San José State University

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Thayer Watkins
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 Using the Properties of the Deuteron to Estimate the Parameters of the Nuclear Force

This material develops a theory of the nuclear force. It arose out of an examination of Hideki Yukawa's theory of the nuclear force being carried by π mesons. Yukawa applied an exponential factor to the Coulomb potential but the exponential factor should be applied to the force because the exponential factor arises from the decay of the force-carrying mesons with distance. The formula for the nuclear force is then

#### −H*e−λ*d/d²

where d is the distance between the nucleon centers, H* and λ* are constants with dimensions of kg·m³/s² and m−1, respectively. It is more convenient to represent λ* as 1/d0 so the force formula is

#### −H*e−d/d0/d²

The value of d0 can be determined from the mass of the particles, the π mesons, which carries the nuclear force.

In previous work an estimate was obtained for the value of H* based upon the energy required to disassociate the proton and neuteron of the deuteron. However in that work approximations and simplifications were used. This material uses the most accurate data available and takes into account the slight difference in the masses of the proton and neutron. Those masses are

#### Mass of proton: mp = 1.6726×10−27 kg Mass of neutron: mn 1.6749×10−27 kg and thus the mass ratio is mn/mp = 1.001375

Let rp be the radius of the proton orbit in a deuteron. The ratio of that radius to the separation distance between the two nucleons is then

#### rp/d = 1.001375/2.001375=0.5003435.

If the nuclear force formula in terms of the separation distance d between the two nucleons is

#### −H*e−d/d0/d²

where H* and d0 are constants with dimensions of kg·m³/s² and m, respectively, then in terms of the radius of the proton orbit rp the force formula is

where

#### H=H*(0.2503436)=H*/3.99451 and r0=(0.5003435)d0=7.617×10−16m=0.7617 fermi.

For convenience let rp/r0 be denoted as ρ. Previous work derived a condition which ρ must satisfy for a given level of H. Also the previous work derived the relationship between the energy V required to disassociate the deuteron and the values of H and ρ. The two conditions to be satisfied are

#### V = (H/r0)E(ρ) where E(ρ)=(∫ρ∞e−s/s²)ds) or, in logarithmic form ln(V) = ln(H) − ln(r0) + ln(E(ρ)) and ρ = ln(γ) + ½ln(1 + ν²ρ²) where γ=H/(hc) and ν=r0/s0 where s0 = (hc)/(mpc²) = 0.21031 fermi, a natural unit of length.

The value of V is 2.22457 MeV, which in joules is 3.5642×10−13.

For numerical solution it is more convenient to express the two conditions as

#### f(ρ, H) = ln(V) − ln(H) + ln(r0) − ln((E(ρ)) = 0 g(ρ, H) = ρ − ln(γ) − ½ln(1 + ν²ρ²) = 0

The numeric procedure is based upon computing f0 and g0 for trial values of ρ and H and then finding adjustments to the trial values that bring f and g closer to zero.

#### f0 + (∂f/∂ρ)dρ + (∂f/∂H)dH = 0 g0 + (∂g/∂ρ)dρ + (∂g/∂H)dH = 0

where the partial derivatives are computed at the trial values of ρ and H.

Let the trial values of f and g be expressed as a column vector F0 and likewise the adjustments are expressed as a column vector dR. Then the adjustments dR to the trial solutions are found as the solutions to the matrix equation

#### MdR = −F0 where M is the matrix of the partial derivatives. That is to say dR = −M−1F0

The elements of the matrix M are given by

#### (∂f/∂ρ) = −(e−ρ/ρ²)/E(ρ) (∂f/∂H) = −1/H (∂g/∂ρ) = 1 −(ν²ρ)/(1+ν²ρ²) (∂g/∂H) = −1/H

The above system converges to the solution ρ=2.711535 and H=4.820872×10−26. This means that the radius of the orbit of the proton in the deuteron is

#### rp=(2.711535)(0.7617)=2.0654 fermis.

Since H=H*/3.99451 it means that H*=1.92570×10−25 kg·m³/s².

The tangential velocity of the nucleons may be obtained from the expression for β; i.e., β = γe−r/r0/l. The parameter l is the angular momentum number for the proton and, as explained in previous work, must necessarily be equal to unity. The parmeter γ is is equal to H/(hc)=1.5249. Since γ/l=1.5249/1 and ρ=2.711535, β=(1.5249)(0.066435)=0.1013. The results indicate that for the deuteron that the velocity of the nucleons in the lowest energy state is 10 percent of the speed of light. Their individual kinetic energies as a ratio of their rest mass energy are given by [(1−β²)−½−1]. This means that in the deuteron the kinetic energy of the nucleons about 0.5 of 1 percent of their rest mass energy.

Since H*=1.92570×10−25 and the constant for the electrostatic force is 2.31×10−27 the nuclear force is greater than the electrostatic force up to the point where eρ=19.257/0.23102 or ρ= ln(83.36)=4.42. This corresponds to a distance 4.42d0 or d0=6.73 fermi. The ratio H*/(hc), which represents a fine structure constant for the nuclear force, is equal to 6.09. Since this value is greater than unity the analysis of the nuclear force cannot be achieved by a series approximation in the way that the electrostatic force can be where the fine structure constant is 1/137.036. This does not mean that the nuclear force cannot be analyzed; it just means that it has to be done by a different scheme than what is used for the electrostatic force.

## Conclusions

The results imply that the formula for nuclear force expressed with respect to particle separation d is:

###### F = −H*e−d/d0/d² whereH* = 1.92570×10−25 kg·m³/s² andd0 = 1.522×10−15 m = 1.522 fermi

The radius of the single feasible orbit for the proton is 2.0654 fermis. The radius of the neutron orbit is a slightly smaller 2.0626 fermi. The separation distance d of the nucleons is then 4.129 fermi. (The accepted value for the diameter of the deuteron is 4.2 fermi.) The potential energy for the nucleons in the deuteron is then 2.22457 MeV. The tangential velocity of the component particles of the deuteron relative to the speed of light is 0.1013. The angular velocity of the nucleons is 1.470×1022 radians per second or 2.34×1021 revolutions per second.

The angular momentum for the proton is

#### mpvprp = (1.6726×10−27 kg)( 3.03974×107 m/s)(2.0654×10−15 m) = 1.0501×10−34 kg m²/s

The ratio of this angular momentum to h = 1.0545715×10−34 kg m²/s is 0.99576. This is amazingly consistent with the quantization of angular momentum in terms of units of h. There is no where in the above model any element of quantum theory, yet the result is consistent with quantum theory.

The ratio of the angular momentum of the neutron to that of the proton is the product of the square of the ratio of the orbit radii and the ratio of masses. Thus the angular momentum of the neutron is (0.9973)(1.001375)(0.99576)=0.99444h. All in all this is a great vindication of the model.

## Update

The above analysis was based upon the assumption that the mass deficit of the deuteron is equal to the energy of the gamma ray emitted upon its formation. If the mass deficit is taken equal to the decrease in potential energy upon its formation and that this energy is divided between the increase in kinetic energy and the energy of the gamma ray then a different value of H* is found.

###### H* = 3.392372×10−26 kg·m³/s² andd0 = 1.522×10−15 m = 1.522 fermi

This estimate is based upon the diameter of the deuteron being 4.2 fermi and the diameters of the proton and neutron both being 1 fermi. The separation distance of the proton and neutron in the deuteron is 4.2-0.5-0.5=3.2 fermi.

Further details are given in the Spectrum of Two-Particle Systems although there H* is denoted as H.