This material shows that stable arrangements of particles such as protons which repel each other can be created with the intermediation of particles to which they are attracted such as electrons. This is the case even when there is a preponderance of the repelling particles.

Consider the simplest of such arrangements, two protons with an electron half way between them. Here this will be referred to as a pseudo-deuteron to emphasize its particle-like character, but else this entity would be referred to as a positively charged ion of a hydrogen molecule, H₂⁺.

Let s be the separation distance of the centers of a proton and the electron. The separation distance of the centers of the protons is then 2s.

The formula for the electrostatic force between two unit charges is of the form

F = ±K/s²

where K is a constant. The negative sign is for attraction and the positive for repulsion.

The force on a proton is then

F = −K/s² + K/(2s)²
which reduces to
F = −(3/4)K/s²

This ignores any nuclear strong force interaction, either attraction or repulsion, between the protons.

The system would have to rotate to counterbalance this attraction.

The potential energy for the two proton-electron interactions is −2K/s. For the proton-proton interaction the potential energy is +K/(2s). Thus the total potential energy is −(3/2)K/s.

The Quantum Mechanics of the Arrangement
of Two Protons and One Electron

Let ω be the angular rate of rotation of the system. The angular momentum L of the system is given by

L = 2mωs²

where m is the mass of the proton. The electron is considered a point particle and hence has no angular momentum.

If this is quantized in units of h, Planck's constant divided by 2π, then

2mωs² = nh
where n is an integer
and hence
ω = nh/(2ms²)
and for future use
ω² = n²h²/(4m²s⁴)

For circular orbits the centrifugal force on a proton is mω²s. The net electrostatic force on a proton is, as indicated above, is (3/4)K/s². Thus for dynamic balance

mω²s = (3/4)K/s²
which implies
ω² = (3/4)K/(ms³)

The Quantization Condition for Separation Distance

When the two expressions for ω² are equated the result is

n²h²/(4m²s⁴) = (3/4)K/(ms³)
which can be expressed as
s = (n²h²)/(3mK)

This is the quantization condition for separation distance.

The value of K is 2.3101×10^-28 kg m³/s², that of m is 1.6726×10^-27 kg. Planck's constant divided by 2π is 1.054572×10^-34 kg m²/s and its square is 1.11×10^-68 kg² m⁴/s². This means that

s = 9.59×10^-15n² meters

The lowest value for s is then 9.59 fermi. The separation distance of the nucleons in a real deuteron is about 2.25 fermi so a real deuteron likely has a different spatial structure than the pseudo-deuteron.

The Quantization Conditions for the Other Characteristics

Since ω=nh/(2ms²) and s²=n⁴h⁴/(9m²K²)

ω = (9/2)mK²/(n³h³)

The tangential velocity of the protons, ωs, then must be given by

ωs = (3/2)K/(nh)

For n=1 this is 3.29×10⁶ m/s, which is only 1.1 percent of the speed of light. The kinetic energy T of the system is then given by

T = 2(½m(ωs)²) = (9/4)mK²/(n²h²)

The potential energy was previously derived as being equal to −(3/2)K/s so its quantization condition is

V = −(9/2)mK²/(n²h²)

As usual for electrostatic force systems the kinetic energy is opposite in sign to the potential energy and one half of its magnitude. For n=1 the potential energy is −2.16×10¹³ joules and the kinetic energy 1.08×10¹³ joules. Thus in the formation of a pseudo-deuteron the particle would go from zero potential and zero kinetic energy at infinite separation to −2.26×10^-14 joules of potential energy and 1.13×10^-14 joules of kinetic energy. The difference between the loss of potential energy and the gain in kinetic energy would go into the emission of a photon of 1.13×10^-14 joules of energy. Since one million electron volts (MeV) is equivalent to 1.6×10^-13 joules the energy of the emitted photon would be 0.113 MeV.

(To be continued.)