There is reason to believe that the nuclear strong force has the form

F = H*exp(−s/s₀)/s²

where H and s₀ are constants.

In previous studies, starting from this assumption concerning the functional form of the strong force, it was found

• That the ground state of the deuteron could not be one with nonzero angular momentum.
• That for the formation of a deuteron with nonzero angular momentum and separation distance of 2.252 fermi between the centers of the proton and neutron the loss of potential energy is divided 30 percent to an increase in kinetic enery and 70 percent to the emission of a gamma photon.

The measured energy of the gamma photon associated with the formation or disassociation of a deuteron is 2.22457 million electron volts (MeV). This means the ground state of the deuteron has a potential energy of −3.17418672 MeV and kinetic energy of 0.94961672 MeV.

For circular orbits for the deuteron of equal mass nucleons and quantized angular momentum the quantization condition for separation distance s relative to the scale parameter s₀ is

σ*exp(−σ) = λn²

where σ=s/s₀, n is the principal quantum number for angular momentum and

λ = (h²/(s₀Hm)

where h is Planck's constant divided by 2π and m is the mass a nucleon.

The value of h² is 1.11212×10^-68 kg²m⁴/s². From the Yukawa relation and the mass of the strong-force-carrying particles π mesons the value of s₀ is 1.522×10_{-15 meters. The value of H found in a previous study is about 3.01×10^-27 kg m³/s². The mass of the proton is 1.67262×10^-27 kg. Thus the value of λ is about 1.45, a pure number.}

The shape of the function σ*exp(−σ) is shown below

The maximum for this function occures at σ=1 and is 1/2.71828 = 0.368.

Since λ=1.45 and the minimum value for n is 1 there is solution for the quantization condition. Thus there is no higher energy state for the deuteron. The only state for the deuteron is the ground state of zero angular momentum.