San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 Further Confirmation that the Nucleonic (Strong Force) Charge of a Neutron is Opposite in Sign and Two Thirds of the Magnitude of that of a Proton

There are three types of interactions between nucleons in nuclei which affect the binding energy. First there is the interaction of spin pair formation. There are the three types of spin pairs; neutron-neutron, proton-proton and neutron-proton. These all involve an attraction. The second type is the strong force interactions. This is a repulsion between like type nucleons and an attraction between unlike type nucleons. The third type is the electrostatic repulsion between protons.

For now in the analysis the electrostatic repulsion between protons is ignored. Below is a tabulation of the number of interactions of the first two types for a nucleus with n neutrons and p protons.

 Interaction Number n-p spinpair min(n,p) n-n spinpair [n/2] p-p spinpair [p/2] n-p n*p n-n n(n-1)/2 p-p p(p-1)/2

The symbols [n/2] and [p/2] denote the integer parts of n/2 and p/2.

Let the strong force charge of a proton be taken as 1 and that of a neutron denoted as −q. The force between particle of charges q1 and q2 is proportional to q1q2. The potential energy is also proportional to q1q2 and therefore also the binding energy. Then the binding energy associated with the strong force interaction of a neutron and a proton is proportional to 1*q, between two protons is proportional to 1*1 and between two neutrons is proportional to q*q=q².

The equation for the binding energy BE should be

#### BE = c1min(n,p) + c2[n/2] + c3[p/2] + Hq*np − Hq²*n(n-1)/2 − H*p(p-1)/2

where H is a constant that depends upon the parameters of the strong force and the separation distance between nucleons.

## The Regression Results

The regression coefficients obtained by regressing the binding energies of the 2931 nuclides on the above variables are

#### BE = 10.31174·min(n,p) + 13.83460·[n/2] + 4.31001·[p/2] + 0.27804·n*p − 0.19254·n(n-1)/2 − 0.48958·p(p-1)/2 [34.7]   [76.9]   [8.3] [35.9]   [-37.3]   [-40.8]

where the numbers in the brackets below the coefficients is the coefficient's t-ratio, the ratio of the coefficient to its standard deviation. The t-ratio for a coefficient must be two or greater in magnitude for the coefficient to be statistically significantly different from zero at the 95 percent level of confidence. The coefficient of determination (R²) for the regression is 0.999882068.

## Implications of the Results

From the regression equation

#### BE = c1min(n,p) + c2[n/2] + c3[p/2] + Hq*np − Hq²*n(n-1)/2 − H*p(p-1)/2

it is seen that the regression coefficient for n*p should be Hq and that of p(p-1)/2 should be −H. Therefore the ratio of the coefficient of n*p to that of p(p-1)/2 should be −q. The value of that ratio is −0.6925. The coefficient of of n(n-1)/2 should be −Hq² while that for n*p is Hq. Therefore the ratio of the coefficient of n(n-1)/2 to that of n*p should also be −q. That ratio is −0.5679. The ratio of the coefficient of n(n-1)/2, −Hq², to that of p(p-1)/2, −H, should be q². That ratio is 0.393289 and its square root is ±0.62713. Other studies concluded that the value of q is 2/3. These results confirm that estimate.

## The Effect of the Electrostatic Repulsion Between Protons

The electrostatic repulsion of protons would only affect the coefficient for p(p-1)/2. This would give a coefficient that is −(H+Δ) where Δ is some positive amount. Thus the regression equation would be

#### BE = c1min(n,p) + c2[n/2] + c3[p/2] + Hq*np − Hq²*n(n-1)/2 − (H+Δ)*p(p-1)/2

Thus the ratio of the coefficient of n*p to that of p(p-1)/2 would be −Hq/(H+Δ) and thus smaller in magnitude than −q. Likewise the ratio of the coefficient of n(n-1)/2 to that of p(p-1)/2 would be Hq²/(H+Δ) and thus smaller than q². The regression results are consistent with these deductions.

## Conclusions

The nucleonic (strong force) charge of the neutron is opposite in sign to that of a proton and two thirds of its magnitude.