﻿ A Revision of the Estimates of the Parameters of the Nuclear Force Formula based upon the Separation Distance of the Centers of the Nucleons in a Deuteron

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 Revision of the Estimates of the Parameters of the Nuclear Force Formula based upon the Separation Distance of the Centers of the Nucleons in a Deuteron

The separation distance of the proton and neutron in a deuteron provides a way of estimating the parameters in a formula for the nuclear strong force. It is the separation distance of the centers which is crucial. This quantity can be obtained by subtracting the sum of the radii of the proton and neutron from the diameter of the deuteron. There are various concepts of the diameters and radii of particles and nuclides. One is the root-mean-square (rms) charge radius. There are others, such as the mass radius and the magnetization radius, but so long as the same concept is used for the deuteron, proton and neutron the separation distance of the centers of the proton and neutron should be the same.

A group of physicists under the editorship of Savely G. Karshenboin published in 2008 a book devoted to the compilation of the best estimates of physical properties of particles, simple atoms and simple molecules (Precision Physics of Simple Atoms and Molecules, Springer-Verlag).

The best estimate of the rms-charge diameter of a deuteron from page 70 of the above mentioned work is 4.260 fermi with a margin of error of ±0.02 fermi. The recommended estimate of the rms-charge radius of the proton, given on page 49 of the above work, is 0.895 fermi. Precision Physics of Simple Atoms and Molecules does not give an estimate for the radius of the neutron. Another source gives the rms-radius of the neutron as 1.11 fermi.

Thus the separation distance of the centers of the nucleons is

## Derivation of Estimates of the Parameters of the Nuclear Strong Force

The nuclear strong force between two particles of nucleonic charge Z1 and Z2 is assumed to be given by the formula

#### F = HZ1Z2e−s/s0/s²

where H and s0 are parameters to be estimated. A value of s0 can be derived from the Yukawa Relation and the mass of the pi mesons. Its value is 1.522 fermi. The sign of F is given by the sign of Z1Z2. It is negative and the force is an attraction only if Z1 and Z2 are of opposite sign. If Z1 and Z2 are both of the same sign, either positive or negative, then the sign of F is positive and it is a repulsion.

Previous analysis was based upon the presumption that the nucleonic charges of a proton and neutron were equal. Now it is known that they are of opposite sign and the value for the neutron is two-thirds of that of a proton. The nucleonic charge of the proton can be defined as +1. This make the nucleonic charge of a neutron −2/3 and the product of the charges also −2/3.

## The Extended Bohr Model for a Deuteronic Particle

Let m1 and m2 be the masses of the two particles and r1 and r2 be their distances for their center of mass. These values satisfy the condition

#### m1r1 = m2r2and thus r2 = (m1/m2)r1

Since the separation distance s is equal to (r1+r2)

#### s = (1 + m1/m2)r1 = [(m1+m2)/m2]r1and hence r1 = [m2/(m1+m2)]s and likewise r2 = [m1/(m1+m2)]s

If ω is the rate of rotation of the system then the tangential velocities are ωr1 and ωr2. and thus the angular momentum L of the system is

#### L = m1ωr1² + m2ωr2

Since m1r1² is equal to s²m1m2²/(m1+m2)² and likewise m2r2² equals s²m1²m2/(m1+m2)², the angular momentum L reduces to

#### L= s²ω(m1m2)/(m1+m2) and this further reduces to L = s²ωμ

where μ is the reduced mass of the system and is given by

#### 1/μ = 1/m1 + 1/m2

The reduced mass of the proton and neutron is 8.368746×10−28 kg.

In the Bohr model the angular momentum of the system is quantized to a multiple of Planck's constant. This is not just a raw speculation; see Bohr for its justification. Thus

#### L = s²ωμ = nh

where n is an integer and h is Planck's constant divided by 2π.

For circular orbits the force on each particle must balance the force required to keep it in that orbit; i.e.,

Since

#### m1r1 = m2r2 = sm1m/(m1m2) = sμ the condition for a circular orbit is HZ1Z2e−s/s0/s² = sμω² or, equivalently ω² = (HZ1Z2/μ)e−s/s0/s³

From the quantization condition for angular momentum it is found that

#### ω² = (nh)²/(μ²s4)

Equating these two expressions for ω² gives

#### (HZ1Z2/μ)e−s/s0/s³ = (nh)²/(μ²s4) which reduces to se−s/s0 = (nh)²/(μHZ1Z2) or more conveniently (s/s0)e−s/s0 = (nh)²/(μHs0Z1Z2)

This is the quantization conditions for the separation distance. Thus the separation distance of the nucleons is a solution of a transcendental equation.

Let s/s0 be denoted as z and let σ denote (nh)²/(μHZ1Z2) so the quantization takes the form

#### z*e-z = (σ/s0)

Note that σ is a natural unit of length for the system.

For a known value of the separation distance s the value of z can be determined and from that σ is determined. From σ, H can be determined.

For a separation distance s=2.252 fermi, z is then equal to 2.252/1.522=1.4796. This means that

#### (σ/s0) = 1.4796e-1.4796=0.33694 and thus σ = (0.336944)(1.522 fermi)= 0.5128 fermi.

Since for the base level n=1, σ = h²/(μHZ1Z2) and for the deuteron Z1Z2=2/3

#### H = h²/(μσ(2/3)) = (3/2)(1.11212132×10-68/8.368746×10−28)/σ = (3/2)1.3288984×10-41/σ = (3/2)2.591310×10-26 = 3.887×10-26 kg*m³/sec²

Thus the nuclear force between two particles of nucleonic charge Z1 and Z2 is given by

###### F = HZ1Z2e−s/s0/s² where H = 3.887×10-26 kg*m³/sec²and s0 = 1.522 fermi

The force constant H can be expressed as a multiple of hc=3.161510-26 , where c is the velocity of light; i.e.,

#### H = 1.224hc

This is an order of magnitude larger than the value derived from the Virial Theorem. See Deuteron15.

## The Potential Energy Function

For a force F which is a function of distance s the potential energy function V(s) is

#### V(s) = −∫s∞F(x)dx

For the force formula given previously this evaluates to

#### V(s) = −∫s∞H*exp(−x/s0)(dx/x²)

If the variable of integration x is replaced by z=x/s0 then the above expression becomes

#### V(s/s0) = −(H/s0)∫s/s0∞H*exp(−z)(dz/z²) or, equivalently V(s/s0) = −(H/s0)W(s/s0)

where W(s/s0)=∫s/s0exp(−z)(dz/z²), a dimensionless function.

The evaluation of (H/s0) is of some interest. It is

#### (H/s0) = 3.887×10-26/1.522×10-15 = 2.55388×10-11 joules

In millions of electron volts (MeV) this is

## Conclusion

The value of (H/s0) obtained through the use of the Virial Theorem is 18.3636 MeV. Since the evidence based upon the Virial Theorem is stronger, the only conclusion possible is that the extension of the Bohr model is flawed as a model of the deuteron, perhaps as a result of neglecting the radial kinetic energy.