Chapter 31, Problem 67GP

To determine

The time taken to release $90\mathrm{\%}$ of ${}_{38}^{90}\text{Sr}$ .

Answer to Problem 67GP

  $\mathrm{t}=96.66\text{ years}$

Explanation of Solution

Given:

Half time, ${\mathrm{t}}_{1/2}=29.1\text{ years}$

Initial amount, ${\mathrm{N}}_0$

Final amount, $0.10{\mathrm{N}}_0$

Formula used:

  $\mathrm{N}\left(\mathrm{t}\right)={\mathrm{N}}_0{\left(0.5\right)}^{\mathrm{t}/{\mathrm{t}}_{1/2}}$

Where, $\mathrm{N}\left(\mathrm{t}\right)$ = final amount remained

  ${\mathrm{N}}_0$ = initial amount

Calculation:

The time ( $\mathrm{t}$ ) is calculated using above formula,

  $\begin{array}{l}\mathrm{N}\left(\mathrm{t}\right)={\mathrm{N}}_0{\left(0.5\right)}^{\mathrm{t}/{\mathrm{t}}_{1/2}}\\ 0.10{\mathrm{N}}_0={\mathrm{N}}_0{\left(0.5\right)}^{\mathrm{t}/29.1}\\ $ 0.1={(0.5)}^{t/29.1}$\end{array}$

Tanking natural log both sides,

  $\begin{array}{l}\frac{\mathrm{t}}{29.1}=\frac{\ln (0.10)}{\ln (0.5)}\\ \frac{\mathrm{t}}{29.1}=3.321\\ \mathrm{t}=96.66\text{ years}\end{array}$

Conclusion:

The taken to release $90\mathrm{\%}$ of ${}_{38}^{90}\text{Sr}$ is $96.66\text{ years}$ .