Chapter 15, Problem 45QP

(a)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

The mass number $\left(A\right)$ of carbon is $14$ .

The atomic number $\left(Z\right)$ of carbon is $6$ .

Hence, the number of neutrons $\left(N\right)$ is calculated as shown below.

$\begin{array}{c}N=A-Z\\ =14-6\\ =8\end{array}$

The value of $\frac{N}{Z}$ ratio using values of $N$ and $Z$ .

$\begin{array}{l}\frac{\text{N}}{\text{Z}}=\frac{8}{6}\\ \frac{\text{N}}{\text{Z}}=1.33\end{array}$

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

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

(b)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

The mass number $\left(A\right)$ of thorium is $234$ .

The atomic number $\left(Z\right)$ of thorium is $90$ .

Hence, the number of neutrons $\left(N\right)$ is calculated as shown below.

$\begin{array}{c}N=A-Z\\ =234-90\\ =144\end{array}$

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

$\begin{array}{c}\frac{\text{N}}{\text{Z}}=\frac{144}{90}\\ =1.6\end{array}$

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

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

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

The mass number $\left(A\right)$ of uranium is $234$ .

The atomic number $\left(Z\right)$ of uranium is $92$ .

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

$\begin{array}{c}N=A-Z\\ =234-92\\ =142\end{array}$

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

$\begin{array}{c}\frac{\text{N}}{\text{Z}}=\frac{142}{92}\\ =1.54\end{array}$

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

$\text{U}\to \text{N}\text{p + }\text{β}{}^{-}$

However, alpha decay of $\text{U}$ may also occurs

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

(d)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

The mass number $\left(A\right)$ of oxygen is $15$ .

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

Hence, the number of neutrons $\left(N\right)$ is calculated as shown below.

$\begin{array}{c}N=A-Z\\ =15-8\\ =7\end{array}$

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

$\begin{array}{c}\frac{\text{N}}{\text{Z}}=\frac{7}{8}\\ =0.86\end{array}$

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

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

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

(e)

Interpretation Introduction

Interpretation:

The type of nuclear decay that 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 raises the ratio.

The mass number $\left(A\right)$ of nitrogen is $16$ .

The atomic number $\left(Z\right)$ of nitrogen is $7$ .

The number of neutrons $\left(N\right)$ is calculated as shown below.

$\begin{array}{c}N=A-Z\\ =16-7\\ =9\end{array}$

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

$\begin{array}{c}\frac{\text{N}}{\text{Z}}=\frac{9}{7}\\ =1.27\end{array}$

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

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

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