Chapter 24, Problem 58A

Interpretation Introduction

Interpretation: The half life needs 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 58A

Half-life of radioisotope is 126.94 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=1250.0 g, ${\text{N}}_{\text{0}}=8510\text{g}$, t=350 min

Put these values in above equation is as follows:

$\begin{array}{c}\text{1250=8450}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{350}{\text{T}}}\\ \frac{\text{1250}}{\text{8450}}={\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{350}{\text{T}}}\end{array}$

Apply logarithm the equation is as follows:

$\begin{array}{c}\log \left(\frac{\text{1250}}{\text{8450}}\right)=\log {\left(\frac{\text{1}}{\text{2}}\right)}^{\frac{350\min }{\text{T}}}\\ \log \left(\frac{\text{1250}}{\text{8450}}\right)=\left(\frac{350\min }{\text{T}}\right)\log \left(0.5\right)\\ -0.8299=\left(\frac{350\min }{\text{T}}\right)\left(-0.3010\right)\\ \text{T=}\frac{350\min }{-0.8299}\times \left(-0.3010\right)\\ =126.94\min \end{array}$

Thus half-life of radioisotope is 126.94 min.

Conclusion

An unknown radioisotope exhibits 8540 decays per second. The half-life of radioisotope is 126.94 min.