Chapter 24, Problem 86A

Interpretation Introduction

Interpretation:

The half life of $\text{radon}-222$ needs to be calculated.

Concept introduction:

Half-life: Itis thetime required for one-half of a radioisotope’s nuclei to decay into its products.

Answer to Problem 86A

Half-life of $\text{radon}-222$ is 3.8 days.

Explanation of Solution

Given information,

$\begin{array}{l}\text{Initial}{\text{amount(N}}_{\text{0}})=\text{150mg}\\ \text{Remaining}\text{amount(N)}=\text{18}\text{.7mg}\\ \text{total}\text{time(t)}=\text{11}\text{.4days}\end{array}$

Relation between initial amount, remaining amount is

$\begin{array}{c}\frac{\text{N}}{{\text{N}}_{\text{0}}}={\left(\frac{1}{2}\right)}^n\\ \text{n}=\frac{\ln \left(\frac{\text{N}}{{\text{N}}_{\text{0}}}\right)}{\ln \left(\frac{1}{2}\right)}\\ \text{n}=\frac{\ln \left(\frac{\text{18}\text{.7}}{150}\right)}{\ln \left(\frac{1}{2}\right)}\\ \text{n}=3.004\text{half}\text{lives}\end{array}$

Since,

$\text{n}=\frac{\text{t}}{{\text{t}}_{\frac{\text{1}}{\text{2}}}}$

where n is the number of half lives

t is the total time passed

$t_{\frac{1}{2}}$ is the half-life.

Therefore,

$\begin{array}{c}{\text{t}}_{\frac{\text{1}}{\text{2}}}\text{=}\frac{\text{t}}{\text{n}}\\ {\text{t}}_{\frac{\text{1}}{\text{2}}}=\frac{11.4}{3.004}\\ {\text{t}}_{\frac{\text{1}}{\text{2}}}=3.8\text{days}\end{array}$

Conclusion

To calculate time first find the number of half-lives and then use the equations to find time.