Chapter 25, Problem 47PS

 (a)

Interpretation Introduction

Interpretation:

Time required for Cobalt-60 to decrease to one eighth of its original activity has to be calculated.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of ${}^{12}\text{C}$ and ${}^{14}\text{C}$ that takes in. when the living organism dies, the intake of ${}^{14}\text{C}$ stops and it’s the ratio no longer exhibits equilibrium; undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{where,}\text{k}\text{is rate constant}\text{.}\\ \left(\text{or}\right)\\ {\text{ln[N]}}_{\text{t}}{\text{- ln[N]}}_0\text{= -kt}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The fraction of the activity of a Cobalt-60 source remains after one year has to be calculated.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of ${}^{12}\text{C}$ and ${}^{14}\text{C}$ that takes in. when the living organism dies, the intake of ${}^{14}\text{C}$ stops and it’s the ratio no longer exhibits equilibrium; undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{where,}\text{k}\text{is rate constant}\text{.}\\ \left(\text{or}\right)\\ {\text{ln[N]}}_{\text{t}}{\text{- ln[N]}}_0\text{= -kt}\end{array}$