﻿ Minimum Energy Nuclear Composition and Nuclear Stability
San José State University

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Thayer Watkins
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USA

Minimum Energy Nuclear
Composition and Nuclear Stability

The stable isotopes of the heavier element have a great surplus of neutrons compared to the number of protons. For example the stable isotope of uranium, U_238, has 92 protons and 146 neutrons, nearly a (3/2) to 1 ratio. This may be a matter of that combination involves a lower energy than other cominations of the same number of nucleons.

Let the nucleonic charge of a proton be taken to 1 and that of a neutron be denoted as q. Then the force due a neutron-neutron interaction is proportional to q². The forces between neutron-proton and proton-proton interactions are proportional to q· and 1·1, respectively. If q is a negative value the the force between a neutron and a proton is an attraction, whereas the forces for the other two interactions are necessarily repulsions. The energies involved for these interactions are in the same proportions: q², q and 1. Ignoring the electrostatic repulsion between protons for the moment, the energy for a nucleus of p protons and n neutrons is proportional to the sum of the products of the numbers of the various interactions and the proportionality factors; i.e.,

#### E = ½(p²−p) + pnq + ½(n²−n)q²

Consider the minimization of E subject to the constraint that (p+n)=a. This can be achieved using the Lagrangian multiplier method. This involves minimizing

#### ½(p²−p) − pnq + ½(n²−n)q² +λ(a − (p+n))

The first order equations for a constrained minimum are therefore:

#### (p−½) + nq − λ = 0 (n−½)q² + pq − λ = 0

These can reduced to

#### (n−½)q² + pq = (p−½) + nq and further to (n−½)q² − nq = (p−½) − pq and n(q²−q) −½)q² = p(1−q) −½ and still further to n(q²−q) = p(1−q) − ½(1 − q²)

Dividing through by (q−1) gives

#### nq = −p + ½(1+;q) and finally n = (−1/q)p + ½(1/q+1)

For q=−2/3 this evaluates to

#### n = (3/2)p − (1/4)

The minimum energy combination of p and n would tend to be the most stable compositional arrangement. For p=92 the formula gives n=229.75 rather than 238.

## The Effect of the Electrostatic Repulsion between Protons

But as noted the above ignores the electrostatic repulsion between protons. That can be represented by making the energy of the interaction between protons as (1+d) rather that 1, where d represents the relative maggnitude of the electrostatic force compared to that of the nucleonic force at the distances involved in the nuclear interactions. This makes the quantity to be minimized equal to

#### ½(p²−p)(1+d) − pnq + ½(n²−n)q² +λ(a − (p+n))

The first order equations for a constrained minimum are then:

#### (p−½)(1+d) + nq − λ = 0 (n−½)q² + pq − λ = 0

These can reduced to

#### (n−½)q² + pq = (p−½)(1+d) + nq and further to (n−½)q² − nq = (p−½)(1+d) − pq and n(q²−q) −½q² = p((1+d)−q) −½(1+d) and still further to n(q²−q) = p((1+d)−q) − ½((1+d) − q²) and finally to n(q²−q) = p(1−q +d) − ½(1− q² + d)

Dividing through by (1−q) gives

#### −nq = p(1+d/(1-q)) −½(1+q + d/(1-q)) and finally n = (−1/q)p(1+d/(1-q)) + ½[(1/q+1) + d/(q(1-q) ]

For q=−2/3 and d=1/10 this evaluates to

#### n = (3/2)p(1+3/50) −(1/2) [(1/2) + (9/100)]

For p=92 the formula gives (p+n)=238.11 rather than 238, a deviation of about 0.05 of 1%. This 0.1 is a plausible value for d, the ratio of the electrostatic repulsion between protons to the nucleonic repulsion.

A regression of n on p for the nuclides with p≥26 (iron) gives

#### n = 1.57054p − 10.83610

The coefficient of determination (R²) for this equation is 0.92345 and the t-ratio for the coefficient of p is 172.4. The ratio of that coefficient to 3/2 is 1.0470273. The previous analysis says that this ratio should be (1+3d/5). This means that d, the ratio of the electrostatic repulsion to the nucleonic force repulsion, should be 0.078.