Chapter 24, Problem 57A

Interpretation Introduction

Interpretation:

Starting with 20.0g of polonium, the time when only 1.0 g remains is to be calculated.

Concept introduction:

The time period that is needed for one-half of a radioisotope’s nuclei to decay into its products is half life.

Answer to Problem 57A

Amount of time that takes for the sample to have only one gram remaining is 12.966 min.

Explanation of Solution

After a given amount of time has elapsed the remaining amount of a radioactive sample is given by the formula is as follows:

${\text{N=N}}_{\text{0}}{\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{\text{t}}{\text{T}}}$

Here N is mass of sample, ${\text{N}}_{\text{0}}$ is original mass, t is the time elapsed and T is duration of samples half life.

Given N=1.0 g, ${\text{N}}_{\text{0}}=20\text{g}$, T=3 min

Put these values in above equation is as follows:

$\begin{array}{c}\text{1}\text{.0g=20}\text{.0g}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{\text{t}}{3\min }}\\ \frac{\text{1g}}{\text{20g}}={\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{\text{t}}{3\min }}\\ 0.05={\left(0.5\right)}^{\frac{\text{t}}{3\min }}\end{array}$

Apply logarithm the equation is as follows:

$\begin{array}{c}\log 0.05=\log {\left(0.5\right)}^{\frac{\text{t}}{3\min }}\\ \log 0.05=\frac{\text{t}}{3\min }\log \left(0.5\right)\\ -1.3010=\frac{\text{t}}{3\min }\left(-0.3010\right)\\ \text{t=}\frac{-1.3010}{-0.3010}\times 3\min \\ =4.3222\times 3\min \\ =12.966\min \end{array}$

Thus amount of time that takes for the sample to have only one gram remaining is 12.966 min.

Conclusion

For the sample the amount of time that takes to have only one gram remaining is 12.966 min or 13 min.