Question

Question #e3336

Answer 1

`"15.5 mg"`

Explanation:

The half-life of a radioactive nuclide, `t_"1/2"`, is simply a measure of the time needed for half of a given sample to undergo radioactive decay.

So if you take `A_0` to be the initial mass of carbon-14, you can say that this sample will be reduced to

• `A_0 * 1/2 = A_0/2 = 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
  `vdots`

and so on.

Now, let's say that `A_t` represents the mass of carbon-14 that remains undecayed after a period of time `t`. You can say that `A_t` will be equal to

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

with

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

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

In your case, the sample has an initial mass of `"52 mg"` and the half-life of carbon-14 is equal to `5730`years, so you can say that after a period of time `t` passes, the sample will be reduced to

`A_t = "52 mg" * (1/2)^(t/"5730 years")`

To find the amount of carbon-14 that remains undecayed after `"10,000"` years, simply plug in this value into the above equation.

You will have

`color(red)(n) = ("10,000" color(red)(cancel(color(black)("years"))))/(5730color(red)(cancel(color(black)("years"))))`

and

`A_t = "52 mg" * (1/2)^("10,000"/5730) = color(darkgreen)(ul(color(black)("15.5 mg")))`

The answer is rounded to the nearest tenth.