Question

Question #a4462

Answer 1

`"11.1 days"`

Explanation:

Your tool of choice here will be the equation

`color(blue)(ul(color(black)(A_t = A_0 * (1/2)^n)))`

Here

• `A_t` is the amount of the radioactive substance that remains undecayed after a time period `t`
• `A_0` is the initial amount of said substance
• `n` is the number of half-lives that pass in the time period `t`

Your first goal here is to figure out the value of `n`. Start by rewriting the equation as

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

This will be equivalent to

`ln(A_t/A_0) = ln[(1/2)^n]`

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

Since

`ln(1/2) = ln(1) - ln(2) = - ln(2)`

you can say that

`n = -ln(A_t/A_0)/ln(2)`

Plug in your values to find

`n = - ln((0.100 color(red)(cancel(color(black)("g"))))/(0.750color(red)(cancel(color(black)("g")))))/ln(2) = 2.907`

Now, you know that `2.907` half-lives must pass in order for the sample to be reduced from `"0.750 g"` to `"0.100 g"` and that radon-222 has a half-life of `3.823` days.

The number of half-lives that pass in a given period of time can be written as

`color(blue)(ul(color(black)(n = "total time that passes"/"half-life" = t/t_"1/2")))`

The total time `t` needed will be

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

In your case, you will have

`color(darkgreen)(ul(color(black)(t = 2.907 * "3.823 days" = "11.1 days")))`

The answer is rounded to three sig figs, the number of sig figs you have for the initial and final mass of the sample.