Chapter 20, Problem 94RQ

Interpretation Introduction

Interpretation:

The number of years strontium-90 $\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=28.1\text{ y}\right)$ will take for its activity to decrease from $0.245{\text{ Ci g}}^{-1}$ to $1.00\times {10}^{-6}{\text{ Ci g}}^{-1}$ is to be calculated. Then, the age of the rock is to be determined.

Concept Information:

The half-life of a first-order reaction is

$\begin{array}{c}{t}_{1/2}=\frac{\ln 2}{\text{k}}\\ =\frac{0.693}{\text{k}}\end{array}$

Where, k is decay constant and $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is half-life.

The rate of reaction that depends upon the concentration of only one reactant is known as a first-order reaction and is given by:

$rate=-\frac{\Delta \left[\text{A}\right]}{\Delta t}=\text{k}\left[\text{A}\right]$

The above expression gives the following relation between concentration and time:

$\left[\text{A}\right]={\left[\text{A}\right]}_{\text{0}}{e}^{-\text{k}t}$

Where, ${\left[\text{A}\right]}_0$ and $\left[\text{A}\right]$ denote, respectively, the initial and final concentrations of reactant A.