Question

Question #d5cfe

Answer 1

`"13 billion years"`

Explanation:

Your go-to equation here looks like this

`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`

Now, you know that `5%` of the initial amount of your element remains undecayed after a period fo time `t`, which you must determine.

In such cases, you can express `A_t` as a function of `A_0`

`A_t = 5/100 * A_0`

This means that the amount left after a period of time `t` is equal to `5% = 5/100` of the initial amount

Plug this into the above equation to get

`5/100 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^n`

Now, you can rewrite this as

`ln(5/100) = ln[(1/2)^n]`

This is done by taking the natural log of both sides of the equation

This will be equivalent to

`ln(5/100) = n * ln(1/2)`

which gets you

`n = ln(5/100)/ln(1/2) = 4.32`

So, you know that in order for `5%` of the initial amount of your element to remain undecayed, i.e. in order for `95%` to decay, a period of time equal to `4.32` half-lives must pass.

Since you have

`n = "total time that passes"/"half life" = t/t_"1/2"`

you can say that

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

In your case, this will get you

`color(darkgreen)(ul(color(black)(t = 4.32 * "3 billion years" = "13 billion years")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the half-life of the element.