Chapter 24, Problem 24.108P

(a)

Interpretation Introduction

Interpretation:

Amount of ${}^{90}Sr$ present in the body after 10 years has to be determined for a woman where who drank contaminated milk 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

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

(b)

Interpretation Introduction

Interpretation:

Time required for $99.9\%$ of ${}^{90}Sr$ ingested by the women to decay 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

  $\begin{array}{c}t\text{ = }\frac{1}{k}\ln \frac{N_0}{N_t}\\ \text{where,}\\ \text{ln}\text{=}\text{natural}\text{logarithm}\\ {\text{N}}_{\text{0}}\text{= Amount}\text{of}\text{subsatnce}\text{at}\text{time}\text{t}\text{=}\text{0}\\ {\text{N}}_{\text{t}}\text{= Amount}\text{of}\text{subsatnce}\text{after}\text{time, t}\\ \text{k}\text{=}\text{rate}\text{constant}\end{array}$