Question

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?

Answer 1

`"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 `"3 kg"`, to `"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 xx "3 kg" = "6 kg"`

How about `2` half-lives ago?

`2 xx "5770 years" = "11,540 years ago"`

The mass of the sample was

`2 xx "6 kg" = "12 kg"`

How about `3` half-lives ago?

`3 xx "5770 years" = "17,310 years ago"`

The mass of the sample was

`2 xx "12 kg" = "24 kg"`

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

`4 xx "5770 years" = "23,080 years ago"`

The mass of the sample was

`2 xx "24 kg" = "48 kg"`

Therefore, `"23,080` years ago, the sample contained `"48 kg"` of carbon-14.

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

`color(blue)(|bar(ul(color(white)(a/a)"A"_t = "A"_0 * 1/2^ncolor(white)(a/a)|)))`

Here

`"A"_t` - the mass of the sample that remains after a period of time `t`
`"A"_0` - the initial mass of the sample
`n` - the number of half-lives that pass in the period of time `t`

In your case, you have

`n = ("23,080" color(red)(cancel(color(black)("years"))))/(5770color(red)(cancel(color(black)("years")))) = 4`

You will thus have

`"A"_0 = "A"_t * 2^4`

`"A"_0 = "3 kg" * 2^4 = color(green)(|bar(ul(color(white)(a/a)color(black)("48 kg")color(white)(a/a)|)))`