Question

In the nuclear reaction \(\mathrm{X}(\eta, \alpha)_{3}^{7}\) Li the atom \(\mathrm{X}\) will be (A) \({ }_{2} \mathrm{He}^{4}\) (B) \(_{5} \mathrm{~B}^{11}\) (C) \(_{5} \mathrm{~B}^{10}\) (D) \({ }_{5} \mathrm{~B}^{9}\)

Short answer

The atom X in the nuclear reaction is \({}_{5}^{11}\)B.

Step-by-step solution

Identify the given information

We are given the nuclear reaction: X(η, α)\({}_{3}^{7}\)Li. We need to determine the element and isotope of atom X.

Interpret the provided reaction

In this nuclear reaction, an atom X interacts with a photon (η) and produces an alpha particle (α) and a lithium atom with atomic number 3 and mass number 7. We can rewrite the reaction as follows: X + η -> α + \({}_{3}^{7}\)Li

Identify the alpha particle

An alpha particle is the nucleus of a helium atom. Therefore, an alpha particle has an atomic number of 2 and a mass number of 4. We can represent α as \({}_{2}^{4}\)He.

Apply the conservation laws

We use the conservation of charge and mass to determine the element and isotope of atom X. According to the conservation laws: 1. The sum of atomic numbers on the left side of the reaction must equal the sum of atomic numbers on the right side. 2. The sum of mass numbers on the left side of the reaction must equal the sum of mass numbers on the right side.

Calculate the atomic and mass number of atom X

From the given reaction, we know that X + η -> α + \({}_{3}^{7}\)Li. Since η is a photon, it has no atomic number and no mass. Hence, we can write the equations for the conservation of charge and mass: Atomic number conservation: Z = 2 + 3 Mass number conservation: A = 4 + 7 By solving these equations, we get: Z = 5 A = 11

Identify the element and isotope of atom X

Using the calculated atomic number and mass number, we can identify the element and isotope of atom X. The atomic number of 5 corresponds to the element Boron (B), and the mass number of 11 corresponds to the isotope \({}_{5}^{11}\)B. So the given nuclear reaction can be written as: X(η, α)\({}_{3}^{7}\)Li -> \({}_{5}^{11}\)B(η, α)\({}_{3}^{7}\)Li Therefore, the correct answer is: (B) \({}_{5}^{11}\)B

Key concepts

Alpha Particle

An alpha particle is a tiny, dense bundle of matter that consists of two protons and two neutrons. This composition makes it essentially a nucleus of the helium atom. When we talk about an alpha particle in nuclear reactions, it is typically represented as \({}_{2}^{4}\)He.

• The '2' refers to the atomic number, which symbolizes the two protons.
• The '4' is the mass number, derived from adding the two protons and two neutrons together.

Engaging with nuclear reactions often involves considering these particles as they are a common product or reactant. Their presence or production can affect the balance of charge and mass in nuclear equations, making their understanding essential for interpreting such reactions.

Conservation of Charge

The principle of the conservation of charge is fundamental in understanding nuclear reactions. It dictates that the total electric charge before and after a reaction must remain the same. This law implies that during any nuclear transformation, the sum of the atomic numbers on the reactant side must equal the sum on the product side. For example:

• In our given exercise, the charge (atomic number) on the reactant side involving X and η must equal the total charge on the resultant α particle and \({}_{3}^{7}\)Li.
• This law helps us not just identify the unknown elements involved but also confirm the balance of nuclear equations.

Hence, meticulous application of this law ensures our computed results align with the physical reality of nuclear interactions.

Conservation of Mass

Similar to the conservation of charge, the conservation of mass is another key concept in nuclear reactions. This principle asserts that the total mass number (or nucleon count) in a nuclear reaction must be preserved.

• The mass number of the reactants and products should be equal.
• In the exercise, the sum of mass numbers on one side must match the sum on the opposing side, ensuring no nucleons are lost or created out of nothing.

This principle is crucial because it guides us in calculating unknowns in reactions by aligning the reactant's nucleon total with the products. Exploring these conservation laws offers a concrete understanding of the behavior of particles in nuclear reactions.

Helium Nucleus

The helium nucleus has a significant role in nuclear chemistry and physics. It is the basis of the alpha particle, making it a common theme in nuclear reactions. A helium nucleus, as contained in an alpha particle, has:

• An atomic number of 2, due to its two protons.
• A mass number of 4, corresponding to its total protons and neutrons.

This particle emerges in many nuclear processes, especially those involving nuclear decay or fusion. Understanding the characteristics of helium nucleus helps when deciphering nuclear reactions, as they act as a predictable element in many scenarios.

Boron Isotope

In this context, the boron isotopes are pertinent to identifying the unknowns in nuclear reactions. Isotopes are versions of elements that have the same atomic number but different mass numbers.

• Boron, symbolized as 'B', has an atomic number of 5.
• The isotope discussed in the exercise is \({}_{5}^{11}\)B, indicating an atomic number 5 and a mass number 11.

These isotopic forms are crucial for balancing nuclear equations. They shine a light on the molecular transformations occurring during these reactions. Enabling a deep understanding of isotopic variations aids in both predicting and verifying the products or reactants in nuclear processes.