Chapter 18, Problem 19QAP

Balance the following equations by filling in the blanks.

(a) ${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{54}^{137}\underset{\_}{}+2_0^1\text{n}+\underset{\_}{}$
(b) ${}_{90}^{232}\text{Th}{+}_6^{12}\underset{\_}{}\to {\underset{\_}{}}_0^{1}n{+}_{96}^{240}\text{Cm}$
(c) ${}_2^4\text{He}{+}_{42}^{96}\text{Mo}{\to }_{43}^{100}\underset{\_}{}+\underset{\_}{}$
(d) $\underset{\_}{}{+}_1^2\text{H}{\to }_{84}^{210}\underset{\_}{}{+}_0^{1}n$

Interpretation Introduction

(a)

$\begin{array}{l}12+0=1+Z\\ Z=12-1=11\end{array}$

Interpretation:

The following reaction should be completed.

$\text{U}\text{+}\text{n}{\to }_{54}^{137}\text{\_\_\_+2}\text{n}\text{+\_\_\_\_}$

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 19QAP

$\text{U}\text{+}\text{n}{\to }_{54}^{137}\text{Xe+2}\text{n}{\text{+}}_{\text{38}}^{\text{97}}\text{Sr}$

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.

The given incomplete reaction is:

$\text{U}\text{+}\text{n}{\to }_{54}^{137}\text{\_\_\_+2}\text{n}\text{+\_\_\_\_}$

According to the periodic table, an element having 54 atomic number is $\text{Xe}$.

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

The reaction is written as:

$\text{U}\text{+}\text{n}{\to }_{54}^{137}\text{Xe+2}\text{n}{\text{+}}_Z^A\text{X}$

Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}235+1=137+A+2\\ A=236-139\\ A=97\end{array}$

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

$\begin{array}{l}92+0=54+Z\\ Z=92-54=38\end{array}$

Thus, the atomic number of strontium is 38 on the basis of periodic table.

Overall reaction is:

$\text{U}\text{+}\text{n}{\to }_{54}^{137}\text{Xe+2}\text{n}{\text{+}}_{\text{38}}^{\text{97}}\text{Sr}$

Interpretation Introduction

(b)

Interpretation:

The following reaction should be completed.

$\text{T}{\text{h+}}_6^{12}\_\_\_\to \text{C}\text{m+\_\_\_}\text{n}$

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 19QAP

$\text{T}{\text{h+}}_6^{12}\text{C}\to \text{C}\text{m+2}\text{n}$

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.

The given incomplete reaction is:

$\text{T}{\text{h+}}_6^{12}\_\_\_\to \text{C}\text{m+\_\_\_}\text{n}$

According to periodic table, an element with atomic number 6 is carbon.

The reaction is written as:

$\text{T}{\text{h+}}_6^{12}\text{C}\to \text{C}\text{m+\_\_\_}\text{n}$

Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}232+12=242+\mathrm{number}ofneutrons\\ \mathrm{number}ofneutrons=2\end{array}$

Overall reaction is:

$\text{T}{\text{h+}}_6^{12}\text{C}\to \text{C}\text{m+2}\text{n}$

Interpretation Introduction

(c)

Interpretation:

The following reaction should be completed.

$\text{H}{\text{e+}}_{42}^{96}\text{Mo}{\to }_{43}^{100}\text{\_\_\_+\_\_\_}$

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 19QAP

$\text{H}{\text{e+}}_{42}^{96}\text{Mo}{\to }_{43}^{100}\text{Tc+}e$

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.

The given incomplete reaction is:

$\text{H}{\text{e+}}_{42}^{96}\text{Mo}{\to }_{43}^{100}\text{\_\_\_+\_\_\_}$

According to the periodic table, an element with atomic number 43 is technetium.

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

The reaction is written as:

$\text{H}{\text{e+}}_{42}^{96}\text{Mo}{\to }_{43}^{100}{\text{Tc+}}_Z^A\text{X}$

Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}4+96=100+A\\ A=100-100=0\end{array}$

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

$\begin{array}{l}2+42=Z+43\\ Z=44-43=1\end{array}$

Thus, the product is $e$

Overall reaction is:

$\text{H}{\text{e+}}_{42}^{96}\text{Mo}{\to }_{43}^{100}\text{Tc+}e$

Interpretation Introduction

(d)

Interpretation:

The following reaction should be completed.

$\_\_\_\_{+}_1^2\text{H}{\to }_{84}^{210}\text{\_\_\_+}\text{n}$

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 19QAP

${}_{83}^{209}\text{Bi}{+}_1^2\text{H}{\to }_{84}^{210}\text{Po+}\text{n}$

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.

The given incomplete reaction is:

$\_\_\_\_{+}_1^2\text{H}{\to }_{84}^{210}\text{\_\_\_+}\text{n}$

According to the periodic table, an element with atomic number 84 is polonium.

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

The reaction is written as:

${}_{\text{Z}}^{\text{A}}\text{X}{+}_1^2\text{H}{\to }_{84}^{210}\text{Po+}\text{n}$

Mass number of the reactant must be equal to the mass number of the product.

$\begin{array}{l}A+2=210+1\\ A=211-2=209\end{array}$

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

$\begin{array}{l}1+Z=84\\ Z=84-1=83\end{array}$

Thus, the atomic number of bismuth is 83 on the basis of periodic table.

Overall reaction is:

${}_{83}^{209}\text{Bi}{+}_1^2\text{H}{\to }_{84}^{210}\text{Po+}\text{n}$