zero and hence the principal quantum number for the fourth and fifth shells would be 3 and 4, repspectively. Thus according to the above

r2/r1 = (4/3)²/1.4 = 1.27

The Yukawa Potential

The Yukawa potential energy function is

V(r) = KZ*exp(−r/r0)/r

where K and r0 are constants. From the Yukawa relation r0 is known to be 1.522 fermi. Let r/r0 be denoted as ρ. Then

2exp(ρ2/(ρ1exp(ρ1) = (Z2/Z1)/(V2/V1)

Since for the comparison between the last particle in the fourth shell and the first particle in the fifth shell the right-hand-side (RHS) of the above equation is 4.133 the solution of the equation involves finding a ζ such that

ζ*exp(ζ) = 4.133
or, equivalently
ln(ζ) + ζ = ln(4.133) = 1.419

The solution to this equation is approximately 1.22. Therefore

ρ21 = r2/r1 = 1.22

The Potential Derived from a
Negative Exponential Weighting of
an Inverse Distance Squared Force

The force function for the nuclear strong force is taken to be

F = −HZ*exp(−r/r0)/r²

where H and r0 are constants. This is based upon the strong force being carried by π mesons which decay over time and hence over distance. The exponential term is the survival factor for the force-carrying particles at a distance r.

The potential function is then

V(r) = −HZ∫r(exp(−s/r0)/s²)ds

If the variable of integration is changed from s to σ=s/r0 the potential function becomes

V(r) = −(HZ/r0)∫r/r0(exp(−σ)/σ²)dσ

Let ρ=r/r0 and

U(ρ) = ∫ρ(exp(−σ)/σ²)dσ

The condition to be satisfied by ρ1 and ρ2 is

U(ρ2)/U(ρ1) = (Z1/Z2)(V2/V1)

Let ζ be such that

exp(−ζ)/ζ² = (Z1/Z2)(V2/V1)


U(ρ2) = [exp(−ζ)/ζ²]∫ρ1(exp(−σ)/σ²)dσ
which reduces to
U(ρ2) = ∫ρ1(exp(−(σ+ζ))/(ζσ)²)dσ

Changing the variable of integration in the integral on the RHH to τ = ζσ gives

U(ρ2) = (1/ζ) ∫ζρ1(exp(−(τ/ζ+ζ))/τ²)dτ

(To be continued.)

The Sum of a Negatively Weighted Exponential and that Due to the Electrostatic Force

(To be continued.)

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