Question

Question #e2e73

Answer 1

`"0.19 Ci"`

Explanation:

The idea here is that you need to find the amount of phosphorus-32 that will decay in five days to leave behind `"0.15 Ci"`.

Notice that problem provides you with the isotope's nuclear half-life, which as you know tells you how much time is needed for a sample of radioactive substance to decay to half of its initial value.

If you take `A_0` to be the initial amount of phosphorus-32 and `A` to be amount of phosphorus-32 that you need for the experiment, i.e. `"0.15 Ci"`, you can say that you have

`color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^n color(white)(a/a)|)))`

Here `n` represents the number of half-lives that pass in a given amount of time and is calculated using

`color(blue)(|bar(ul(color(white)(a/a)n = "period of time"/"half-life"color(white)(a/a)|)))`

You know that the sample must travel for `5` days, and that phosphorus-32 has a half-life of `14.3` days, which means that `n` will be equal to

`n = (5 color(red)(cancel(color(black)("days"))))/(14.3color(red)(cancel(color(black)("days")))) = 0.34965035`

Rearrange the first equation to solve for `A_0`, the amount of phosphorus-32 that is needed in order to have `"0.15 Ci"` left after `n` half-lives pass

`A = A_0 * 1/2^n implies A_0 = A * 2^n`

Plug in your values to find

`A_0 = "0.15 Ci" * 2^0.34965035 = color(green)(|bar(ul(color(white)(a/a)"0.19 Ci"color(white)(a/a)|)))`

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