Chapter 21, Problem 21.77QP

(a)

Interpretation Introduction

Interpretation:

The type of decay and the isotope which undergoes that particular decay should be identified.

Concept Introduction:

Nuclear reaction is a physical process in which there is a change in identity of an atomic nucleus. Natural radioactive decays, artificial radioactive decays... are considered as nuclear reactions because these processes make changes in the identity of an atomic nucleus.

Common particles in radioactive decay and nuclear transformations are mentioned below,

    $\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

There are various types of nuclear processes. The changes in atomic number and mass number accompanying radioactive decay are mentioned below,

$\begin{array}{cccccc}\text{Type}\text{of}\text{Decay}& \text{Symbol}& \text{charge}& \text{Mass}& \begin{array}{l}\text{Change}\\ \text{in}\text{atomic}\text{number}\end{array}& \begin{array}{l}\text{Change}\\ \text{in}\text{mass}\text{number}\end{array}\\ \text{Beta}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}& \text{-1}& \text{0}& \text{+1}& \text{none}\\ \text{Positron}& {}_{\text{+1}}^{\text{0}}\text{e}& \text{+1}& \text{0}& \text{-1}& \text{none}\\ \text{Alpha}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}& \text{+2}& \text{4}& \text{-2}& \text{-4}\\ \text{Gamma}& {}_{\text{0}}^{\text{0}}\text{γ}& \text{0}& \text{0}& \text{none}& \text{none}\end{array}$

An isotope can be represented using ${}_Z^{A}X$ symbol

$\begin{array}{l}\text{Z}\to \text{Atomic}\text{number}=\text{Number}\text{of}\text{proton}\\ \text{A}\to \text{Mass}\text{number}=\text{Number}\text{of}\text{protons}\text{+}\text{Number}\text{of}\text{neutrons}\end{array}$

Explanation of Solution

Given diagram for the decay is,

Figure 1

From the diagram the isotope can be identified,

$As$ is the Atom with atomic number 33here it has $39$ neutrons.

$Ge$ is the Atom with atomic number $32$ here it has $40$ neutrons.

Atomic number is decreased from $33to32$  and there is no change in the mass number this means the atom undergoes $positron\left({}_{+1}^0\beta \right)$ emission.

The nuclear reaction can be represented as,

${}_{33}^{72}As\to {}_{32}^{72}Ge+{}_{+1}^0\beta$

(b)

Interpretation Introduction

Interpretation:

The type of decay and the isotope which undergoes that particular decay should be identified.

Concept Introduction:

Nuclear reaction is a physical process in which there is a change in identity of an atomic nucleus. Natural radioactive decays, artificial radioactive decays... are considered as nuclear reactions because these processes make changes in the identity of an atomic nucleus.

Common particles in radioactive decay and nuclear transformations are mentioned below,

$\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

There are various types of nuclear processes. The changes in atomic number and mass number accompanying radioactive decay are mentioned below,

$\begin{array}{cccccc}\text{Type}\text{of}\text{Decay}& \text{Symbol}& \text{charge}& \text{Mass}& \begin{array}{l}\text{Change}\\ \text{in}\text{atomic}\text{number}\end{array}& \begin{array}{l}\text{Change}\\ \text{in}\text{mass}\text{number}\end{array}\\ \text{Beta}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}& \text{-1}& \text{0}& \text{+1}& \text{none}\\ \text{Positron}& {}_{\text{+1}}^{\text{0}}\text{e}& \text{+1}& \text{0}& \text{-1}& \text{none}\\ \text{Alpha}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}& \text{+2}& \text{4}& \text{-2}& \text{-4}\\ \text{Gamma}& {}_{\text{0}}^{\text{0}}\text{γ}& \text{0}& \text{0}& \text{none}& \text{none}\end{array}$

Explanation of Solution

Given diagram for the decay is,

Figure 2

From the diagram the isotope can be identified,

$Ag$ is the Atom with atomic number 47 here it has 113 neutrons.

$Cd$ is the Atom with atomic number 48 here it has 113neutrons.

Atomic number is increased from $47to48$  and there is no change in the mass number this means the atom undergoes $beta\left({}_{-1}^0\beta \right)$ emission.

The nuclear reaction can be represented as,

${}_{47}^{113}Ag\to {}_{48}^{113}Cd+{}_{-1}^0\beta$

(c)

Interpretation Introduction

Interpretation:

The type of decay and the isotope which undergoes that particular decay should be identified.

Concept Introduction:

Nuclear reaction is a physical process in which there is a change in identity of an atomic nucleus. Natural radioactive decays, artificial radioactive decays... are considered as nuclear reactions because these processes make changes in the identity of an atomic nucleus.

Common particles in radioactive decay and nuclear transformations are mentioned below,

$\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

There are various types of nuclear processes. The changes in atomic number and mass number accompanying radioactive decay are mentioned below,

$\begin{array}{cccccc}\text{Type}\text{of}\text{Decay}& \text{Symbol}& \text{charge}& \text{Mass}& \begin{array}{l}\text{Change}\\ \text{in}\text{atomic}\text{number}\end{array}& \begin{array}{l}\text{Change}\\ \text{in}\text{mass}\text{number}\end{array}\\ \text{Beta}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}& \text{-1}& \text{0}& \text{+1}& \text{none}\\ \text{Positron}& {}_{\text{+1}}^{\text{0}}\text{e}& \text{+1}& \text{0}& \text{-1}& \text{none}\\ \text{Alpha}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}& \text{+2}& \text{4}& \text{-2}& \text{-4}\\ \text{Gamma}& {}_{\text{0}}^{\text{0}}\text{γ}& \text{0}& \text{0}& \text{none}& \text{none}\end{array}$

Explanation of Solution

Given diagram for the decay is,

Figure 3

From the diagram the isotope can be identified.

$Rn$ is the Atom with atomic number 86 here it has 136 neutrons.

$Pb$ is the Atom with atomic number 84 here it has $134$ neutrons.

Atomic number is decreased from $86to84$  and the mass number is reduced to by 4, this means the atom undergoes $alpha\left({}_2^4\alpha \right)$ emission.

The nuclear reaction can be represented as,

${}_{86}^{222}Rn\to {}_{84}^{218}Pb+{}_2^4\alpha$