Chapter 15, Problem 119QP

(a)

Interpretation Introduction

Interpretation:

In the conversion of $\text{U}$ to thorium- $230$ , the number of alpha emission and beta emission are to be determined.

Explanation of Solution

The nuclear reaction that represents the conversion of $\text{U}$ to thorium- $230$ is shown below.

$\text{U}\stackrel{}{\to }\text{T}\text{h}+\text{H}\text{e}$

Therefore, in the conversion of $\text{U}$ to thorium- $230$ , there will be $1$ alpha emission.

(b)

Interpretation Introduction

Interpretation:

The produced product, when $\text{U}$ undergoes decay by the emission of five alpha particles and two beta particles, is to be identified.

Explanation of Solution

In the nuclear reaction that represents the conversion of $\text{U}$ to polonium- $214$ shown below, the total mass number is reduced by $20$ and the total atomic number is reduced by $8$ .

$\begin{array}{l}{}_{92}^{234}\text{U}{\stackrel{\alpha }{\to }}_{90}^{230}\text{Th}{\stackrel{\alpha }{\to }}_{88}^{226}\text{Ra}{\stackrel{\alpha }{\to }}_{86}^{222}\text{Rn}{\stackrel{\alpha }{\to }}_{84}^{218}\text{Po}{\stackrel{\alpha }{\to }}_{82}^{214}\text{Pb}{\stackrel{{\beta }^{-}}{\to }}_{83}^{214}\text{Bi}\\ {}_{83}^{214}\text{Bi}{\stackrel{{\beta }^{-}}{\to }}_{84}^{214}\text{Po}\end{array}$

Therefore, in the conversion of $\text{U}$ to polonium- $214$ , there will be $5$ alpha emissions and $2$ beta emissions.

(c)

Interpretation Introduction

Interpretation:

In the conversion of $\text{U}$ to polonium- $218$ , the number of alpha emission and beta emission are to be determined.

Explanation of Solution

In the nuclear reaction that represents the conversion of $\text{U}$ to polonium- $218$ shown below, the total mass number is reduced by $16$ and the total atomic number is reduced by $8$ .

${}_{92}^{234}\text{U}{\stackrel{\alpha }{\to }}_{90}^{230}\text{Th}{\stackrel{\alpha }{\to }}_{88}^{226}\text{Ra}{\stackrel{\alpha }{\to }}_{86}^{222}\text{Rn}{\stackrel{\alpha }{\to }}_{84}^{218}\text{Po}$

Therefore, in the conversion of $\text{U}$ to polonium- $218$ , there will be $4$ alpha emissions.

(d)

Interpretation Introduction

Interpretation:

In the conversion of $\text{U}$ to lead- $214$ , the number of alpha emission and beta emission are to be determined.

Explanation of Solution

In the nuclear reaction that represents the conversion of $\text{U}$ to lead- $214$ shown below, the total mass number is reduced by $20$ and the total atomic number is reduced by $10$ .

${}_{92}^{234}\text{U}{\stackrel{\alpha }{\to }}_{90}^{230}\text{Th}{\stackrel{\alpha }{\to }}_{88}^{226}\text{Ra}{\stackrel{\alpha }{\to }}_{86}^{222}\text{Rn}{\stackrel{\alpha }{\to }}_{84}^{218}\text{Po}{\stackrel{\alpha }{\to }}_{82}^{214}\text{Pb}$

Therefore, in the conversion of $\text{U}$ to lead- $214$ , there will be $5$ alpha emissions.

(e)

Interpretation Introduction

Interpretation:

In the conversion of $\text{U}$ to radon- $222$ , the number of alpha emission and beta emission are to be determined.

Explanation of Solution

In the nuclear reaction that represents the conversion of $\text{U}$ to radon- $222$ shown below, the total mass number is reduced by $12$ and the total atomic number is reduced by $6$ .

${}_{92}^{234}\text{U}{\stackrel{\alpha }{\to }}_{90}^{230}\text{Th}{\stackrel{\alpha }{\to }}_{88}^{226}\text{Ra}{\stackrel{\alpha }{\to }}_{86}^{222}\text{Rn}$

Therefore, in the conversion of $\text{U}$ to radon- $222$ , there will be $3$ alpha emissions.