Chapter 25, Problem 30PS

(a)

Interpretation Introduction

Interpretation: Probable mode of decay of manganese-54 and the equation for this radioactive decay has to be given

Concept introduction:

Radioactive decay: It is the process in which a radioactive nucleus loses its energy by the emission of radiations namely, beta particle, alpha particle

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

The radioactive isotope of manganese-54 is transmuted to iron-56 along with the emission of a proton.

The equation for this radioactive decay is given below:

  $Mn+\alpha \to H+Fe$

(b)

Interpretation Introduction

Interpretation: Probable mode of decay of silver-110 and the equation for this radioactive decay has to be given

Concept introduction:

Radioactive decay: It is the process in which a radioactive nucleus loses its energy by the emission of radiations namely, beta particle, alpha particle

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

The radioactive isotope of silver-110 is transmuted to cadmium-110 along with the emission of a beta particle.

The equation for this radioactive decay is given below:

  $Ag\to Cd+\beta$.

(c)

Interpretation Introduction

Interpretation: Probable mode of decay of americium-241 and the equation for this radioactive decay has to be given

Concept introduction:

Radioactive decay: It is the process in which a radioactive nucleus loses its energy by the emission of radiations namely, beta particle, alpha particle

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

The radioactive isotope of americium-241 is transmuted to neptunium-237 along with the emission of an alpha particle.

The equation for this radioactive decay is given below:

  $Am\to Np+\alpha$.

(d)

Interpretation Introduction

Interpretation: Probable mode of decay of mercury-197m and the equation for this radioactive decay has to be given

Concept introduction:

Radioactive decay: It is the process in which a radioactive nucleus loses its energy by the emission of radiations namely, beta particle, alpha particle

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

The radioactive isotope mercury-197m is first converted to mercury-197. Then the radioactive isotope mercury-197 is transmuted to gold-197.

The equation for this radioactive decay is given below:

  $Hg\to Au$.