Chapter 5, Problem 96RE

To determine

To find: The radioactive cobalt present after $20\text{years}$ and $40\text{years}$ .

Answer to Problem 96RE

The radioactive cobalt present after $20\text{years}$ and $40\text{years}$ is $7.2\text{gram}$ and $0.5\text{gram}$ respectively.

Explanation of Solution

Given information:

The half life of radioactive cobalt is $5.27\text{years}$ and the initial amount of the radioactive cobalt is $100\text{grams}$ .

The rate of decay of the radioactive cobalt is calculated as,

  $\begin{array}{l}A=A_0{e}^{kt}\\ \frac{1}{2}{A}_0=A_0{e}^{k\times 5.27}\\ k=-0.132\end{array}$

The radioactive cobalt present after $20\text{years}$ is calculated as,

  $\begin{array}{c}A=A_0{e}^{kt}\\ =\left(100\right)e^{\left(-0.132\right)\left(20\right)}\\ =7.2\text{gram}\end{array}$

The radioactive cobalt present after $40\text{years}$ is calculated as,

  $\begin{array}{c}A=A_0{e}^{kt}\\ =\left(100\right)e^{\left(-0.132\right)\left(40\right)}\\ =0.5\text{gram}\end{array}$

Thus, the radioactive cobalt present after $20\text{years}$ and $40\text{years}$ is $7.2\text{gram}$ and $0.5\text{gram}$ respectively.