# You have found a fossilized leg bone of some unknown mammal. Based on the size of the bone, you determine that it should have contained about 100 g of carbon-14 when the animal was alive. The bone now contains 12.5 g of carbon-14. How old is the bone?

Jan 11, 2016

$\text{17,190 years}$

#### Explanation:

Nuclear half-life is simply a measure of how much time must pass in order for a sample of a radioactive substance to decrease to half of its initial value.

Simply put, in one nuclear half-life, half of the atoms in the initial sample undergo radioactive decay and the other half do not.

Since the problem doesn't provide the nuclear half-life of carbon-14, you will have to do a quick search.

You'll find it listed as

${t}_{\text{1/2" = "5730 years}}$

https://en.wikipedia.org/wiki/Carbon-14

So, what does that tell you?

An initial sample of carbon-14, ${A}_{0}$, will be halved with the passing of every half-life, which in your case is $5730$ years. You can thus say that you'll be left with

${A}_{0} \cdot \frac{1}{2} \to$ after the passing of one half-life

${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 \to$ after the passing of two half-lives

${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 \to$ after the passing of three half-lives

${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} / 16 \to$ after the passing of four half-lives

$\vdots$

and so on.

You can thus say that $A$, the mass of the radioactive substance that remains undecayed, will be equal to

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

$n$ - the number of half-lives that pass in a given period of time

So, you know that you start with $\text{100.0 g}$ of carbon-14, and end up with $\text{12.5 g}$ after the passing of an unknown amount of time.

This means that you can say

overbrace(12.5 color(red)(cancel(color(black)("g"))))^(color(orange)("remaining mass")) = overbrace(100.0 color(red)(cancel(color(black)("g"))))^(color(purple)("initial mass")) * 1/2^n

Rearrange to get

$\frac{12.5}{100.0} = \frac{1}{2} ^ n$

$\frac{1}{8} = \frac{1}{2} ^ n \implies {2}^{n} = 8$

Since $8 = {2}^{3}$, you will have

${2}^{n} = {2}^{3} \implies n = 3$

So, three half-lives must pass in order for your sample of carbon-14 to be reduced from $\text{100.0 g}$ to $\text{12.5 g}$. Since

$\textcolor{b l u e}{n = \text{period of time"/"half-life" = t/t_"1/2}}$

you can say that

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

t = 3 xx "5730 years" = color(green)("17,190 years")