Chapter 25, Problem 14PS

(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,

${}_{10}^{19}Ne\to {}_{+1}^0\beta +?$

This reaction is completed by the emission of a positron from ${}_{10}^{19}Ne$ nuclei. Positron emission does not make any change in the mass number of product but the atomic number of product will be decreased by 1.

Thus,

The complete reaction is,

${}_{10}^{19}Ne\to {}_{+1}^0\beta +{}_9^{19}F$

(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,

${}_{26}^{59}Fe\to {}_{-1}^0\beta +?$

${}_{26}^{59}Fe$ 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 ${}_{27}^{59}Co$

Thus,

The complete reaction is,

${}_{26}^{59}Fe\to {}_{-1}^0\beta +{}_{27}^{59}Co$

(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,

${}_{19}^{40}K\to {}_{-1}^0\beta +?$

${}_{19}^{40}K$ Undergoes nuclear reaction, eliminating a ${}_{-1}^0\beta$; there will be no changes in mass number but the atomic number of product will be increased by one after beta decay. Therefore the probable product after this reaction is ${}_{20}^{40}Ca$

Thus,

The complete reaction is,

${}_{19}^{40}K\to {}_{-1}^0\beta +{}_{20}^{40}Ca$

(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,

${}_{18}^{37}Ar+{}_{-1}^{0}e\left(electroncapture\right)\to ?$

After an electron capture process, product has an atomic number less than to the atomic number of reactant and there is no change in the atomic mass number. Therefore the probable product is ${}_{17}^{37}Cl$

Thus,

The complete reaction is,

${}_{18}^{37}Ar+{}_{-1}^{0}e\left(electroncapture\right)\to {}_{17}^{37}Cl$

(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,

${}_{26}^{55}Fe+{}_{-1}^{0}e\left(electroncapture\right)\to ?$

After an electron capture process, product has an atomic number less than to the atomic number of reactant and there is no change in the atomic mass number. Therefore the probable product is ${}_{25}^{55}Mn$

Thus,

The complete reaction is,

${}_{26}^{55}Fe+{}_{-1}^{0}e\left(electroncapture\right)\to {}_{25}^{55}Mn$

(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,

${}_{13}^{26}Al\to {}_{12}^{25}Mg+?$

After nuclear reaction, atomic number and mass number are decreased by one in product. So the probable decay is ${}_1^{1}H$ decay.

Thus,

The complete reaction is,

${}_{13}^{26}Al\to {}_{12}^{25}Mg+{}_1^{1}H$