# Question 116c1

Mar 5, 2017

$\text{5620 years}$

#### 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 $t$

In your case, you know that the relic contains 27.3% of the initial amount of the radioactive substance. You can thus say that

${A}_{t} = \frac{27.3}{100} \cdot {A}_{0}$

The equation becomes

$\frac{27.3}{100} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} \cdot {\left(\frac{1}{2}\right)}^{n}$

$\frac{27.3}{100} = {\left(\frac{1}{2}\right)}^{n}$

This can be rewritten as

$\ln \left(\frac{27.3}{100}\right) = \ln \left[{\left(\frac{1}{2}\right)}^{n}\right]$

$\ln \left(\frac{27.3}{100}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

which gets you

$n = \ln \frac{\frac{27.3}{100}}{\ln} \left(\frac{1}{2}\right) = 1.873$

Before moving on, take a second to check if this value makes sense.

As you know, the half-life of a radioactive nuclide tells you the time needed for half of an initial mass of said nuclide to decay.

In other words, the mass of the nuclide will be halved with every passing half-life. This implies that you have

• A_0 * 1/2 = A_0/2 = 50/100 * A_0 = 50% A_0 -> after one half-life
• A_0/2 * 1/2 = A_0/4 = 25/100 * A_0 = 25% A_0 -> after two half-lives

In your case, you are left with 27.3%, a little over 25%, which means that the time needed for the sample to decay is very close to $2$ half-lives, i.e. the number of half-lives that passed is $< 2$.

So the fact that $1.873$ half-lives passed confirms that you are on the right track.

Now, you can say that

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

This means that you will have

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

which will get you

t = 1.873 * "3000 years" = color(darkgreen)(ul(color(black)("5620 years")))#

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