San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 A Quantum Analysis of a Simple Model of the Alpha Particle

The alpha particle will be considered as a neutron-neutron pair and a proton-proton pair revolving about their center of mass. For simplification the masses of the neutron and the proton will be consider equal and their common value denoted as m. Likewise the separation distances of the nucleon centers will be considered equal to the common value of s0. The neutron pair is situated at a right angle to the proton pair and the separation distance of the midpoints of the nucleon pairs will be denoted as s. A front and side view of the system is shown below. The blue spheres are the neutrons and the red ones are the protons. The center of mass is shown as a black dot. The axis of rotation could be any line passing through the center of mass. For the analysis below the axis of rotation will be parallel to the axis of the neutron pair.

The radius of the orbits of the neutrons is r and r is equal to s/2. The radius of the orbits of the protons is larger. It is such that its square is equal to r²+(s0/2)².

## Angular Momentum

If the system rotates at a rate of ω then the angular momentum L is given by

#### L = 2mωr² + 2mω(r²+s0²/4) = 2mω(2r²+s0²/4) or, equivalently L = 4mω(r²+s0²/8)

Let n be the quantum number for the system. Then

## The Geometry of the Alpha Particle

Let the origin of the three dimensional coordinate system be the center of mass of the particle. The distances of both neutrons to each proton are the same so only one has to be considered. A front and side view of the system is shown below. Let the coordinates for the center of one neutron be (s0/2, 0, r) and that of one proton (0, s0/2, −r). The distance z between the center of a neutron and that of a proton is then given by

## Angular Momentum Once Again

Since r² + s0²/8 = (z/2)² the formula for angular momentum takes the form

#### L = 4mω(z/2)² = mωz² = nh

For later use note that

## Dynamic Balance

The force between the neutron and the proton is assumed to be given by

#### F = qnqpH·exp(−z/z0)/z²

where qn and qp are the nucleonic (strong force) charges of a neutron and proton, respectively. If qp is taken to be +1 then the estimate of qn determined elsewhere is −2/3. The symbol H denotes a positive constant, z is the separation distance of the centers of the two particles and z0 is a positive constant.

The direction cosines of the force are given by:

#### ((s0/2)/z, (s0/2)/z, 2r/z)

The relevant direction cosine for the force on the neutron is (2r/z. Thus the radial component of the force on a neutron is

#### Fr = −((2/3)H·exp(−z/z0)/z²)(2r/z) which reduces to Fr = −(2/3)H·exp(−z/z0)2r/z³

The centrifugal force for a neutron is mrω². Thus for dynamic balance

#### mrω² + (−(2/3)H·exp(−z/z0)2r/z³) = 0 and hence ω² = (4/3)H·exp(−z/z0)/(mz³)

If this expression for ω² is equated with the expression for it previously derived the result is

#### n²h²/(m²z4) = (4/3)H·exp(−z/z0)/(mz³) which reduces to n²h²/(mz4) = (4/3)H·exp(−z/z0)/z³ and further to z·exp(−z/z0) = (3/4)(n²h²)/(mH)

This is the quantization condition for z. It may be put into a dimensionless form by dividing by z0 and letting z/z0 be denoted by ζ. Thus

#### ζ·exp(−ζ) = σαn²

where σα is equal to (3/4)h²/(mz0H). Let ζ·exp(−ζ) be denoted as Γ(ζ), so the quantization condition for ζ is

## Quantization of the Other Characteristics

From the quantization of ζ the separation distance z is given by z=ζz0. The quantization of the radius of rotation is then given by

#### r² = (z/2)² − s0²/8 or, equivalently r² = (ζ²z0² − s0²/2)/4 = z0²(ζ² − (s0²/z0²)/2)/4 and hence (r/z0)² = (ζ² − ½ζ0²)/4 or, since s=2r, (s/z0)² = (ζ² − ½ζ0²)

That of the rate of rotation ω is given by

#### ω = nh/(mz²) which means that ω² = n²h²/(m²z4)

The kinetic energy K is given by

#### K = 2(½mr²ω²) + 2(½m(r²+s0²/4) = 2mω²(2r²+s0²/4) which can be expressed as K = mz²(n²h²/(m²z4)) which reduces to K = n²h²/(mz²)

The potential energy for one neutron-proton strong force interaction is given by

#### V(z) = −∫z∞[((2/3)H·exp(−x/z0)/x²]dx or, equivalently V(z) = −(2/3)H·∫z∞[(exp(−x/z0)/x²]dx

Let y=x/z0 so dx=z0dy. Then the potential energy can be expressed as

#### V(z) = −(2/3)(H/z0)·∫∞z/z0[exp(−y)/y²]dy

Let the mathematical function ∫s(exp(−t)/t²)dt be denoted as W(s). Thus the potential energy is

#### V(z) = −(2/3)(H/z0)·W(z/z0) or, in terms of ζ=z/z0 V(ζ) = −(2/3)(H/z0)·W(ζ)

Thus the quantization of the separation distance z establishes the quantization of potential energy and kinetic energy, hence also of total energy T=K+V.

## Comparison with Case of Quasi-Alpha Particles

nucleons can form two spin pairs; one with a nucleon of the same type and one with a nucleon of the opposite type. This can lead to chains of linkages of the form n-p-p-n-n-p-p-n … as shown below. These chains are made up of modules involving two neutrons and two protons. Such modules are like alpha particles and can be called quasi-alpha particles. The balance of the forces in a chain of quasi-alpha particles is the same as those in a chain of alpha particles. Therefore the binding energies in nuclei made up of chains of quasi-alpha particles are essentially the same as those nuclei made up of alpha particles.

Most of the binding energy of a nucleus can be accounted for by computing the binding energy which would result from the formation of alpha particles within it. The binding energy of an alpha particle is about 28.3 million electron volts (MeV). The problem with the alpha particle model of nuclei is that alpha particles repel each other not only through the electrostatic force but also through the nucleonic (strong force). The linkages described above explain how the nucleons in shells are held together. The notion of quasi-alpha particles explains why the binding energies are about what would be expected if the nucleons in a nucleus formed alpha particles whenever possible.

## Comparison with Case of a Deuteron

The conditions stemming from the quantization of angular momentum are

#### 2mωr² = nh and hence ω = nh/(2mr²) and ω² = n²h²/(4m²r4)

Dynamic balance requires

#### mrω² − (2/3)H·exp(−2r/z0)/(2r)² = 0 and hence ω² = (2/3)H·exp(−2r/z0)/(4mr3)

Equating the two expressions for ω² gives

#### ω² = n²h²/(4m²r4) = (2/3)H·exp(−2r/z0)/(4mr3) which reduces to 2r·exp(−2r/z0) = 3n²h²/(mH) or, in terms of the separation distance z=2r z·exp(−z/z0) = 3n²h²/(mH) or, equivalently (z/z0)exp(−z/z0) = 3n²h²/(mz0H)

In terms of ζ=z/z0 this is

#### ζ·exp(−ζ) = σdn²

where σd is equal to (3h²/(mz0H))n².

Note that this is the essentially the same as for an alpha particle but

## Approximate Solutions

The function Γ(ζ) is approximately equal to γζ over its lower range. Therefore

#### ζ = ζ0 + (σ/γ)n²

where σ is ph²/(mz0H) and p is 3 for a deuteron and (3/4) for an alpha particle.

Thus for the same quantum number n, ζ and z for an alpha particle is one fourth of what they are for a deuteron.

Therefore

#### z = ζz0 = ph²/(γmH)

Since the rate of rotation ω is n²h²/(mz²), its value for an alpha particle will be 16 times that for a deuteron. That is also true for the kinetic energies.

It is harder to establish the ratio for potential energies. That ratio depends upon the value of ζ for the deuteron. The separation distance of the centers of the nucleons in a deuteron is approximately 2.25 fermi. The value of z0 established from the Yukawa relation is about 1.5 fermi. Thus the quantized value of ζ for a deuteron is about 1.5. This makes the value of ζ for an alpha particle 0.375. A numerical evaluation of the function W(ζ) gives the values W(1.5)=0.04923163 and W(0.375)=1.111677393. Thus the ratio of the potential energies of a neutron-proton interaction in an alpha particle to the one in a deuteron is 22.58. But there are two such interactions in an alpha particle so the ratio of the potential energies for neutron-proton interactions is 45.16. There must be a deduction for the interactions involved in the neutron-neutron pair and the proton-proton pair.

The binding energy of a nuclide is closely related to the loss of potential energy in its formation. Therefore it is not surprising that the binding energy for an alpha particle is so much greater than that of a deuteron.

(To be continued.)