# Question 36005

Jul 25, 2017

$\text{1060 years}$

#### Explanation:

Your tool of choice here will be the following equation

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

Here

• ${A}_{t}$ is the mass of $\text{^14"C}$ that remains undecayed after a period of time $t$
• ${A}_{0}$ is the initial mass of $\text{^14"C}$
• $\textcolor{red}{n}$ represents the number of half-lives that pass in a given time period $t$

Now, you know that your fossil contains 88.0% of the initial mass of $\text{^14"C}$, so you can say that you have

${A}_{t} = \frac{88.0}{100} \cdot {A}_{0}$

Plug this into the above equation to get

$\frac{88.0}{100} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

$\frac{88.0}{100} = {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

Rearrange to solve for $n$

$\ln \left(\frac{88.0}{100}\right) = \ln \left[{\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}\right]$

You will end up wtih

$\textcolor{red}{n} \cdot \ln \left(\frac{1}{2}\right) = \ln \left(\frac{88.0}{100}\right)$

which will get you

$\textcolor{red}{n} = \ln \frac{\frac{88.0}{100}}{\ln} \left(\frac{1}{2}\right) = 0.1844$

This tells you that $0.1844$ half-lives have passed from the moment the initial sample of $\text{^14"C}$ started to undergo radioactive decay to the moment you took your measurements.

If you take ${t}_{\text{1/2}}$ to be the half-live of $\text{^14"C}$, you can say that you have

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

which, in your case, is equal to

$t = 0.1844 \cdot {t}_{\text{1/2}}$

Since $\text{^14"C}$ has a half-life of about $5730$ years (see here), you can say that

t = 0.1844 * "5730 years" = color(darkgreen)(ul(color(black)("1060 years")))#

have passed, i.e. the fossil is about $1060$ years old.

The answer is rounded to three sig figs.