# Question d5cfe

Mar 5, 2017

$\text{13 billion years}$

#### Explanation:

Your go-to equation here looks like this

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

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 $t$

Now, you know that 5% of the initial amount of your element remains undecayed after a period fo time $t$, which you must determine.

In such cases, you can express ${A}_{t}$ as a function of ${A}_{0}$

${A}_{t} = \frac{5}{100} \cdot {A}_{0}$

This means that the amount left after a period of time $t$ is equal to 5% = 5/100 of the initial amount

Plug this into the above equation to get

$\frac{5}{100} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} \cdot {\left(\frac{1}{2}\right)}^{n}$

Now, you can rewrite this as

$\ln \left(\frac{5}{100}\right) = \ln \left[{\left(\frac{1}{2}\right)}^{n}\right]$

This is done by taking the natural log of both sides of the equation

This will be equivalent to

$\ln \left(\frac{5}{100}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

which gets you

$n = \ln \frac{\frac{5}{100}}{\ln} \left(\frac{1}{2}\right) = 4.32$

So, you know that in order for 5% of the initial amount of your element to remain undecayed, i.e. in order for 95%# to decay, a period of time equal to $4.32$ half-lives must pass.

Since you have

$n = \text{total time that passes"/"half life" = t/t_"1/2}$

you can say that

$t = n \cdot {t}_{\text{1/2}}$

In your case, this will get you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{t = 4.32 \cdot \text{3 billion years" = "13 billion years}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the half-life of the element.