# Question cf933

May 28, 2015

The artifact would b 17 190 years old.

You calculate the number of half-lives and multiply by the length of one half-life.

The number of half-lives is $n = \frac{t}{t} _ \left(\frac{1}{2}\right)$, so $t = n {t}_{\frac{1}{2}}$.

For each half-life, you divide the total amount of the isotope by 2, so

$\frac{\text{Amount remaining" = "original amount}}{2} ^ n$ or

$A = {A}_{0} / {2}^{n}$

You can rearrange this to

${A}_{0} / A = {2}^{n}$

If original amount was 100 %, and 12.5 % of the nuclide remains undecayed, we have

$\frac{100}{12.5} = {2}^{n}$

$8 = {2}^{n}$

$n = 3$

$t = n {t}_{\frac{1}{2}} = \text{3 × 5730 yr" = "17 190 yr}$

May 28, 2015

The artifact would be 17,190 years old.

The half-life of a radioactive isotope expresses the time needed for a sample of that isotope to reach half of its original mass.

So, if you're dealing with a sample of carbon-14, you know that it's going to take 5,730 years for that sample to reach half of the mass it started with.

Moreover, after another 5,730 years, the sample will once again reach half of the mass it started with. But this time, the mass it started with is half of the initial mass, so that means that you're going to be left with one quarter of the initial mass after 2 half-lives pass.

Notice that the problem doesn't provide an initial mass, which means that it's not important in the calculations. Think of it like this: regardless of how much carbon-14 you start with, you'll be left with

• 50% of the initial mass $\to$ after 1 half-life;
• 25% of the initial mass $\to$ after 2 half-lives;
• 12.5% of the initial mass $\to$ after 3 half-lives;
• 6.25% of the initial mass $\to$ after 4 half-lives.

And so on.

In your case, you know that you're left with 12.5% of the initial mass of carbon-14, which means that 3 half-lives must have passed. This means that

t = 3 * "5,730" = color(green)("17,190 years")#

have passed since the artifact was made.