# A parchment fragment was discovered that had about 74% as much Carbon 14 radioactivity as does plant material on the earth today. How would you estimate the age of the parchment?

## The half life of carbon-14 is about 5730 years.

Dec 5, 2015

$\text{2490 years}$

#### Explanation:

It's important that you don't let the wording of the problem confuse you.

"74% as much carbon-14 as does plant material on the Earth today"

is equivalent to

"of the initial amount of carbon-14 present in the parchment, only 74% remains undecayed"

Now, a radioactive isotope's nuclear half-life is defined as the time needed to an initial sample of this isotope to be reduced to half of its initial value.

Simply put, any amount of carbon-14 you start with will be halved with the passing of each half-life. Mathematically, you can write this as

$\textcolor{b l u e}{A = {A}_{0} \cdot \frac{1}{2} ^ n} \text{ }$, where

$A$ - the amount remaining after a given period of time
${A}_{0}$ - the initial amount
$n$ - the number of half-lives that pass in that given period of time

In your case, you know that 74% of the initial mass of carbon-14 present in the parchment remains undecayed. This means that you can write

$A = \frac{74}{100} \cdot {A}_{0}$

The above equation becomes

$\frac{74}{100} \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{A}_{0}}}} \cdot \frac{1}{2} ^ n$

Solve this equation for $n$

$\ln \left(\frac{74}{100}\right) = \ln \left({\left(\frac{1}{2}\right)}^{n}\right)$

$\ln \left(\frac{74}{100}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

$n = \ln \frac{\frac{74}{100}}{\ln} \left(\frac{1}{2}\right) = 0.434403$

Since $n$ represents the number of half-lives that must pass in order for the sample to be reduced to that percentage of the initial mass

$\textcolor{b l u e}{n = \text{period of time"/"half-life}}$

you can say that

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

Therefore

t = 0.434403 * "5730 years" = color(green)("2490 years")

I'll leave the answer rounded to three sig figs.