Chapter 30, Problem 63GP

To determine

To Calculate:The time (in half-lives) for a radioactive sample to drop to given percentage of its original activity.

Explanation of Solution

Given:

Activity, $A=0.01A_o$

Here, $A_o$ is the original activity.

Formula Used:

Activity in terms of half-life $\left(t_{1/2}\right)$ , original activity $\left(A_o\right)$ and time (t):

  $A=A_o{e}^{-\frac{0.693t}{t_{1/2}}}$

Calculations:

Substitute $A=0.01A_o$ :

  $\begin{array}{l}0.01A_o=A_o{e}^{-\frac{0.693t}{t_{1/2}}}\\ \Rightarrow 0.01=e^{-\frac{0.693t}{t_{1/2}}}\\ \Rightarrow \ln 0.01=-\frac{0.693t}{t_{1/2}}\\ \Rightarrow -4.61=-\frac{0.693t}{t_{1/2}}\\ \Rightarrow t=6.65t_{1/2}\end{array}$

Conclusion:

Thus, at $t=6.65t_{1/2}$ , the activity of the radioactive sample would be $1.00\%$ of the original sample.