Chapter 24, Problem 24.102P

(a)

Interpretation Introduction

Interpretation:

Half-life of ${}^{99m}Tc$ has to be determined using the given information.

Concept Introduction:

Radioactive decay is a first order reaction.

The half–life period for first order reaction is as follows:

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\ln \text{ 2}}{\text{k}}\\ \text{Where,}\\ \text{k = Rate constant}\\ {\text{t}}_{\text{1/2}}\text{= half-life}\end{array}$

Integrated rate law for a first order reaction is given below

  $\begin{array}{c}\ln \left(\frac{N_t}{N_0}\right)=-kt\\ \text{where,}\\ \text{ln}\text{=}\text{natural}\text{logarithm}\\ {\text{N}}_{\text{0}}\text{= Amount}\text{of}\text{subsatnce}\text{at}\text{time}\text{t}\text{=}\text{0}\\ {\text{N}}_{\text{t}}\text{= Amount}\text{of}\text{subsatnce}\text{after}\text{time, t}\\ \text{k}\text{=}\text{Rate}\text{constant}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Percentage of isotope that got lost if $2.0\text{ h}$ is taken to prepare and administer the dosage has to be determined using the given information.

Concept Introduction:

Radioactive decay is a first order reaction.

The half–life period for first order reaction is as follows:

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\ln \text{ 2}}{\text{k}}\\ \text{Where,}\\ \text{k = Rate constant}\\ {\text{t}}_{\text{1/2}}\text{= half-life}\end{array}$

Integrated rate law for a first order reaction is given below

  $\begin{array}{c}\ln \left(\frac{N_t}{N_0}\right)=-kt\\ \text{where,}\\ \text{ln}\text{=}\text{natural}\text{logarithm}\\ {\text{N}}_{\text{0}}\text{= Amount}\text{of}\text{subsatnce}\text{at}\text{time}\text{t}\text{=}\text{0}\\ {\text{N}}_{\text{t}}\text{= Amount}\text{of}\text{subsatnce}\text{after}\text{time, t}\\ \text{k}\text{=}\text{Rate}\text{constant}\end{array}$