Chapter 15, Problem 46QP

(a)

Interpretation Introduction

Interpretation:

The type of nuclear decay, the unstable nuclide $\text{O}$ undergoes, is to be determined.

Explanation of Solution

If the $\frac{\text{N}}{\text{Z}}$ ratio is too high, nuclide undergoes beta decay to lower the ratio. If the $\frac{\text{N}}{\text{Z}}$ ratio is too low, nuclide undergoes positron emission or electron capture which increases the ratio.

The mass number $\left(A\right)$ of oxygen is $19$ .

The atomic number $\left(Z\right)$ of oxygen is $8$ .

The number of neutrons $\left(N\right)$ is calculated using the following expression.

$\begin{array}{c}N=A-Z\\ =19-8\\ =11\end{array}$

The $\frac{N}{Z}$ ratio using the values of $N$ and $Z$ calculated above is,

$\begin{array}{l}\frac{\text{N}}{\text{Z}}=\frac{11}{8}\\ \frac{\text{N}}{\text{Z}}=1.38\end{array}$

Since the $\frac{\text{N}}{\text{Z}}$ ratio is high, beta decay may occur.

$\text{O}\to \text{F}\text{ + }\text{β}{}^{-}$

Therefore, the unstable nuclide, $\text{O}$ will undergo beta decay.

(b)

Interpretation Introduction

Interpretation:

The type of nuclear decay, the unstable nuclide $\text{P}\text{a}$ undergoes, is to be determined.

Explanation of Solution

If the $\frac{\text{N}}{\text{Z}}$ ratio is too high, nuclide undergoes beta decay to lower the ratio. If the $\frac{\text{N}}{\text{Z}}$ ratio is too low, nuclide undergoes positron emission or electron capture, which increases the $\frac{\text{N}}{\text{Z}}$ ratio.

The mass number $\left(A\right)$ of $\text{Pa}$ is $230$ .

The atomic number $\left(Z\right)$ of $\text{Pa}$ is $91$ .

The number of neutrons $\left(N\right)$ is calculated using the following expression.

$\begin{array}{c}N=A-Z\\ =230-91\\ =139\end{array}$

The $\frac{N}{Z}$ ratio using the values of $N$ and $Z$ calculated above is,

$\begin{array}{l}\frac{\text{N}}{\text{Z}}=\frac{139}{91}\\ \frac{\text{N}}{\text{Z}}=1.53\end{array}$

Since the $\frac{\text{N}}{\text{Z}}$ ratio is too high, beta decay may occur.

$\text{P}\text{a }\to \text{U}\text{ + }\text{β}{}^{-}$

Therefore, the unstable nuclide, $\text{P}\text{a}$ will undergo beta decay.

(c)

Interpretation Introduction

Interpretation:

The type of nuclear decay, the unstable nuclide $\text{C}$ undergoes, is to be determined.

Explanation of Solution

If the $\frac{\text{N}}{\text{Z}}$ ratio is too high, nuclide undergoes beta decay to lower the ratio. If the $\frac{\text{N}}{\text{Z}}$ ratio is too low, nuclide undergoes positron emission or electron capture, which increases the ratio.

The mass number $\left(A\right)$ of $\text{C}$ is $10$ .

The atomic number $\left(Z\right)$ of $\text{C}$ is $6$ .

The number of neutrons $\left(N\right)$ is calculated using the following expression.

$\begin{array}{c}N=A-Z\\ =10-6\\ =4\end{array}$

The $\frac{N}{Z}$ ratio using values of $N$ and $Z$ as calculated above is

$\begin{array}{c}\frac{\text{N}}{\text{Z}}=\frac{4}{6}\\ =0.67\end{array}$

Since the $\frac{\text{N}}{\text{Z}}$ ratio is too low, positron emission may occur.

$\text{C}\to \text{B}\text{ + }\text{β}{}^{+}$

Therefore, the unstable nuclide $\text{C}$ , will undergo positron emission.

(d)

Interpretation Introduction

Interpretation:

The type of nuclear decay, the unstable nuclide $\text{N}$ undergoes, is to be determined.

Explanation of Solution

If the $\frac{\text{N}}{\text{Z}}$ ratio is too high, nuclide undergoes beta decay to lower the ratio. If the $\frac{\text{N}}{\text{Z}}$ ratio is too low, nuclide undergoes positron emission or electron capture, which increases the ratio.

The mass number $\left(A\right)$ of $\text{N}$ is $13$ .

The atomic number $\left(Z\right)$ of $\text{N}$ is $7$ .

The number of neutrons $\left(N\right)$ is calculated using the following expression.

$\begin{array}{c}N=A-Z\\ =13-7\\ =6\end{array}$

The $\frac{N}{Z}$ ratio using the values of $N$ and $Z$ calculated above is,

$\begin{array}{l}\frac{\text{N}}{\text{Z}}=\frac{6}{7}\\ \frac{\text{N}}{\text{Z}}=0.86\end{array}$

Since the $\frac{\text{N}}{\text{Z}}$ ratio is low, positron emission may occur.

$\text{N}\to \text{C}\text{ + }\text{β}{}^{+}$

Therefore, the unstable nuclide, $\text{N}$ will undergo positron emission.

(e)

Interpretation Introduction

Interpretation:

The type of nuclear decay, the unstable nuclide $\text{P}\text{u}$ undergoes, is to be determined.

Explanation of Solution

If the $\frac{\text{N}}{\text{Z}}$ ratio is too high, nuclide undergoes beta decay to lower the ratio. If the $\frac{\text{N}}{\text{Z}}$ ratio is too low, nuclide undergoes positron emission or electron capture which increases the ratio.

The mass number $\left(A\right)$ of $\text{Pu}$ is $244$ .

The atomic number $\left(Z\right)$ of $\text{Pu}$ is $94$ .

The number of neutrons $\left(N\right)$ is calculated using the following expression.

$\begin{array}{c}N=A-Z\\ =244-94\\ =150\end{array}$

The $\frac{N}{Z}$ ratio using values of $N$ and $Z$ calculated above is,

$\begin{array}{l}\frac{\text{N}}{\text{Z}}=\frac{150}{94}\\ \frac{\text{N}}{\text{Z}}=1.6\end{array}$

Since the $\frac{N}{Z}$ ratio is too high, beta decay may occur.

$\text{P}\text{u}\to \text{A}\text{m + }\text{β}{}^{-}$

Therefore, the unstable nuclide, $\text{P}\text{u}$ will undergo beta decay.