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

Jun 21, 2016

$\text{48 kg}$

#### Explanation:

The nuclear half-life is simply the time needed for half of the atoms present in a sample of a radioactive substance to undergo radioactive decay.

In your case, carbon-14 is said to have a half-life of $5770$ years, which means that the mass of any sample of carbon-14 will be halved with the passing of $5770$ years.

Now, if $1$ half-life from now the mass of the sample will be halved, it follows that $1$ half-life ago it was twice its current value.

An interesting technique to use here is to backtrack from current time, when the sample is down to $\text{3 kg}$, to $\text{23,080}$ years ago.

You can thus say that $5770$ years before this point, the mass of the sample was twice the current mass

$2 \times \text{3 kg" = "6 kg}$

How about $2$ half-lives ago?

$2 \times \text{5770 years" = "11,540 years ago}$

The mass of the sample was

$2 \times \text{6 kg" = "12 kg}$

How about $3$ half-lives ago?

$3 \times \text{5770 years" = "17,310 years ago}$

The mass of the sample was

$2 \times \text{12 kg" = "24 kg}$

Finally, how about $4$ half-lives ago?

$4 \times \text{5770 years" = "23,080 years ago}$

The mass of the sample was

$2 \times \text{24 kg" = "48 kg}$

Therefore, "23,080 years ago, the sample contained $\text{48 kg}$ of carbon-14.

When the values don't allow for such simple calculations, you can use the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{A"_t = "A}}_{0} \cdot \frac{1}{2} ^ n \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${\text{A}}_{t}$ - the mass of the sample that remains after a period of time $t$
${\text{A}}_{0}$ - the initial mass of the sample
$n$ - the number of half-lives that pass in the period of time $t$

n = ("23,080" color(red)(cancel(color(black)("years"))))/(5770color(red)(cancel(color(black)("years")))) = 4
${\text{A"_0 = "A}}_{t} \cdot {2}^{4}$
"A"_0 = "3 kg" * 2^4 = color(green)(|bar(ul(color(white)(a/a)color(black)("48 kg")color(white)(a/a)|)))