Question

Question #7bdb7

Answer 1

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life tells you the time needed for half of an initial sample to undergo radioactive decay.

In your case, you know that it takes `3` minutes for half of any amount of polonium-218 that you have to undergo radioactive decay.

If you take `A_0` to be the initial mass of polonium-238 and `A_t` to be the amount that remains undecayed after a given period of time `t`, you can say that you have

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

Here

• `n` is the number of half-lives that pass in the given period of time `t`

In your case, you know that it takes `30` minutes to transport the sample of polonium-238, which implies that you have

`n = (30 color(red)(cancel(color(black)("minutes"))))/(3color(red)(cancel(color(black)("minutes")))) = 10`

So if `10` half-lives pass in `30` minutes and you know that you must end up with `"0.10 g"` of polonium-238, you can say that you have

`"0.10 g" = A_0 * (1/2)^10`

Rearrange to solve for `A_0`

`A_0 = "0.10 g" * 2^10 = "102.4 g"`

Now, I'll leave the answer rounded to two sig figs, but you could round it to one significant figure based on the value you have for the half-life of the isotope.

`color(darkgreen)(ul(color(black)("amount needed" = 1.0 * 10^2color(white)(.)"g")))`