Question

Question #b3120

Answer 1

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life, `t_"1/2"`, tells you the time needed for half of the atoms present in a sample of said isotope to undergo radioactive decay.

This means that if you start with `A_0` radioactive isotope, you can say that you will end up with

• `A_0 * 1/2 = A_0/2^color(red)(1) ->` after `color(red)(1)` half-life
• `A_0/2 * 1/2 = A_0/4 = A_0/2^color(red)(2) ->` after `color(red)(2)` half-lives
• `A_0/4 * 1/2 = A_0/8 = A_0/2^color(red)(3) ->` after `color(red)(3)` half-lives
• `A_0/8 * 1/2 = A_0/16 = A_0/2^color(red)(4) ->` after `color(red)(4)` half-lives
  `vdots`

and so on. You can thus say that the number of radioactive isotopes that remain undecayed after a period of time `t` is equal to

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

Here `color(red)(n)` represents the number of half-lives that pass in the given time `t`.

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

In your case, you know that you start with `100` radioactive isotopes and that `88` have already decayed to their daughter nuclides. This, of course, implies that you're left with `12` radioactive isotopes after an unknown time `t` passes.

You can thus say that

`12 = 100 * (1/2)^color(red)(n)`

Rearrange to get

`12/100 = (1/2)^color(red)(n)`

This will be equivalent to

`ln(12/100) = ln[(1/2)^color(red)(n)]`

which gets you

`color(red)(n) = ln(12/100)/ln(1/2) = 3.06`

So, you know that `3.06` half-lives must pass in order for the number of radioactive isotopes to go from `100` to `12`.

You now have

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

and since you know that

`t_"1/2" = "1000 years"`

you can say that the rock is

`"age of rock" = 3.06 * "1000 years" = color(darkgreen)(ul(color(black)("3060 years")))`

I'll leave the answer rounded to three sig figs, but keep in mind that your values do not justify this value.