Question

A parchment fragment was discovered that had about 74% as much Carbon 14 radioactivity as does plant material on the earth today. How would you estimate the age of the parchment?

The half life of carbon-14 is about 5730 years.

Answer 1

`"2490 years"`

Explanation:

It's important that you don't let the wording of the problem confuse you.

The part about

"74% as much carbon-14 as does plant material on the Earth today"

is equivalent to

"of the initial amount of carbon-14 present in the parchment, only 74% remains undecayed"

Now, a radioactive isotope's nuclear half-life is defined as the time needed to an initial sample of this isotope to be reduced to half of its initial value.

Simply put, any amount of carbon-14 you start with will be halved with the passing of each half-life. Mathematically, you can write this as

`color(blue)(A = A_0 * 1/2^n)" "`, where

`A` - the amount remaining after a given period of time
`A_0` - the initial amount
`n` - the number of half-lives that pass in that given period of time

In your case, you know that `74%` of the initial mass of carbon-14 present in the parchment remains undecayed. This means that you can write

`A = 74/100 * A_0`

The above equation becomes

`74/100 color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * 1/2^n`

Solve this equation for `n`

`ln(74/100) = ln( (1/2)^n)`

`ln(74/100) = n * ln(1/2)`

`n = ln(74/100)/ln(1/2) = 0.434403`

Since `n` represents the number of half-lives that must pass in order for the sample to be reduced to that percentage of the initial mass

`color(blue)(n = "period of time"/"half-life")`

you can say that

`n = t/t_"1/2" implies t = n * t_"1/2"`

Therefore

`t = 0.434403 * "5730 years" = color(green)("2490 years")`

I'll leave the answer rounded to three sig figs.