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