# Question #a4462

Feb 27, 2017

$\text{11.1 days}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{n}}}}$

Here

• ${A}_{t}$ is the amount of the radioactive substance that remains undecayed after a time period $t$
• ${A}_{0}$ is the initial amount of said substance
• $n$ is the number of half-lives that pass in the time period $t$

Your first goal here is to figure out the value of $n$. Start by rewriting the equation as

${A}_{t} / {A}_{0} = {\left(\frac{1}{2}\right)}^{n}$

This will be equivalent to

$\ln \left({A}_{t} / {A}_{0}\right) = \ln \left[{\left(\frac{1}{2}\right)}^{n}\right]$

$\ln \left({A}_{t} / {A}_{0}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

Since

$\ln \left(\frac{1}{2}\right) = \ln \left(1\right) - \ln \left(2\right) = - \ln \left(2\right)$

you can say that

$n = - \ln \frac{{A}_{t} / {A}_{0}}{\ln} \left(2\right)$

Plug in your values to find

$n = - \ln \frac{\left(0.100 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(0.750color(red)(cancel(color(black)("g}}}}\right)}{\ln} \left(2\right) = 2.907$

Now, you know that $2.907$ half-lives must pass in order for the sample to be reduced from $\text{0.750 g}$ to $\text{0.100 g}$ and that radon-222 has a half-life of $3.823$ days.

The number of half-lives that pass in a given period of time can be written as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = \text{total time that passes"/"half-life" = t/t_"1/2}}}}$

The total time $t$ needed will be

$n = \frac{t}{t} _ \text{1/2" implies t = n * t_"1/2}$

In your case, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{t = 2.907 \cdot \text{3.823 days" = "11.1 days}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the initial and final mass of the sample.