Chapter 18, Problem 47AE

(a)

Interpretation Introduction

Interpretation:

 The balanced alpha emission ${}_5^{11}\text{B}$ has to be written.

Concept Introduction:

Alpha-emission involves liberation of a ${}_2^4\text{He}$. The general equation for alpha emission process is written as follows:

  ${}_{\text{Z}}^{\text{A}}\text{Y}{\to }_{Z-2}^{\text{A}}\text{X}{+}_2^4\text{He}$

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

Explanation of Solution

The nuclear reaction alpha emission is as follows:

  ${}_5^{11}\text{B}{\to }_{Z-2}^{\text{A}}\text{X}{+}_2^4\text{He}$        (1)

The nuclei that should come for X in equation (1) must have two atomic number more than 5 and mass number that can add up 4 to yield 11 as net mass number, so the nucleus must be ${}_3^7\text{Li}$ as indicated below.

  ${}_5^{11}\text{B}{\to }_3^7\text{Li}{+}_2^4\text{He}$

(b)

Interpretation Introduction

Interpretation:

The balanced beta emission ${}_{38}^{88}\text{Sr}$ has to be written.

Concept Introduction:

Beta-emission involves liberation of a beta-particle that is represented as ${}_{-1}^0\text{e}$. The general equation for beta emission process is written as follows:

  ${}_{\text{Z}}^{\text{A}}\text{Y}{\to }_{\text{Z+1}}^{\text{A}}\text{X}{+}_{-1}^0\text{e}$

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

Explanation of Solution

The nuclear reaction for beta emission is as follows:

  ${}_{12}^{29}\text{Mg}{\to }_{\text{Z+1}}^{\text{A}}\text{X}{+}_{-1}^0\text{e}$        (2)

There are 29 mass number and 12 atomic number on left side of equation (2) and thus to get same net mass numbers and atomic number the nuclei X must possess 11 as atomic number and 29 as mass number. Hence the nuclei that should come is ${}_{39}^{88}\text{Y}$.

  ${}_{38}^{88}\text{Sr}{\to }_{39}^{88}\text{Y}{+}_{-1}^0\text{e}$

(c)

Interpretation Introduction

Interpretation:

 The balanced neutron absorption ${}_{47}^{107}\text{Ag}$ has to be written.

Concept Introduction:

Neutron absorption involves capture of neutron that is represented as ${}_0^1\text{n}$. The general equation neutron absorption process is written as follows:

  ${}_{\text{Z}}^{\text{A}}\text{Y}{+}_0^1\text{n}{\to }_{\text{Z}}^{\text{A+1}}\text{X}$

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

Explanation of Solution

The nuclear reaction for neutron absorption is as follows:

  ${}_{47}^{107}\text{Ag}{+}_0^1\text{n}{\to }_{\text{Z}}^{\text{A+1}}\text{X}$        (3)

The nuclei that should come for X in equation (3) must have same atomic number and one more than mass number 107, so the nucleus X must be ${}_{47}^{108}\text{Ag}$ as indicated below.

  ${}_{47}^{107}\text{Ag}{+}_0^1\text{n}{\to }_{47}^{108}\text{Ag}$

(d)

Interpretation Introduction

Interpretation:

The balanced proton emission ${}_{39}^{41}\text{K}$ has to be written.

Concept Introduction:

Proton-emission involves liberation of a ${}_1^1\text{p}$. The general equation for proton emission process is written as follows:

  ${}_{\text{Z}}^{\text{A}}\text{Y}{\to }_{Z-1}^{\text{A}-\text{1}}\text{X}{+}_1^1\text{p}$

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

Initial nuclide is denoted as Y.

Product nuclide is Y.

Explanation of Solution

The nuclear reaction of proton emission is as follows:

  ${}_{19}^{41}\text{K}{\to }_{Z-1}^{\text{A}-\text{1}}\text{X}{+}_1^1\text{p}$        (4)

The nuclei that should come for X in equation (4) must have one atomic number less than 39 and one mass number less than 41, so the nucleus must be ${}_{18}^{40}\text{Ar}$ as indicated below.

  ${}_{19}^{41}\text{K}{\to }_{18}^{40}\text{Ar}{+}_1^1\text{p}$

(e)

Interpretation Introduction

Interpretation:

 The balanced for electron absorption of ${}_{51}^{116}\text{Sb}$ has to be written.

Concept Introduction:

Electron absorption involves capture of electron that is represented as ${}_{-1}^0\text{e}$. The general equation electron absorption process is written as follows:

  ${}_{\text{Z}}^{\text{A}}\text{Y}{+}_{-1}^0\text{e}{\to }_{\text{Z}-\text{1}}^{\text{A}}\text{X}$

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

Explanation of Solution

The nuclear reaction electron neutron absorption is as follows:

  ${}_{51}^{116}\text{Sb}{+}_{-1}^0\text{e}{\to }_{\text{Z}-\text{1}}^{\text{A}}\text{X}$        (5)

The nuclei that should come for X in equation (5) must have same atomic number and on more than mass number 107, so the nucleus X must be ${}_{47}^{108}\text{Ag}$ as indicated below.

  ${}_{51}^{116}\text{Sb}{+}_{-1}^0\text{e}{\to }_{50}^{116}\text{Sn}$