Chapter 15, Problem 98QP

(a)

Interpretation Introduction

Interpretation:

The type of radioactive decay for an unstable nuclide $\text{F}$ 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.

Calculate the $\frac{N}{Z}$ ratio for the given unstable nuclide.

$\frac{\text{N}}{\text{Z}}=\frac{17-9}{9}=\frac{8}{9}=0.89$

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

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

Calculate the $\frac{N}{Z}$ ratio for $\text{O}$ .

$\frac{\text{N}}{\text{Z}}=\frac{17-8}{8}=\frac{9}{8}=1.13$

Thus, positron emission raises the ratio.

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

(b)

Interpretation Introduction

Interpretation:

The type of radioactive decay for an unstable nuclide $\text{F}$ 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.

Calculate the $\frac{N}{Z}$ ratio for the given unstable nuclide.

$\frac{\text{N}}{\text{Z}}=\frac{21-9}{9}=\frac{12}{9}=1.33$

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

$\text{F}\to \text{N}\text{e + }\text{β}{}^{-}$

Calculate the $\frac{N}{Z}$ ratio for $\text{N}\text{e}$ .

$\frac{\text{N}}{\text{Z}}=\frac{21-10}{10}=\frac{11}{10}=1.1$

Thus, beta decay lowers the ratio.

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

(c)

Interpretation Introduction

Interpretation:

The type of nuclear decay for an unstable nuclide $\text{P}\text{o}$ 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.

Calculate the $\frac{N}{Z}$ ratio for the given unstable nuclide.

$\frac{\text{N}}{\text{Z}}=\frac{216-84}{84}=\frac{132}{84}=1.57$

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

$\text{P}\text{o }\to \text{A}\text{t + }\text{β}{}^{-}$

Calculate the $\frac{N}{Z}$ ratio for $\text{A}\text{t}$ .

$\frac{\text{N}}{\text{Z}}=\frac{216-85}{85}=\frac{131}{85}=1.54$

Thus, beta decay lowers the ratio. The $\frac{N}{Z}$ ratio of $\text{A}\text{t}$ is also high, so, it should undergo further beta decay.

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

(d)

Interpretation Introduction

Interpretation:

The type of nuclear decay of unstable nuclide $\text{N}\text{e}$ 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.

Calculate the $\frac{N}{Z}$ ratio for the given unstable nuclide.

$\frac{\text{N}}{\text{Z}}=\frac{19-10}{10}=\frac{9}{10}=0.9$

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

$\text{N}\text{e }\to \text{F}\text{ + }\text{β}{}^{+}$

Calculate the $\frac{N}{Z}$ ratio for $\text{F}$ .

$\frac{\text{N}}{\text{Z}}=\frac{19-9}{9}=\frac{10}{9}=1.11$

Thus, positron emission raises the ratio.

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

(e)

Interpretation Introduction

Interpretation:

The type of nuclear decay for an unstable nuclide $\text{C}$ 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.

Calculate the $\frac{N}{Z}$ ratio for the given unstable nuclide.

$\frac{\text{N}}{\text{Z}}=\frac{15-6}{6}=\frac{9}{6}=1.5$

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

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

Calculate the $\frac{N}{Z}$ ratio for $\text{N}$ .

$\frac{\text{N}}{\text{Z}}=\frac{15-7}{7}=\frac{8}{7}=1.14$

Thus, beta decay lowers the ratio.

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