Question

The binding energy of \({ }_{2}^{3}\) He is lower than that of \({ }_{1}^{3} \mathrm{H} .\) Provide a plausible explanation, considering the Coulomb interaction between two protons in \({ }_{2}^{3}\) He.

Short answer

Answer: The binding energy of ${}_{2}^{3} \mathrm{He}$ is lower than that of ${}_{1}^{3} \mathrm{H}$ due to the repulsive Coulomb interaction between the two protons in ${}_{2}^{3} \mathrm{He}$. This interaction makes the nucleus less stable and easier to disassemble compared to ${}_{1}^{3} \mathrm{H}$, which has no such interaction.

Step-by-step solution

Examine the given nuclei

Examine the two nuclei in question: \({ }_{2}^{3} \mathrm{He}\) and \({ }_{1}^{3} \mathrm{H}\). The nucleus of \({ }_{2}^{3} \mathrm{He}\) consists of 2 protons and 1 neutron, while the nucleus of \({ }_{1}^{3} \mathrm{H}\) consists of 1 proton and 2 neutrons.

Understand the binding energy

Understand what binding energy is: Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. A higher binding energy means the nucleus is more stable, while a lower binding energy means the nucleus is less stable.

Compare the Coulomb interaction in \({ }_{2}^{3} \mathrm{He}\) and \({ }_{1}^{3} \mathrm{H}\)

Compare the Coulomb interaction between the protons in the two nuclei: In \({ }_{2}^{3} \mathrm{He}\), there are two protons that experience a repulsive Coulomb interaction which is proportional to the inverse square of the distance between them (Coulomb's law). In \({ }_{1}^{3} \mathrm{H}\), however, there is only one proton and therefore no Coulomb repulsion between protons takes place.

Relate Coulomb interaction to the binding energy

Relate the Coulomb interaction to the binding energy of the nuclei: As mentioned earlier, the binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. In \({ }_{2}^{3} \mathrm{He}\), the repulsive Coulomb interaction between the two protons decreases the stability of the nucleus. This repulsion makes it easier (i.e., requires less energy) to disassemble the nucleus, thereby resulting in a lower binding energy than that of \({ }_{1}^{3} \mathrm{H}\).

Provide the explanation

Now that we have analyzed the role of the Coulomb interaction and its influence on the binding energies, we can provide the plausible explanation: The binding energy of \({ }_{2}^{3} \mathrm{He}\) is lower than that of \({ }_{1}^{3} \mathrm{H}\), due to the repulsive Coulomb interaction between the two protons in \({ }_{2}^{3} \mathrm{He}\) making the nucleus less stable and easier to disassemble, when compared to \({ }_{1}^{3} \mathrm{H}\), which has no such interaction.

Key concepts

Coulomb Interaction

The Coulomb interaction is a fundamental aspect of nuclear physics, describing the force between two charged particles. For instance, in a helium-3 nucleus ( ({ }_{2}^{3}He) ), there are two positively charged protons that repel each other. This repulsion is a direct consequence of the electrostatic force, also known as Coulomb force, which operates between particles with electrical charge.

According to Coulomb's law, the magnitude of this force is directly proportional to the product of the electrical charges and inversely proportional to the square of the distance between the particles. Therefore, in the case of ({ }_{2}^{3}He), the two protons are being pushed apart by this force, which affects the overall stability of the nucleus. To understand this influence, one must consider the delicate balance of forces within the nucleus, where the strong nuclear force acts to hold the particles together, counterbalanced by the Coulomb force pushing them apart.

Nuclear Stability

Nuclear stability is defined by how strongly the constituent particles within an atomic nucleus are bound to each other. The more tightly these particles are held together, the more energy is required to split the nucleus apart, indicating a higher binding energy and thus greater stability.

This stability is determined by a delicate interplay between the attractive strong nuclear force, which acts between all nucleons (protons and neutrons), and the repulsive electrostatic force due to the Coulomb interaction that affects only the protons. Neutrons play a critical role in adding stability to the nucleus by providing additional strong force without contributing to the Coulomb repulsion. Hence, a helium-3 nucleus, with its two protons, is less stable compared to a tritium nucleus ( ({ }_{1}^{3}H) ), which has only one proton and does not experience significant Coulombic repulsion. This relative lack of stability is reflected in the helium-3's lower binding energy.

Coulomb's Law

Coulomb's Law is a quantitative formulation that describes the electric force between two point charges. This principle is key to explaining the repulsive interaction in atomic nuclei containing more than one proton. The law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. It can be expressed mathematically as:
\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]
where \( F \) is the magnitude of the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant. In the context of nuclear physics, Coulomb's Law helps us understand why two protons within a nucleus like ({ }_{2}^{3}He) repel each other and how this repulsion affects the binding energy and stability of the nucleus. The repulsive force is a key factor in determining the energy required to disassemble the nucleus into its constituent protons and neutrons.