Question

Obtain a semi empirical binding energy per nucleon formula. Using this as a guide, explain why the Coulomb force, which is only about \(\frac{1}{100}\) as strong as the internucleon attraction for two protons "in contact" (cf. Table 11.2), would eventually have to become a dominant factor in large nuclei. Assume that \(Z . N\), and \(A\) increase in rough proportion to one another.

Short answer

The semi empirical binding energy per nucleon formula is \( B/A = a_V - a_S A^{-1/3} - a_C Z(Z-1)A^{-4/3} - a_A (A - 2Z)^2/A^2 + δ(A,Z)/A \). Though the Coulomb force is weaker than the internucleon attractions, it becomes dominant in large nuclei because the number of proton-proton pairs and hence the Coulomb term in the binding energy increases quadratically with nucleon number, while attractive internucleon interactions increase linearly.

Step-by-step solution

Understand the Semi empirical binding energy per nucleon formula

Firstly, we start with the empirical formula for the binding energy \( B \) of a nucleus, which is given by \( B = a_V A - a_S A^{2/3} - a_C Z(Z-1)A^{-1/3} - a_A (A - 2Z)^2/A + δ(A,Z) \), where \( a_V, a_S, a_C, a_A \) are constants and \( δ(A,Z) \) is the pairing term.

Divide the formula by A to get binding energy per nucleon

We then simplify this formula to get the binding energy per nucleon, which is \( B/A = a_V - a_S A^{-1/3} - a_C Z(Z-1)A^{-4/3} - a_A (A - 2Z)^2/A^2 + δ(A,Z)/A \).

Understand how the Coulomb term varies with A

Next, we focus on the third term of the formula - herafter referred to as the Coulomb term - which represents the energy contribution from the electromagnetic interaction between the protons. As the nucleon number \( A \) increases, the volume of the nucleus, proportional to \( A^{1/3} \), also increases. So the distance between the protons also increases, meaning the Coulomb repulsive force decreases. But the number of interactions, which is proportional to \( Z(Z-1) \), increases faster. So overall, the Coulomb term becomes larger in magnitude as \( A \) increases.

Explain why the Coulomb force becomes dominant in large nuclei

Eventually for large nuclei, despite being originally only about 1/100 as strong as the internucleon attractions, the Coulomb term can become dominant because the number of proton-proton pairs (hence Coulomb repulsion interactions) increases quadratically with number of protons, while attractive internucleon interactions only increase linearly.

Key concepts

Nuclear Physics

Nuclear physics is the field that deals with the constituents and structure of atomic nuclei as well as the interactions that hold these particles together. At the heart of a nucleus, there are two types of baryons—protons and neutrons—collectively known as nucleons.

Protons carry a positive electric charge, while neutrons are neutral, and together they work in balance to form stable nuclei through the strong nuclear force. This force is incredibly powerful but operates over a very short range. The number of protons in the nucleus determines the element and is denoted by the symbol 'Z,' the number of neutrons is given as 'N,' and the total number of nucleons is the mass number 'A'.

Understanding the binding energy, which is the energy required to disassemble a nucleus into its individual protons and neutrons, is crucial in nuclear physics. It provides insight into the stability of an atom and the energy that can be released during nuclear reactions, such as in nuclear fission or fusion.

Coulomb Force

The Coulomb force, also known as the electrostatic force, is a fundamental interaction between charged particles. In the context of the nucleus, it refers to the repulsive force between protons due to their positive charges.

According to Coulomb's law, the electrostatic force between two stationary, point-like charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In a nucleus, as more protons are added, the repulsion among them increases because the number of proton-proton pairs increases. However, the average distance between them also grows as the nucleus becomes larger.

Interestingly, for small nuclei, the Coulomb force is not the dominant factor in determining nuclear stability due to the strength of the strong nuclear force. But as nuclei become larger and contain more protons, the effects of the Coulomb repulsion become increasingly significant, leading to greater instability in heavy nuclei.

Semi-Empirical Mass Formula

The semi-empirical mass formula is a theoretical approach that allows physicists to estimate the binding energy of nuclei. The formula considers several key components of nuclear structure, each with corresponding terms that depend on the mass number 'A' and the atomic number 'Z'.

These major components include the volume term, which is proportional to the total number of nucleons and represents the energy from the strong nuclear force; the surface term, as nucleons at the surface have fewer neighbors and hence a lower binding energy; the Coulomb term, representing the electrostatic repulsion among protons; the asymmetry term, accounting for the energy differences between protons and neutrons; and the pairing term, which provides corrections based on whether the nucleons form pairs.

By dividing the total binding energy formula by the mass number, scientists can determine the binding energy per nucleon. This calculation helps explain why, as nuclei become larger, the Coulomb term grows in significance and impacts the overall stability of the nucleus—particularly for large nuclei where the volume effect of the strong force can't sufficiently counterbalance the increasing electrostatic repulsion between the growing number of protons.