Chapter DY, Problem 6PQ

Interpretation Introduction

Interpretation:

The activity of a radioisotope is 3000 counts per minute at one time and 2736 counts per minute 48 hours later.  The half-life of the radioisotope has to be calculated.

Concept Introduction:

The emission of elementary particles by atoms is known as radioactivity.  The relation between radioactivity and half-life is:

    $\begin{array}{l}\text{ln}\left(\frac{\text{A}}{{\text{A}}_{\text{o}}}\right)\text{=-kt}\\ \text{A-activity at time 't'}\text{.}\\ {\text{A}}_{\text{o}}\text{-initial activity}\\ \text{k-decay constant}\end{array}$

Half-life is given by the equation,

    $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{k-decay constant}\end{array}$