Previous work developed and the estimated the parameters for a formula of the nuclear force. This force can now be compared with the electrostatic force the operates between two protons in a nucleus.

The results of that previous work concerning the deuteron imply that the formula for nuclear force expressed with respect to particle separation d is:

F = −H*e^−d/d₀/d²
where
H* = 1.92570×10⁻²⁵ kg·m³/s²
and
d₀ = 1.522×10⁻¹⁵ m.

The radius of the single feasible orbit for the proton with respect to the center of mass of the nucleon pair 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 potential energy for the two-nucleon system of 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.

Force Comparisons

Let d be the separation of two nucleons. If the nucleons are both protons then they experience a force due to electrostatic repulsion given by

F = −ke²/d²
where
k = 8.988×10⁹ N m²/ C²
and
e = 1.602×10⁻¹⁹ C
and hence
|F| = 2.30668×10⁻²⁸/d²

Now the two forces can be compared at the range of distances involved in the nucleus.

At distances of a few fermi the nuclear force is overwhelmingly larger than the electrostatic force, but at 10 fermi they are of the same magnitude. Beyond 10 fermi the nuclear force is insignificant compared with the electrostatic force. Thus in a nucleus that is so large that some protons are separated by distances greater than 10 fermi an element of instability is created.

There are other forces involved such as the gravitational attraction between the nucleons, but the gravitation attraction is so weak that even at a distance of 0.5 fermi it is less than 10⁻³⁰ newtons.

Potentials

The potential energy for the electrostatic force is very simple to compute; i.e.,

P(d) = ke²/d
= 2.30668×10⁻²⁸/d

The potential for the nuclear force may be expressed as a formula but its evaluation is more complex

P∞_d[H*exp(−s/d₀)/s²]ds
with a change of the variable of integration to z=s/d₀
this becomes
P(d) = −(H*/d₀)∫^∞_d/d0[exp(−z)/z²]dz

Integration-by-parts may be applied to the integral by letting U=exp(-z) and dV=(1/z²)dz. Thus V=−1/z and dU=−exp(-z)dz. Hence

∫^∞_xexp(-z)/z² = [UV]^∞_x − ∫^∞_xVdU
which reduces to
∫^∞_xexp(-z)/z² = exp(-x)/x − ∫^∞_x[exp(-z)/z]dz

This latter integral is closely related to a function which has a name. The named function is the exponential integral, Ei(x), but differs slightly in definition from the required integral. The potential for the nuclear force can be computed using numerical integration. The results are shown below. The potential is plotted in units of 10^-10 joules; i.e. 10^-10 kg m²/s².

At a distance of 1.0 fermi the electrostatic potential is equal to 2.307×10⁻¹³ joules but the nuclear potential is equal to -9.89×10⁻¹² joules, a value about 43 times larger in magnitude. Expressed in electron volts these potentials are 1.44 Mev and -61.73 Mev, respectively.

At a distance of 4.0 fermi the electrostatic potential is equal to 5.077×10⁻¹⁴ joules. At that distance the nuclear potential is equal to -6.943×10⁻¹⁴ joules, a value only about 37 percent larger in magnitude. Expressed in electron volts these potentials are 0.32 Mev and -0.43 Mev, respectively.

(To be continued.)