Chapter 15, Problem 120QP

(a)

Interpretation Introduction

Interpretation:

The required time to diminish $100\text{mg}$ of sample to $50\text{mg}$ is to be determined.

Explanation of Solution

The half-life of technetium- $99$ is $6.0\text{ h}$ . Thus, after $6.0\text{ h}$ , the amount of technetium- $99$ gets half and as the initial amount of technetium- $99$ was $100\text{mg}$ , so it takes $6\text{h}$ to diminish this amount to $50\text{mg}$ .

(b)

Interpretation Introduction

Interpretation:

The required time to diminish $100\text{mg}$ of sample to $25\text{mg}$ is to be determined.

Explanation of Solution

As the half of $100\text{mg}$ is $50\text{mg}$ and the half of $50\text{mg}$ is $25\text{mg}$ . So, the total number of half-life cycle required to diminish the $100\text{mg}$ of technetium- $99$ to $25\text{mg}$ of technetium- $99$ is $2$ . The total time required to diminish $100\text{mg}$ of technetium- $99$ to $25\text{mg}$ of technetium- $99$ is calculated as shown below:

$t=2t_{1/2}$

Where $t$ is the required time and $t_{1/2}$ is the half-life.

The value for $t$ is substituted in the above equation to calculate $t$ .

$\begin{array}{c}t=2\times 6\text{h}\\ \text{=12}\text{h}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The amount of technetium- $99$ left after $18.0\text{h}$ is to be determined.

Explanation of Solution

The half-life of technetium- $99$ is $6.0\text{ h}$ . Thus, after $6.0\text{ h}$ , the amount of technetium- $99$ gets half and $18.0\text{ h}$ is equal to three half-lives for technetium- $99$ . The amount left after $18.0\text{ h}$ is calculated as shown below:

$\text{N}={\left(\frac{1}{2}\right)}^3\times {\text{N}}_{\text{o}}$

Where $\text{N}$ and ${\text{N}}_{\text{o}}$ is the amount after $18.0\text{ h}$ and the initial amount of technetium- $99$ , respectively.

The value for ${\text{N}}_{\text{o}}$ is substituted to calculate $\text{N}$ .

$\begin{array}{c}\text{N}={\left(\frac{1}{2}\right)}^3\times 100\text{mg}\\ \text{=12}\text{.5}\text{mg}\end{array}$

Hence, the amount left after $18.0\text{ h}$ is $\text{12}\text{.5 mg}$ .

(d)

Interpretation Introduction

Interpretation:

The required time to diminish $100\text{mg}$ of sample to $6.25\text{mg}$ is to be determined.

Explanation of Solution

As the half of $100\text{mg}$ is $50\text{mg}$ , the half of $50\text{mg}$ is $25\text{mg}$ , the half of $25\text{mg}$ is $12.5\text{mg}$ , and the half of $12.5\text{mg}$ is $6.25\text{mg}$ . So, the total number of half-life cycle required to diminish $100\text{mg}$ of technetium- $99$ to $6.25\text{mg}$ of technetium- $99$ is $4$ . The total time required to diminish $100\text{mg}$ of technetium- $99$ to $6.25\text{mg}$ of technetium- $99$ is calculated as shown below:

$t=4t_{1/2}$

Where $t$ is the required time and $t_{1/2}$ is the half-life.

The value for $t$ is substituted in the above equation to calculate $t$ .

$\begin{array}{c}t=4\times 6\text{h}\\ \text{=24}\text{h}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The required time to diminish $100\text{mg}$ of sample to $3.12\text{mg}$ is to be determined.

Explanation of Solution

As the total time required to diminish $100\text{mg}$ of technetium- $99$ to $6.25\text{mg}$ of technetium- $99$ is $24\text{h}$ and the half of $6.25\text{mg}$ is $3.12\text{mg}$ . So, after $5$ half-life cycle, $100\text{mg}$ of technetium- $99$ diminishes to $3.12\text{mg}$ of technetium- $99$ . The total time required to diminish $100\text{mg}$ of technetium- $99$ to $3.12\text{mg}$ of technetium- $99$ is calculated as shown below:

$t=5t_{1/2}$

Where $t$ is the required time and $t_{1/2}$ is the half-life.

The value for $t$ is substituted in the above equation to calculate $t$ .

$\begin{array}{c}t=5\times 6\text{h}\\ \text{=30}\text{h}\end{array}$