San José State University

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The Functional Form of the Incremental
Binding Energies of Nucleons in Nuclides

The Bohr model of hydrogen-like atoms indicates that the ionization energy E for an outer electron is of the form

#### E = −AZ²/n²

where A is a constant, Z is the net positive charge experienced by the electron and n is the principal quantum number for the electron. The net charge Z is the number of positive charges in the nucleus less the shielding by the inner shell electrons. The constant A depends upon the constant in the formula for the electrostatic force and also upon Planck's constant. Its value is approximately 13.6 electron volts (eV).

An equation of the above form explains the ionization energies not only for hydrogen-like atoms and ions but atoms and ions in general if the net charge Z takes into account not only the shielding by inner shell electrons but also the shielding by electrons in the same shell. See Electron Ionization.

The purpose of the material is to derive the functional form for the incremental binding energies of nucleons (neutrons and protons) in nuclei.

Consider a nucleon of charge z interacting with a nucleonic cluster of net nucleonic charge Z. The mass of the single nucleon is m and that of the nucleonic cluster is M. The system rotates about the center of mass and the distances from that center of mass are given by:

#### mrm = MrM and hence rm/rM = M/m

The separation distance s of the centers of the nucleon and the nucleonic cluster is given by

#### s = rm + rM = rM[rmM + 1] or, equivalently s = rM[M/m + 1] = rMM[1/m + 1/M]

The expression [1/m+ 1/M] is the reciprocal of the reduced mass μ. Thus

Thus

## Angular Momentum

If the system is rotating at a rate ω then its angular momentum L is given by

#### L = mωrm² + MωrM² = mωrm[rm+rM] L = mωrms

But mrm is equal to μs so

#### L = ωμs²

The angular momentum is quantized; i.e.,

#### L = nhand thus ωμs² = nh

where n is a positive integer (known as the principal quantum number) and h is Planck's constant divided by 2π.

This means that

## Kinetic Energy

The rotational kinetic energies of the two bodies are

#### Km = ½mω²rm² and KM = ½Mω²rM²

These can be expressed as

#### Km = ½(mωrm)²/m and KM = ½(MωrM)²/M

Since both mωrm and MωrM are equal to ωμs the total kinetic energy K is given by

#### K = ½(ωμs)²/m + ½ (ωμs)²/M = ½(ωμs)²(1/m + 1/M) = ½(ωμs)²/μ and hence K = ½ω²μs²

Since the angular momentum L is equal to μωs² and it is quantized as nh

#### K = ½Lω = ½nhω

But it was previously found that ω is equal to nh/(μs²) so

#### K = (nh)²/(2μs²)

This formula can be examined for the case of the deuteron. Twice the reduced mass for the neutron and proton in a deuteron is 1.67374921×10-27 kilograms. The separation distance of the centers of the nucleons in a deuteron is 2.252×10-15 meters. Planck's constant divided by 2π in the MKS system is 1.054571×10-34 and squared is 1.112122×10-68. Thus for n=1

#### K = 1.112122×10-68/(1.67374921×10-27*(5.071504×10-30)) K = 1.31016295×10-12 joules K = 8.177390 million electron volts (MeV).

When a deuteron is formed there is an emission of a gamma ray of energy 2.224573 MeV. This means that when a deuteron is formed there is a loss of 10.401963 MeV, 8.17739 MeV of which goes into its rotational kinetic energy and 2.224573 MeV of which goes into the emission of a gamma photon.

## Dynamic Balance

The attractive strong force between nucleons is carried by particles which decay. This means that the force between nucleonic charges of z and Z is of the form

#### F = HzZ·exp(−s/s0)/s²

The centrifugal force on the single nucleon is mω²rm. On the nucleonic cluster it is by virtue of the equality of mm and MrM the centrifugal forces on the two bodies are equal. Since both mm and MrM are equal to μs this means the centrifugal force is equal to μsω².

For equilibrium the attractive force must balance the centrifugal force; i.e.,

This means that

#### ω² = HzZ·exp(−s/s0)/(μs³ω²)

From the quantization of angular momentum it is found that

#### ω² = n²h²/(μ²s4)

Equating the two expressions for ω² yields

#### HzZ·exp(−s/s0)/μs³ω² = n²h²/(μ²s4) which reduces to s·exp(−s/s0) = n²h²/(HzZμ) or, equivalently (s/s0)·exp(−s/s0) = n²h²/(HzZμs0)

Thus if (s/s0) is denoted as ζ then the above equation is

#### ζ·exp(−ζ) = n²h²/(HzZμs0)

The function ζ·exp(−ζ) is 0 for ζ=0 and rises to a maximum of 1/e=1/2.72818… at ζ=1 and declines asymptotically toward 0 as ζ increases without bound. This means that there is an upper limit to the principal quantum number n. For now the only relevant interval for ζ is [0, 1].

For convenience let g(x) be the inverse function ζ·exp(−ζ). Thus g() is over the domain [0, 1/e] and has a range of [0, 1]. #### g−1(ζ) = n²h²/(HzZμs0) and hence ζ = g(n²h²/(HzZμs0))

This is the quantization condition for ζ and hence for s=ζs0. From the quantum values of s the quantum values of ω can be derived via

#### L = μs²ω = nhand hence ω = nh/(μs²)

From the quantum levels of s and ω the quantum levels of kinetic and potential energy can be derived.

(To be continued.)