Chapter 25, Problem 89A

Interpretation Introduction

Interpretation : The question is to find the year when the decay rate of Cesium-137 will be 17 counts per minute.

Concept Introduction: The exponential decay equation involving half-life can be written as,

  $A=A_0{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$h$}\right.}$

Here, A is the final amount, $A_0$ is the initial amount, t is the time, and h is the half-life time.

Answer to Problem 89A

The year at which the decay rate of Cesium-137 will be 17 counts per minute is 2135.

Explanation of Solution

Given information: The half-life of the radioisotope Cesium-137 is 30 years. The decay rate of a sample in the year 1985 is 544 counts per minute.

Using the exponential decay equation and substituting the given values,

  $\begin{array}{c}A=A_0{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$h$}\right.}\\ 17=544{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$30$}\right.}\\ 0.03125={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$30$}\right.}\\ {\log }_{1/2}0.03125=\frac{t}{30}\\ 5=\frac{t}{30}\\ \Rightarrow t=150\end{array}$

  $\begin{array}{l}\text{Required year}=1985+150\\ =2135\end{array}$

Conclusion

Hence, the decay rate of Cesium-137 will be 17 counts per minute in the year 2135.