Chapter 18, Problem 70QRT

(a)

Interpretation Introduction

Interpretation:

Completed nuclear equation has to be given for the below equation,

    $\boxed{}\text{+}\text{neutron}\stackrel{}{\to }\text{2}\text{neutrons}+{}^{\text{137}}\text{Ti}+{}^{\text{97}}\text{Zr}$

Concept Introduction:

Nuclear reaction is the one where the nucleus of the atom is involved in the reaction.  This can be represented in form of a nuclear equation.  Missing particle in a nuclear equation can be identified by using the mass number and atomic number balance.

Answer to Problem 70QRT

The completed nuclear equation is,

  $\text{B}\text{i}+\text{n}\stackrel{}{\to }2\text{n}+\text{T}\text{c}+\text{Z}\text{r}$

Explanation of Solution

Given nuclear equation is,

    $\boxed{}\text{+}\text{neutron}\stackrel{}{\to }\text{2}\text{neutrons}+{}^{\text{137}}\text{Tc}+{}^{\text{97}}\text{Zr}$

Neutron can be represented as $\text{n}$.  Atomic number of technetium is 43 and that of zirconium is 40.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{X}+\text{n}\stackrel{}{\to }2\text{n}+\text{T}\text{c}+\text{Z}\text{r}$

Mass number of the particle formed:

    $\begin{array}{l}\text{A}+\text{1}\text{=}\left(2\times 1\right)+\text{137}+97\\ \text{A}\text{=}\text{236}\text{-}\text{1}\\ \text{A}\text{=}\text{235}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}Z+0\text{=}\text{0}\text{+}\text{43}\text{+}\text{40}\\ \text{Z}\text{=}83\end{array}$

Therefore, the atomic number of the particle is 83 and mass number is 235.  The particle with atomic number 83 is bismuth.  Therefore, the complete nuclear equation can be given as,

    $\text{B}\text{i}+\text{n}\stackrel{}{\to }2\text{n}+\text{T}\text{c}+\text{Z}\text{r}$

(b)

Interpretation Introduction

Interpretation:

Completed nuclear equation has to be given for the below equation,

    ${}^{\text{45}}\text{Ti}\stackrel{}{\to }\boxed{}+\text{positron}$

Concept Introduction:

Refer part (a).

Answer to Problem 70QRT

The completed nuclear equation is,

  $\text{T}\text{i}\stackrel{}{\to }\text{S}\text{c}+\text{e}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{45}}\text{Ti}\stackrel{}{\to }\boxed{}+\text{positron}$

Atomic number of titanium is 22.  Positron can be represented as $\text{e}$.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{T}\text{i}\stackrel{}{\to }\text{X}+\text{e}$

Mass number of the particle formed:

    $\begin{array}{l}45\text{=}\text{A}\text{+}\text{0}\\ \text{A}\text{=}\text{45}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}22\text{=}\text{Z}\text{+}\text{1}\\ \text{Z}\text{=}22-1\\ \text{=}\text{21}\end{array}$

Therefore, the atomic number of the particle formed is 21 and mass number is 45.  The particle with atomic number 21 is scandium.  Therefore, the complete nuclear equation can be given as,

    $\text{T}\text{i}\stackrel{}{\to }\text{S}\text{c}+\text{e}$

(c)

Interpretation Introduction

Interpretation:

Completed nuclear equation has to be given for the below equation,

    $\boxed{}\stackrel{}{\to }\text{beta}\text{particle}+{}^{\text{59}}\text{Co}$

Concept Introduction:

Refer part (a).

Answer to Problem 70QRT

The completed nuclear equation is,

  $\text{F}\text{e}\stackrel{}{\to }\text{e}+\text{C}\text{o}$

Explanation of Solution

Given nuclear equation is,

    $\boxed{}\stackrel{}{\to }\text{beta}\text{particle}+{}^{\text{59}}\text{Co}$

Atomic number of cobatle is 27 and beta particle is represented as $\text{e}$.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{X}\stackrel{}{\to }\text{e}+\text{C}\text{o}$

Mass number of the particle formed:

    $\begin{array}{l}\text{A}\text{=}\text{59}\text{+}\text{0}\\ \text{A}\text{=}\text{59}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}\text{Z}\text{=}\text{-1}\text{+}\text{27}\\ \text{Z}\text{=}26\end{array}$

Therefore, the atomic number of the particle formed is  26 and mass number is 59.  The particle with atomic number 26 is iron.  Therefore, the complete nuclear equation can be given as,

    $\text{F}\text{e}\stackrel{}{\to }\text{e}+\text{C}\text{o}$

(d)

Interpretation Introduction

Interpretation:

Completed nuclear equation has to be given for the below equation,

    ${}^{\text{24}}\text{Mg}+\text{neutron}\stackrel{}{\to }\boxed{}+\text{proton}$

Concept Introduction:

Refer part (a).

Answer to Problem 70QRT

The completed nuclear equation is,

  $\text{M}\text{g}\text{+}\text{n}\stackrel{}{\to }\text{N}\text{a}+\text{H}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{24}}\text{Mg}+\text{neutron}\stackrel{}{\to }\boxed{}+\text{proton}$

Atomic number of magnesium is 12.  Neutron is represented as $\text{n}$ and proton is represented as $\text{p}$. The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{M}\text{g}\text{+}\text{n}\stackrel{}{\to }\text{X}+\text{H}$

Mass number of the particle formed:

    $\begin{array}{l}\text{24}\text{+}\text{1}\text{=}\text{A}\text{+}\text{1}\\ \text{A}\text{=}\text{25}\text{-}\text{1}\\ \text{=}\text{24}\end{array}$

Atomic number of the particle formed:

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

Therefore, the atomic number of the particle formed is 11 and mass number is 24.  The particle with atomic number 11 is sodium.  Therefore, the complete nuclear equation can be given as,

    $\text{M}\text{g}\text{+}\text{n}\stackrel{}{\to }\text{N}\text{a}+\text{H}$

(e)

Interpretation Introduction

Interpretation:

Completed nuclear equation has to be given for the below equation,

    ${}^{\text{131}}\text{Cs}+\boxed{}\stackrel{}{\to }{}^{\text{131}}\text{Xe}$

Concept Introduction:

Refer part (a).

Answer to Problem 70QRT

The completed nuclear equation is,

  $\text{C}\text{s}\text{+}\text{e}\stackrel{}{\to }\text{X}\text{e}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{131}}\text{Cs}+\boxed{}\stackrel{}{\to }{}^{\text{131}}\text{Xe}$

Atomic number of xenon is 54 and that of cesium is 55.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{C}\text{s}\text{+}\text{X}\stackrel{}{\to }\text{X}\text{e}$

Mass number of the particle formed:

    $\begin{array}{l}\text{131}\text{+}\text{A}\text{=}\text{131}\\ \text{A}\text{=}\text{0}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}\text{55}\text{+}\text{Z}\text{=}\text{54}\\ \text{Z}\text{=}54-55\\ =-1\end{array}$

Therefore, the atomic number of the particle reacted is -1 and mass number is 0.  The particle with atomic number -1 is electron.  Therefore, the complete nuclear equation can be given as,

    $\text{C}\text{s}\text{+}\text{e}\stackrel{}{\to }\text{X}\text{e}$