# Question 43a81

Oct 5, 2015

$\frac{1}{4096}$

#### 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

• ${A}_{0} / 2 \to$ 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

$\frac{{A}_{0} / 2}{2} = {A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / {2}^{2} \to$ after two half-lives

What about after another half-time passes?

${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / {2}^{3} \to$ after three half-lives

The pattern is very clear at this point - to get the amount of a radioactive isotope that remains after $\textcolor{b l u e}{n}$ half-lives pass, you need to have

${A}_{n} = {A}_{0} / {2}^{\textcolor{b l u e}{n}}$

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

${A}_{\text{1 minute}} = {A}_{0} / {2}^{12} = {A}_{0} \cdot \frac{1}{4096}$