Since the repulsive electrostatic barrier force (F) = Z1*Z2 * kq^2/R^2 = 2.304e-28 newtons (if
Z1 and Z2 = 1.0, protons and deuterons) at 1.0 meter separation (R) = 1.0e-14 meters it will be 2.304 newtons.
However, 2He4 + 2He4 -----> 4Be8 - 93 KeV = 1.488e-14 newton-meter (joule) repulsive force
at separation distance R = 1.0e-14 meters the repulsive force will be 1.488e-14/1.0e-14 = 1.488 newtons..
2He4 + 2He4 + 2He4 -----> 6C12 + 7.262 MeV = - 1.16e-12 newton-meter (joule) attractive force
equal 116.2/2 newtons at separation distance R = 1.0e-14 meters.
Or, 2He4 + 2He4 + 2He4 + 2He4 -----> 8O16 + 14.42 MeV = 2.272e-12 newton-meter (joule) attractive
force equal 227.2/3 newtons at separation distance R = 1.0e-13 meters.
This indicates that the magnetic "solenoid effect" (or strong force) of multiple quarks/solenoid turns
counteracts the electrostatic repulsive force (when the magnetic dipole fields happen to be aligned
to attract. There are three of these "turns/coils" in each proton (Z1) of charge q, seven in each deuteron (Z1)
10 in 2He3, (Z2) and 14 in each 2He4 (Z2).
This suggests that heavier elements (or protons or deuterons -heavier elements) can undergo fusion easier
than the P-P or D-D reactions, for instance the D + 2He3 reaction cross section is close to the D +T both are
higher than D-D.
I'll figure it out in the Spring. :-)
Frederick
