Chapter 24, Problem 24.134P

(a)

Interpretation Introduction

Interpretation:

Time taken for ${}^{242}Cm$ to be safe has to be determined.

Concept introduction:

Half-life: The time required for the concentration of a reactant to decrease to one half of its initial value.

  $\begin{array}{l}{\text{[R]}}_{\text{t}}\text{=}\frac{{\text{[R]}}_{\text{0}}}{\text{2}}\text{,}\\ \\ \text{First}\text{order}\text{reaction:}\\ \\ {\text{ln[R]}}_{\text{t}}\text{=}\text{- kt}\text{+}{\text{ln[R]}}_{\text{0}}\\ \\ \text{ln}\frac{{\text{[R]}}_{\text{t}}}{{\text{[R]}}_{\text{0}}}\text{=}\text{- kt}\left(\text{or}\right)\text{ln}\text{2 = k}{\text{t}}_{\text{1/2}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Time taken for ${}^{\text{214}}\text{Po}$ to be safe has to be determined.

Concept introduction:

Half-life: The time required for the concentration of a reactant to decrease to one half of its initial value.

  $\begin{array}{l}{\text{[R]}}_{\text{t}}\text{=}\frac{{\text{[R]}}_{\text{0}}}{\text{2}}\text{,}\\ \\ \text{First}\text{order}\text{reaction:}\\ \\ {\text{ln[R]}}_{\text{t}}\text{=}\text{- kt}\text{+}{\text{ln[R]}}_{\text{0}}\\ \\ \text{ln}\frac{{\text{[R]}}_{\text{t}}}{{\text{[R]}}_{\text{0}}}\text{=}\text{- kt}\left(\text{or}\right)\text{ln}\text{2 = k}{\text{t}}_{\text{1/2}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Time taken for ${}^{\text{232}}\text{Th}$ to be safe has to be determined.

Concept introduction:

Half-life: The time required for the concentration of a reactant to decrease to one half of its initial value.

  $\begin{array}{l}{\text{[R]}}_{\text{t}}\text{=}\frac{{\text{[R]}}_{\text{0}}}{\text{2}}\text{,}\\ \\ \text{First}\text{order}\text{reaction:}\\ \\ {\text{ln[R]}}_{\text{t}}\text{=}\text{- kt}\text{+}{\text{ln[R]}}_{\text{0}}\\ \\ \text{ln}\frac{{\text{[R]}}_{\text{t}}}{{\text{[R]}}_{\text{0}}}\text{=}\text{- kt}\left(\text{or}\right)\text{ln}\text{2 = k}{\text{t}}_{\text{1/2}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\end{array}$