Chapter 25, Problem 13PS

(a)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

${}_{47}^{111}Ag\to {}_{48}^{111}Cd+?$

${}_{47}^{111}Ag$ Undergoes nuclear reaction, producing ${}_{48}^{111}Cd$; there is no changes in products and reactant’s mass number while the atomic number is increased by one. Therefore according to the above mentioned table, probable decay in this reaction is $\text{Beta}\text{decay}\left({}_{\text{-1}}^{\text{0}}\text{β}\right)$

Thus,

The complete reaction is,

${}_{47}^{111}Ag\to {}_{48}^{111}Cd+{}_{-1}^0\beta$

(b)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

${}_{36}^{87}Kr\to {}_{-1}^0\beta +?$

${}_{36}^{87}Kr$ Undergoes nuclear reaction, eliminating a ${}_{-1}^0\beta$; there will be no changes in mass number but the atomic number will be increased by one after beta decay. Therefore the probable product is ${}_{37}^{87}Rb$

Thus,

The complete reaction is,

${}_{36}^{87}Kr\to {}_{-1}^0\beta +{}_{37}^{87}Rb$

(c)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

${}_{91}^{231}Pa\to {}_{89}^{227}Ac+?$

Comparing both side of the equation, the atomic number of product is reduced by two than in reactant which means the reactant undergoes an alpha decay.

Thus,

The complete reaction is,

${}_{91}^{231}Pa\to {}_{89}^{227}Ac+{}_2^4\alpha$

(d)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

${}_{90}^{230}Th\to {}_2^{4}He+?$

An alpha decay occurred in this reaction. Therefore the product should have an atomic number of $88$ and a mass number of $226$

Thus,

The complete reaction is,

${}_{90}^{230}Th\to {}_2^{4}He+{}_{88}^{226}Ra$

(e)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

${}_{35}^{82}Br\to {}_{36}^{82}Kr+?$

Comparing to the atomic number of reactant, atomic number of product is increased by one and there is no change in the mass number. Therefore probable decay in this reaction is $\text{Beta}\text{decay}\left({}_{\text{-1}}^{\text{0}}\text{β}\right)$

Thus,

The complete reaction is,

${}_{35}^{82}Br\to {}_{36}^{82}Kr+{}_{-1}^0\beta$

(f)

Interpretation Introduction

Interpretation:

Given nuclear equation has to be completed and then the mass number, atomic number, and symbol for the remaining particle should be determined.

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 incomplete nuclear reaction is,

$?\to {}_{12}^{24}Mg+{}_{-1}^0\beta$

In this reaction beta decay was occurred. The reactant nuclei should have an atomic number less than to the atomic number of ${}_{12}^{24}Mg$ .

Thus,

The complete reaction is,

${}_{11}^{24}Na\to {}_{12}^{24}Mg+{}_{-1}^0\beta$