Chapter 18, Problem 69QRT

(a)

Interpretation Introduction

Interpretation:

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

    ${}^{\text{214}}\text{Bi}\stackrel{}{\to }\boxed{}+{}^{\text{214}}\text{Po}$

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 69QRT

The completed nuclear equation is,

  $\text{B}\text{i}\stackrel{}{\to }\text{e}+\text{P}\text{o}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{214}}\text{Bi}\stackrel{}{\to }\boxed{}+{}^{\text{214}}\text{Po}$

Atomic number of bismuth is 83 and atomic number of polonium is 84.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{B}\text{i}\stackrel{}{\to }\text{X}+\text{P}\text{o}$

Mass number of the particle formed:

    $\begin{array}{l}\text{214}\text{=}\text{A}\text{+}\text{214}\\ \text{A}\text{=}\text{214}\text{-}\text{214}\\ \text{=}\text{0}\end{array}$

Atomic number of the particle formed:

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

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

    $\text{B}\text{i}\stackrel{}{\to }\text{e}+\text{P}\text{o}$

(b)

Interpretation Introduction

Interpretation:

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

    $4\text{H}\stackrel{}{\to }\boxed{}+\text{2}\text{positrons}$

Concept Introduction:

Refer part (a).

Answer to Problem 69QRT

The completed nuclear equation is,

  $4\text{H}\stackrel{}{\to }\text{H}\text{e}+2\text{e}$

Explanation of Solution

Given nuclear equation is,

    $4\text{H}\stackrel{}{\to }\boxed{}+\text{2}\text{positrons}$

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,

  $4\text{H}\stackrel{}{\to }\text{X}+2\text{e}$

Mass number of the particle formed:

    $\begin{array}{l}\text{4}\times \text{1}\text{=}\text{A}\text{+}\text{0}\\ \text{A}\text{=}\text{4}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}\text{4}\times \text{1}\text{=}\text{Z}\text{+}\text{2}\\ \text{Z}\text{=}4-2\\ \text{=}\text{2}\end{array}$

Therefore, the atomic number of the particle formed is $2$ and mass number is $4$.  The particle with atomic number 2 is helium.  Therefore, the complete nuclear equation can be given as,

    $4\text{H}\stackrel{}{\to }\text{H}\text{e}+2\text{e}$

(c)

Interpretation Introduction

Interpretation:

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

    ${}^{\text{249}}\text{Es}\text{+}\text{neutron}\stackrel{}{\to }\text{2}\text{neutrons}\text{+}\boxed{}+{}^{\text{161}}\text{Gd}$

Concept Introduction:

Refer part (a).

Answer to Problem 69QRT

The completed nuclear equation is,

  $\text{E}\text{s}+\text{n}\stackrel{}{\to }2\text{n}+\text{B}\text{r}+\text{G}\text{d}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{249}}\text{Es}\text{+}\text{neutron}\stackrel{}{\to }\text{2}\text{neutrons}\text{+}\boxed{}+{}^{\text{161}}\text{Gd}$

Neutron can be represented as $\text{n}$.  Atomic number of einsteinium is 99 and that of gadolinium is 64.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{E}\text{s}+\text{n}\stackrel{}{\to }2\text{n}+\text{X}+\text{G}\text{d}$

Mass number of the particle formed:

    $\begin{array}{l}\text{249}+\text{1}\text{=}\left(2\times 1\right)+\text{A}+161\\ \text{250}\text{=}\text{A}\text{+}\text{163}\\ \text{A}\text{=}\text{250}\text{-}\text{163}\\ \text{=}\text{87}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}99+0\text{=}\text{Z}\text{+}\text{64}\\ \text{Z}\text{=}99-64\\ \text{=}\text{35}\end{array}$

Therefore, the atomic number of the particle formed is 35 and mass number is 87.  The particle with atomic number 35 is bromine.  Therefore, the complete nuclear equation can be given as,

    $\text{E}\text{s}+\text{n}\stackrel{}{\to }2\text{n}+\text{B}\text{r}+\text{G}\text{d}$

(d)

Interpretation Introduction

Interpretation:

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

    ${}^{\text{220}}\text{Rn}\stackrel{}{\to }\boxed{}+\text{alpha particle}$

Concept Introduction:

Refer part (a).

Answer to Problem 69QRT

The completed nuclear equation is,

  $\text{R}\text{n}\stackrel{}{\to }\text{P}\text{o}+\text{H}\text{e}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{220}}\text{Rn}\stackrel{}{\to }\boxed{}+\text{alpha particle}$

Atomic number of radon is 86.  Alpha particle is represented as $\text{H}\text{e}$.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{R}\text{n}\stackrel{}{\to }\text{X}+\text{H}\text{e}$

Mass number of the particle formed:

    $\begin{array}{l}\text{220}\text{=}\text{A}\text{+}\text{4}\\ \text{A}\text{=}\text{220}\text{-}\text{4}\\ \text{=}\text{216}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}\text{86}\text{=}\text{Z}\text{+}\text{2}\\ \text{Z}\text{=}86\text{-}\text{2}\\ \text{=}\text{84}\end{array}$

Therefore, the atomic number of the particle formed is 84 and mass number is 216.  The particle with atomic number 84 is polonium.  Therefore, the complete nuclear equation can be given as,

    $\text{R}\text{n}\stackrel{}{\to }\text{P}\text{o}+\text{H}\text{e}$

(e)

Interpretation Introduction

Interpretation:

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

    ${}^{\text{68}}\text{Ge}+\text{electron}\stackrel{}{\to }\boxed{}$

Concept Introduction:

Refer part (a).

Answer to Problem 69QRT

The completed nuclear equation is,

  $\text{G}\text{e}\text{+}\text{e}\stackrel{}{\to }\text{G}\text{a}$

Explanation of Solution

Given nuclear equation is,

    ${}^{\text{68}}\text{Ge}+\text{electron}\stackrel{}{\to }\boxed{}$

Atomic number of germanium is 32.  Electron is represented as $\text{e}$.  The unknown particle can be assigned as $\text{X}$.  Therefore, the given nuclear equation can be rewritten as,

  $\text{G}\text{e}\text{+}\text{e}\stackrel{}{\to }\text{X}$

Mass number of the particle formed:

    $\begin{array}{l}\text{68}\text{+}\text{0}\text{=}\text{A}\\ \text{A}\text{=}\text{68}\end{array}$

Atomic number of the particle formed:

    $\begin{array}{l}\text{32}\text{-}\text{1}\text{=}\text{Z}\\ \text{Z}\text{=}31\end{array}$

Therefore, the atomic number of the particle formed is 31 and mass number is 68.  The particle with atomic number 31 is gallium.  Therefore, the complete nuclear equation can be given as,

    $\text{G}\text{e}\text{+}\text{e}\stackrel{}{\to }\text{G}\text{a}$