Chapter 18, Problem 9QAP

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Am-241 through $\beta -$ emissions.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 9QAP

$\text{P}\text{u}\to \text{A}{\text{m +}}_{-1}^{0}e+\overline{\nu }$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Am-241 through $\beta -$ emissions.

Am-241 is the isotope of americium-241, atomic number of americium is 95.

A radioactive decay in which a neutron presents inside a nucleus is changed into a proton by emission of electron that is accompanied by antineutrino ( $\overline{\nu }$). The resulting nucleus has atomic number which is increased by one and mass number will remain same.

The reaction can be written as:

To identify the missing species: Let the species be $\text{X}$

${}_Z^A\text{X}\to \text{A}{\text{m +}}_{-1}^{0}e+\overline{\nu }$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}A=241+0\\ A=241\end{array}$

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}Z=95-1\\ Z=94\end{array}$

Thus, the reactant must have mass number and atomic number equal to 241 and 94. From the periodic table, the nucleus is $\text{P}\text{u}$.

Overall reaction is:

$\text{P}\text{u}\to \text{A}{\text{m +}}_{-1}^{0}e+\overline{\nu }$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Kr-81 by K-electron capture.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 9QAP

$\text{R}{\text{b+}}_{-1}^{0}e{\to }_{36}^{81}\text{Kr}$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Kr-81 by K-electron capture.

Kr-81 is the isotope of krypton, atomic number of krypton is 36.

A radioactive decay in which an unstable nucleus captured an electron from inner shell and during this process proton is converted into a neutron and result in the emission of electron neutrino and gamma radiation is known as K-electron capture. The resultant product has atomic number which is reduced by one and mass number will remain same.

The reaction can be written as:

To identify the missing species: Let the species be $\text{X}$

${}_Z^A{\text{X+}}_{-1}^{0}e{\to }_{36}^{81}\text{Kr}$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}A+0=81\\ A=81\end{array}$

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}Z-1=36\\ Z=36+1=37\end{array}$

Thus, the reactant must have mass number and atomic number equal to 81 and 37. From the periodic table, the nucleus is $\text{R}\text{b}$.

Overall reaction is:

$\text{R}{\text{b+}}_{-1}^{0}e{\to }_{36}^{81}\text{Kr}$

Interpretation Introduction

(c)

Interpretation:

The balanced equation for the following nuclear reaction should be written.

The production of Ra-223 by $\alpha$ -emission.

Concept introduction:

A reaction in which a nucleus of an atom or two nuclei and a subatomic particle collide to form one or more nuclide which is different from the starting nuclide is known as nuclear reaction. Nuclear reactions can be classified as nuclear fission and nuclear fusion.

Answer to Problem 9QAP

$\text{T}\text{h}\to \text{R}{\text{a+}}_2^4\alpha$

Explanation of Solution

The reaction in which the change in atomic numbers, energy states and mass number of nuclei takes place is known as nuclear reaction.

Given that, the production of Ra-223 by $\alpha$ -emission.

Ra-223 is the isotope of radium, atomic number of radium is 88.

Alpha emission is a radioactive decay in which the nucleus (unstable) emits an alpha particle

( ${}_2^4\alpha$) and thus, results in the formation of different nucleus having atomic number which is reduced by 2 and mass number which is reduced by 4.

The reaction can be written as:

To identify the missing species: Let the species be $\text{X}$

$\text{X}\to \text{R}{\text{a+}}_2^4\alpha$

To balance the reaction: Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}A=223+4\\ A=227\end{array}$, thus mass number on both sides are equal

Now, atomic number or charges must be balanced, therefore:

$\begin{array}{l}Z=88+2\\ Z=90\end{array}$, thus charges are equal on both sides

Thus, the reactant must have mass number and atomic number equal to 227 and 90. From the periodic table, the nucleus is $\text{T}\text{h}$.

Overall reaction is:

$\text{T}\text{h}\to \text{R}{\text{a+}}_2^4\alpha$