Question

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July 16, 1945. What percentage of the strontium-90 (${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}$28.9 yr) originally produced by that explosion remains as of July 16, 2016?

Short answer

The percentage of strontium-90 originally produced is 20%.

Step-by-step solution

Analyzing the rate constant.

The formula for evaluating the rate constant is:

$\mathbf{k}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{{\mathbf{t}}_{{\displaystyle \frac{1}{2}}}}$

The known value is substituted.

$\begin{array}{c}\mathrm{k}=\frac{0.693}{28.9}\\ =0.024{\text{year}}^{-1}\end{array}$

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{years}\left(t\right)& =& 2012-1945\\ & =& 67\mathrm{years}\end{array}$

Determining the percentage of strontium-90 originally produced by first atomic explosion

Using the integrated rate law for first-order reaction:

$\mathbf{ln}\frac{{\left[\mathbf{N}\right]}_{\mathbf{t}}}{{\left[\mathbf{N}\right]}_{\mathbf{0}}}\mathbf{=}\mathbf{-}\mathbf{kt}$

The known values are substituted.

$\begin{array}{c}\ln \frac{{\left[N\right]}_t}{{\left[N\right]}_0}=-0.024yea{r}^{-1}\times 67year\\ ln\frac{{\left[N\right]}_t}{{\left[N\right]}_0}=-1.608\\ \frac{{\left[N\right]}_t}{{\left[N\right]}_0}=e^{-1.608}\\ =0.20\end{array}$

$\begin{array}{rcl}\mathrm{In}\mathrm{percentage},0.2& =& 0.2\times 100\\ & =& 20\%\end{array}$

Hence, the required percentage is 20%.