Question

Question #43a81

Answer 1

`1/4096`

Explanation:

The idea with nuclear half-life calculations is that the initial mass of the isotope will be halved with every passing of the isotope's half-life.

If you initial amount of fluorine-21 is `A_0`, the you know that you will be left with

• `A_0/2 ->` after one half-life;

What will happen after a second half-life passes?

Well, the amount will be halved again, so you know that you will be left with

`(A_0/2)/2 = A_0/2 * 1/2 = A_0/2^2 ->` after two half-lives

What about after another half-time passes?

`A_0/4 * 1/2 = A_0/2^3 ->` after three half-lives

The pattern is very clear at this point - to get the amount of a radioactive isotope that remains after `color(blue)(n)` half-lives pass, you need to have

`A_n = A_0/2^color(blue)(n)`

In your case, the half-life of fluorine-21 isotope is 5 seconds. How many half-lives must pass for one minute to pass?

`1color(red)(cancel(color(black)("min"))) * (60color(red)(cancel(color(black)("s"))))/(1color(red)(cancel(color(black)("min")))) * "1 half-life"/(5color(red)(cancel(color(black)("s")))) = "12 half-lives"`

This means that after one minute passes you will be left with

`A_"1 minute" = A_0/2^12 = A_0 * 1/4096`