Question

Question #36005

Answer 1

`"1060 years"`

Explanation:

Your tool of choice here will be the following equation

`A_t = A_0 * (1/2)^color(red)(n)`

Here

• `A_t` is the mass of `""^14"C"` that remains undecayed after a period of time `t`
• `A_0` is the initial mass of `""^14"C"`
• `color(red)(n)` represents the number of half-lives that pass in a given time period `t`

Now, you know that your fossil contains `88.0%` of the initial mass of `""^14"C"`, so you can say that you have

`A_t = 88.0/100 * A_0`

Plug this into the above equation to get

`88.0/100 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^color(red)(n)`

`88.0/100 = (1/2)^color(red)(n)`

Rearrange to solve for `n`

`ln(88.0/100) = ln[(1/2)^color(red)(n)]`

You will end up wtih

`color(red)(n) * ln(1/2) = ln(88.0/100)`

which will get you

`color(red)(n) = ln(88.0/100)/ln(1/2) = 0.1844`

This tells you that `0.1844` half-lives have passed from the moment the initial sample of `""^14"C"` started to undergo radioactive decay to the moment you took your measurements.

If you take `t_"1/2"` to be the half-live of `""^14"C"`, you can say that you have

`t = color(red)(n) * t_"1/2"`

which, in your case, is equal to

`t = 0.1844 * t_"1/2"`

Since `""^14"C"` has a half-life of about `5730` years (see here), you can say that

`t = 0.1844 * "5730 years" = color(darkgreen)(ul(color(black)("1060 years")))`

have passed, i.e. the fossil is about `1060` years old.

The answer is rounded to three sig figs.