What came out of a previous investigation in which the clusters were identical and each contained q nucleons is that the energy levels of such systems are roughly proportional to the fourth and fifth powers of q. This is the investigation of the more general case in which the number of nucleons in the clusters can be different.

The Model

Consider a system of two clusters in which one cluster has q₁ nucleons and a mass of M₁ and the other q₂ nucleons and a mass of M₂. This is a deuteron of clusters. The alpha particle might be a deuteron of deuterons. On the other hand it might not be. It could be a neutron pair and a proton pair revolving about their center of mass. Or, it could be something entirely different. An accurate explanation of the binding energy of the alpha particle would have to take into account the moderation of the strong force attraction by the electrostatic repulsion of its two protons. What is not taken into account here is the electrostatic repulsion of the protons.

The Reduced Mass of the System

The reduced mass μ is such that

1/μ = 1/M₁ + 1/M₂
which is equivalent to
μ = M₁M₂/(M₁+M₂)

The Orbit Radii

Let r₁ and r₂ be the distances of the centers of the two clusters from the center of mass of the system. Then

M₁r₁ = M₂r₂
and thus
r₂/r₁ = M₁/M₂

The separation distance s for the centers of the two clusters is then

s = r₁ + r₂
which is then equal to
s = (1 + M₁/M₂)r₁
and hence
r₁ = s/(1 + M₁/M₂)
and likewise
r₂ = s/(1 + M₂/M₁)

The above expressions for r₁ and r₂ can be rewritten as

r₁ = [M₂/(M₁+M₂)]s
r₂ = [M₁/(M₁+M₂)]s

Angular Momentum

Let ω be the rate of rotation of the system. The angular momentum of one cluster is M₁ωr₁² and that of the other is M₂ωr₂². The total angular momentum of the system, p_θ can then be expressed as

p_θ = [M₁ωr₁]ωr₁ + [M₂ωr₂]ωr₂

But both [M₁r₁] and [M₂ r₂] are equal to [M₁M₂/(M₁+M₂)]s which is equal to μs. Therefore

p_θ = μωs(r₁+r₂) = μωs²

The system angular momenum is quantized so

p_θ = μωs² = nh

where n is an integer, called the principal quantum number, and h is Planck's constant divided by 2π. Thus

ω = nh/(μs²)
and, for later use,
ω² = n²h²/(μ²s⁴)

Dynamic Balance

Without any loss of generality the nuclear force between clusters of size q₁ and q₂ may be expressed as

F = Hq₁q₂f(s)/s²

where H is a constant and f(s) is a function such that f(0)=1. Later it will be assumed that f(s)=exp(-s/s₀), but for now the form of the force formula will be kept general.

The attractive nuclear force on each cluster must balance the centrifugal force on that cluster. The centrifugal force on the first cluster is equal to

M₁ω²r₁ = [M₁M₂/(M₁+M₂)]ω²s.
which reduces to
μω²s

Thus

μω²s = Hq₁q₂f(s)/s²
and therefore
ω² = Hq₁q₂f(s)/(μs³)

Equating the two expressions for ω² gives

Hq₁q₂f(s)/(μs³) = n²h²/(μ²s⁴)
which reduces to
sf(s) = n²h²/(μHq₁q₂)

This is the quantization condition for the separation distance of the centers of the two clusters. Note that if M₁=mq₁ and M₂=mq₂ then μ is equal to mq₁q₂/(q₁+q₂). This would mean that

sf(s) = n²h²(q₁+q₂)/(mHq₁²q₂²)

In effect, this would make sf(s) inversely proportional to the third power of the average cluster size.

One can define a reduced cluster size ν as

1/ν = 1/q₁ + 1/q₂
which can be expressed as
ν = q₁q₂/(q₁+q₂)

This means that for cluster masses proportional to cluster size μ=mν. It also means that the previous equation for sf(s) can be expressed as

sf(s) = n²h²/(mHν²(q₁+q₂))

Quantization of the Kinetic Energy

A quantization condition for ω may be obtained from

ω² = Hq₁q₂f(s)/(μs³) = Hq₁q₂sf(s)/(μs⁴)
which, by replacing sf(s) with n²h²/(μHq₁q₂)
can be reduced to
ω² = n²h²/(μs⁴)

Since s is a function of n, the above equation quantizes ω.

The kinetic energy K of the system is given by

K = ½M₁ω²r₁² + ½M₂ω²r₂²
which reduces to
K = ½ω²[M₁r₁² + M₂r₂²]
and further to
K = ½ω²μs²

Replacing ω² by n²h²/(μs⁴) gives

K = ½n²h²/s²

This quantizes kinetic energy but the dependence of K on n and q is obscure. A simple approximation helps reveal that dependence.

At least over some range the function sf(s) can be approximated by γs, where γ is a constant. This follows from sf(s) being zero at s=0. Thus

γs = n²h²/(μHq₁q₂)
and hence
s = n²h²/(γμHq₁q₂)
and therefore
s² = n⁴h⁴/(γ²μ²H²q₁²q₂²)

Since K = ½n²h²/s²

K = ½(γ²μ²H²q₁²q₂²/h²)/n²)

Note again that if M₁=mq₁ and M₂=mq₂ then μ is equal to mq₁q₂/(q₁+q₂). This would mean that

s² = n⁴h⁴(q₁+q₂)²/(γ²m²H²q₁⁴q₂⁴)
and hence
K = ½(γ²m²H²q₁⁴q₂⁴/((q₁+q₂)²(h²)/n²)

Thus for q₁=2 and q₂=2, a deuteron of deutrons, would have kinetic energy equal to 2⁴2⁴/4²=16 times the kinetic energy of a deuteron for the same principal quantum number.

Likewise if q₁=2 and q₂=1 then kinetic energy is equal to 2⁴/3²=16/9=1.78 times the energy of a deuteron.

The Quantization of Potential Energy

The potential energy of the system is a function only of the separation distance of the centers of the clusters is given by

V(s) = ∫_s^+∞[−Hq₁q₂f(p)/p²]dp
which reduces to
V(s) = −Hq₁q₂∫_s^+∞(f(p)/p²)dp

Over some range the integral ∫_s^+∞(f(p)/p²)dp can be approximated by α/s^ζ, where ζ≥1. Let q denote the geometric mean as an average cluster size. It was previously noted that if M₁=mq₁ and M₂=mq₂ it would mean that sf(s) and hence s would be inversely proportional to the third power of the average cluster size q. Then given the inverse dependence of s on q³ this means that the above integral and hence potential energy is proportional to q^3ζ. From the above expression for V(s) this means that the potential energy and the hence the binding energy of a cluster deuteron has potential energy is proportional to q^3ζ−2.

Some Interesting Rough Computations

For f(s)=1, ζ is equal to 1 and hence V(s) is proportional to q. Thus a deuteron of deuterons when the nuclear strong force is simply an inverse distance squared force would have two times the potential energy of a deuteron and its binding energy would be only twice as large.

For f(s)=exp(-s/s₀) the value of ζ depends upon the value of s/s₀. For s<s₀ the value of ζ is about 2. In that case V(s) would be proportional to q⁴. Thus the potential energy and hence the binding energy of a deuteron of deuterons would be about 16 times that of a deuteron. With the binding energy of the deuteron of 2.225 MeV this would make the binding energy of a deuteron of deuterons 35 MeV.

Consider a triteron as deuteron and neutron revolving about their center of mass. The geometric mean q of the cluster sizes of 1 and 2 is √2. With the potential energy and binding energy proportions to q⁴ this would make the binding energy of the triteron equal to (√2)⁴=4 times that of the deuteron. With binding energy of the deuteron being 2.225 MeV this would make the binding energy of the triteron equal to 8.9 MeV whereas the conventional estimate is 8.5 MeV.

More Detailed Analysis

Here it is assumed that ∫_s^+∞(f(p)/p²)dp is can be adequately approximated by K/s^ζ over some appropriate range where K and ζ are constants. From previous analysis when the masses of the clusters are proportional to cluster size

s = n²h²/(γμHq₁q₂)
and
μ = mq₁q₂/(q₁+q₂)
and thus
s = n²h²(q₁+q₂)/(γmH(q₁²q₂²)
which for convenience can
be expressed as
s = φ(q₁+q₂)/(q₁q₂)²

where φ=n²h²/(γmH)

Thus

V = −Hq₁q₂(K/φ^ζ)(q₁q₂)^2ζ/(q₁+q₂)^ζ
which reduces to
V = −H(K/φ^ζ)(q₁q₂)^2ζ+1/(q₁+q₂)^ζ

Conclusions

It is not surprising that an alpha particle has a binding energy of 28.29 MeV compared to a binding energy of 2.225 MeV for a deuteron, or that the binding energy of the triteron is 8.48 MeV. It may be just a matter of scale; i.e., cluster size, given the nature of the dependence of the nuclear force on distance.