|San José State University|
& Tornado Alley
Binding Energy of the Helium 4 Nuclide
(Alpha Particle) Among the Smallest Nuclides
One of the major enigmas of nuclear physics is the relatively high value of the binding energy of the Helium 4 nuclide, the alpha particle, compared to the smaller nuclides such as the Hydrogen 2 nuclide (deuteron) and the Hydrogen 3 nuclide (triteron).
According to the conventional estimates, the binding energy of the deuteron is about 2.225 million electron volts (MeV), that of the triteron 8.48 MeV and the alpha particle an enormous 28.29 MeV. To a degree the differences could be due to the number of nucleon-nucleon interactions (bonds) of the different nuclides. The deuteron has only one bond; the triteron has three and the alpha particle has six. The number of bonds, however, is at best only a partial explanation of the differences.
There is a problem with the conventional estimates of binding energy. Those estimates are derived from an estimate of the mass of a neutron which is based upon the assumption that the binding energy of a deuteron is equal to the energy of the gamma ray photon associated with the formation or disassociation of a deuteron. This leaves out the change in kinetic energy involved in the change in potential energy in a change in state. For electrons in atoms a change in state results in a change in potential energy that is exactly divided between the change in kinetic energy and the energy of the photon emitted or absorbed. At the nuclear level the potential energy based upon the nuclear strong force would not necessarily result in an equal division between kinetic energy and photon energy but there would be some allocation to kinetic energy. A previous study utilizing the Virial Theorem estimated that the mass of the neutron has been underestimated by 0.98638 MeV. The binding energy of a nuclide must be increased by this amount times the number of neutrons it contains. Thus the binding energy of a deuteron is 3.210953 MeV and that of an alpha particle is 30.268434 MeV.
The binding energy of a nuclide comes from the formation of spin pairs of nucleons as well as the potential energy derived from the strong force between nucleons. Another previous study estimates that the binding energy due to the formation of a neutron-proton spin pair is 1.98671 MeV. This means that the binding energy in a deuteron due to the strong force is 1.22424 MeV.
In an alpha particle there are one neutron-neutron spin pair, one proton-proton spin pair and two neutron-proton spin pairs. From estimates of the binding energies associated with the formation of the three different spin pairs the binding energy in an alpha particle due to spin pair formations is 10.8 MeV. That means that the binding energy in the alpha particle due to the strong force is 19.46834 MeV. This is 15.9 times the corresponding amount in the deuteron. The problem is to explain this ratio.
The approach used below treats the alpha particle like a deuteron made up of two clusters; i.e., a neutron pair and a proton pair. What comes out of this analysis is that the energy levels of such systems are highly nonlinearly dependent upon the size of the clusters;i.e., the number of particles in the clusters.
Consider first the simplest case, the one in which two clusters revolve about their center of mass, a deuteron of clusters. Here is a depiction of an alpha particle.
Here the separation distances of the nucleons in the pairs is shown as small compared to the distance between the pairs. This is only a visual convenience. Nucleons of the same type repel each other but the binding due to the pair formation holds them together at the limits at which the spin pairs can form. This is depicted below for a case in which the separation distance limit of pair formation is 1.5.
Nucleons of opposite type attract each other and a stable distance can only be maintained by their revolving around their center of mass. This is depicted below.
The equilibrium position of the system would be at the lowest point of the blue curve.
It is convenient to consider the case of clusters of arbitrary size, although the case of size 2 is the only relevant one. Let q be the number of nucleons is a cluster; in effect, the strong force charge of the cluster. Let m be the mass of a single nucleon. The mass of a cluster is then mq.
Let r be the radius of the orbits of the clusters and s the separation distance of their centers. Thus r=s/2.
Let ω be the rate of rotation of the system. The angular momentum of the system is 2(mq)(ωr)r, which is equal to mqωs²/2. This angular momentum is quantized so
h, h-bar, is Planck's constant divided by 2π. Thus
The nuclear force between clusters of total nucleonic charge of q1 and q2 may be expressed as
where H and s0 are constants. In the following q1 and q2 are equal and are denoted as q.
The attractive nuclear force on a cluster must balance the centrifugal force, which is equal to (mq)ω²r=(mq)ω²s/2. Thus
Equating the two expressions derived for ω² gives
This is the quantization condition for the separation distance s.
It can be expressed more succinctly as
with z=s/s0. The shape of the function z*exp(-z) is shown below.
A quantization condition for ω may be obtained from
Since s is a function of n, the above equation quantizes ω.
The kinetic energy K of the system
is 2(½(mq)(ωr)²). This reduces to mqω²s²/4. Replacing ω²
This quantizes kinetic energy but the full dependence of K on n and q is obscure. A simple approximation helps reveal that dependence.
At least over some range the function z*exp(−z) can be approximated by γz, where γ is a constant. This follows from z*exp(−z) being zero at z=0. Thus
where σ was previously defined as 2
So, since K = n²
Thus an alpha particle as deuteron of clusters of size two would have kinetic energy equal to 25=32 times the kinetic energy of a deuteron for the same principal quantum number n.
The potential energy of the system, which is a function only of the separation distance of the centers of the clusters, is given by
By a change in the variable of integration from p to x=p/s0 the integral ∫s+∞(exp(−p/s0)/p²)dp can be put into the form such that potential energy now expressed as a function of z is given by
For convenience let W(z)=∫z+∞(exp(−x)/x²)dx and hence
Over some range the function W(z) can be approximated by α/zζ, where ζ≥1. The parameter ζ can be approximated by considering the relationship between ln(W(z)) and ln(z), as shown below
The data for this relationship runs from z=0.5 to z=2. The slope of this relationship, which is the value of ζ, varies with the value of z.
Given the inverse dependence of z on q³ this means that this integral W(z) proportional to q3ζ. From the above expression for V(z) this means that the potential energy and the hence the binding energy of a deuteron of clusters of size q has a potential energy which is proportional to q3ζ−2.
The value of ζ depends upon the value of z=s/s0. For s<s0 the value of ζ is about 2. In that case V(z) would be proportional to q3·2−2=q4. Thus the potential energy and hence the binding energy of a deuteron of clusters of size 2 would be about 16 times that of a deuteron. With the strong force binding energy of the deuteron being 1.22424 MeV this would make the binding energy of a deuteron of clusters of size 2 equal to 19.58784 MeV. With the binding energy in the alpha particle due to nucleonic spin pairs being 10.8 MeV this would mean the total binding energy of the alpha particle would be 30.38784 MeV. If the correction due to the error in the neutron mass is subtracted to convert the binding enery to the conventional form the result is 28.41516 MeV. This differs from the conventional estimate by only 0.4 of 1 percent.
The values of the binding energies are summarized in the following table.
|Estimates of the Binding Energies
and Their Components of the Smallest Nuclides
(Millions of Electron Volts MeV)
Thus it is not surprising that an alpha particle has a conventional binding energy of 28.29 MeV compared to a binding energy of 2.225 MeV for a deuteron. It is just a matter of scale; i.e., cluster size, given the nature of the dependence of the nuclear force on distance as exp(−s/s0)/s².
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