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Chapter 18: Problem 2469

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

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

Step by step solution

01

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.
02

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
03

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.
04

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.
05

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
06

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

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Most popular questions from this chapter

As the electron in Bohr is orbit of hydrogen atom Passes from state \(\mathrm{n}=2\) to \(\mathrm{n}=1\), the \(\mathrm{K} . \mathrm{E}\). and Potential energy changes as (A) Two fold, also two fold (B) four fold, two fold (C) four fold, also four fold (D) two fold, four fold

It the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus is $3.6 \mathrm{fm}\( the radius of \){ }^{125}{ }_{52} \mathrm{Te}$ nucleus is nearly equal to (A) \(8 \mathrm{fm}\) (B) \(6 \mathrm{fm}\) (C) \(4 \mathrm{fm}\) (D) \(5 \mathrm{fm}\)

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what Percent of original radioactive substance is left after 5 half life time (A) \(3 \%\) (B) \(5 \%\) (C) \(6 \%\) (D) \(12 \%\)

In Rutherford experiment, the number of Particles scattered at \(90^{\circ}\) angle are 28 Per min. then the number of Particles at the angle \(120^{\circ}\) in Per min will be (A) 25 (B) \(12.0\) (C) 50 (D) 112
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