# Carbon-14 has a half-life of 5770 years. If a fossil is 23,080 years old and it has 3kg of Carbon-14, how much carbon-14 did it originally have?

Jan 6, 2016

$\text{48 kg}$

#### Explanation:

A radioactive isotope's nuclear half-life tells you how much time must pass in order for a sample of this isotope to the reduced to half of its initial size.

In essence, any sample you start with will be halved with every passing of a half-life. So if you start with a mass ${A}_{0}$ of your isotope, you can say that you'll be left with

${A}_{0} \cdot \frac{1}{2} = {A}_{0} / 2 \to$ after one half-life

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

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

$\vdots$

and so on. You can thus find a relationship between how many half-lives have passed in a given period of time and how much of your initial sample is left undecayed

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

$A$ - the amount left undecayed
$n$ - the number of half-lives that have passed

In your case, you know that the fossil is $\text{23,080}$ years old and that carbon-14, the isotope of interest, has half-life of $\text{5,770}$ years.

This means that you determine how many half-lives have passed in this time period

$n = \left(23080 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{years"))))/(5770color(red)(cancel(color(black)("years}}}}\right) = 4$

So, you know that $\text{3.0 kg}$ of carbon-14 are left undecayed after the passing of four half-lives, which means that the sample originally contained

$A = {A}_{0} \cdot \frac{1}{2} ^ n \implies {A}_{0} = A \cdot {2}^{n}$

A_0 = "3.0 kg" * 2^4 = color(green)("48 kg")