The tritium nucleus,  ₁H₂, is the third simplest nucleus, consisting of three nucleons (one proton and two neutrons). The other three-nucleon nucleus,  ₂He₁, is a bit more complicated because of the electrostatic repulsion between the two protons. The binding energy of a tritium nucleus is 8.482 million electron volts (Mev) but that of the Helium 3 nucleus is only 7.718 Mev. The 0.764 Mev difference is due to the mutual repulsion of the two protons in the He 3 nucleus.

This last bit of information is valuable because it can tell us how far apart are the protons in the  ₂He₁ nucleus. The formula for the relation between the potential energy in Mev and the distance in fermi is

V_Mev = 0.6944/d_fermi
d = 1.44/V_Mev (fermi)

(For the derivation see Electrostatic Potential.)

Thus the separation distance is d=1.884 fermi. This corresponds in order of magnitude with the estimate of the distance between the nucleons in a deuteron; i.e., 4.2 fermi. Although the 1.884 fermi distance is the right order of magnitude it is surprisingly small. It would be expected that the distance between two repelling protons in He3 would be larger than the distance between a proton and a neutron in a deuteron which are only subject to the attractive strong nuclear force. It is likely the 4.2 fermi separation distance of the nucleons in a deuteron to be the same order of magnitude of the distances between the nucleons in the tritium nucleus. However the structural arrangement of the nucleons in He₃ may be different from that of tritium and the difference in binding energy could be due to a difference in structural arrangement as well as the repulsion between the two protons in He₃.

Preliminary Analysis

The geometry of the model of the tritium system is given below, in which it is assumed that the masses of the neutron and proton are equal.

For this analysis it is additionally hypothesized that the three nucleons lie in a straight line with equal spacing between them.

The separation distance between the protons in ₂He₁ of 1.884 fermi corresponds to the distance r in the model between the outer neutrons and the proton at the center of mass. The diameter of the deuteron is about 4.2 fermi. If the neutron and proton have a diameter of 0.4 fermi then the separation distance between the centers of the neutron and proton would be 3.8 fermi and the radii of their orbits would be 1.9 fermi. Since the binding energy of a deuteron is 2.224573 MeV, the binding energy for the model of tritium would be approximately 2(2.22457)+(1/4)(2.224573)=5.005289 MeV, far short of the actual binding energy of tritium of 8.482 MeV. If the basic model is correct then the separation distances must be smaller than the 1.9 fermi derived previously.

The nucleons are attracted to each other by a central force which is a function of their separation distance r. For now this force will be represented as F(r).

Let ω be the rate of rotation (angular velocity) about the center of the middle nucleon. A balance of forces on one of the end nucleon requires that

F(r) + F(2r) = mrω²

where m is the mass of a nucleon. This expression can be solved for ω; i.e.,

ω = [F(r)+F(2r)]^½/(mr)^½

The angular momentum of each outer nucleon is mrω so the angular momentum of the system is

2mrω = 2(mr)^½[F(r)+F(2r)]^½

The formula for the force between two nucleons is

F(r) = Hexp(−r/q)/r²

where H = 1.92570×10−25 kg·m³/s² and q = 1.522×10−15 m = 1.522 fermi.

The angular momentum of a nucleon is quantized to integral multiples of Planck's constant divided by 2π (h). Thus

(mr)^½[Hexp(−r/q)/r²+Hexp(−2r/q)/(2r)²]½ = n(h)

where n is a positive integer. This can be reduced to

exp(−r/(2q))[1 + exp(−r/q)]^½/r^½ = n(h)/(mH)^½

For the numerical solution of the above equation it is convenient to express the left-hand side in terms of z=r/q. Thus the equations becomes

exp(−z/2)[1 + exp(−z)/4]^½/z^½ = n(h)(q/(mH))^½

If the average of the masses of the proton and neutron is used for m then the value of (h)((q/mH))^½ is 0.2291625.

Let n=1.

Using a value of 2.959139 for z gives a value LHS of the above equation is equal to (1.0002)(0.2291625) so the value 2.959139 is a bit low but close enough. This means the value of r is equal to 2.959139(1.522)=4.5 fermi. This is on the same order of magnitude as the separation distance of 4.2 fermi for the nucleons in the deuteron but at variance with the separation distance of 1.88 fermi for the protons in the He₃ nucleus. For higher values of the quantum number for the angular momentum the value of r would be smaller.

(To be continued.)

The Coulomb Law for the force of repulsion between two charges separated by a distance r is

F(r) = k_eq₁₂/r²

Thus the potential energy V(r) is given by

V(r) = −∫_r^∞F(s)ds = k_eq₁₂/r

The value of k_e is 9×10⁹ newton-meters²/coulomb². For the charges being those of protons

q₁ = q₂ = 1.6×10^-19 coulombs
and thus
q₁q₂ = 2.56×10^-38 coulomb²

Thus

k_eq₁₂ = 2.304×10^-28 newton-meters²

One newton-meter is equal to 6.2415×10¹² Million electron-volts (MeV).

Thus the potential energy of two protons separated by a distance r (in meters) is

V(r) = (6.2415×10¹²)(2.304×10^-28)/r
= 1.43804×10^-15/r

If r is measured in fermi (10^-15 meters) the formula becomes

v(r) in Mev = 1.43804/r

A computation using more precise values of the figures gives

v(r) in Mev = 1.4399644/r
or essentially
v(r) in Mev = 1.44/r

Therefore,

r in fermi = 1.44/V in Mev