# The half-life of ""^131"I" is 8.07 days. What fraction of a sample of ""^131"I" remains after 24.21 days?

Dec 23, 2015

$\frac{1}{8}$

#### Explanation:

As you know, an isotope's nuclear half-life tells you how much time must pass in order for half of an initial sample of this isotope to undergo radioactive decay.

In other words, an isotope's half-life tells you how much must pass in order for a sample to be reduced to half of its initial value.

If you take ${A}_{0}$ to be the initial sample of an isotope, and $A$ to be the sample remaining after a period of time $t$, then you can say that

• $A = {A}_{0} \cdot \frac{1}{2} \to$ after one half-life
• $A = {A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 \to$ after two half-lives
• $A = {A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 \to$ after three half-lives
$\vdots$

and so on. This means that you can express $A$ in terms of ${A}_{0}$ and the number of half-lives that pass using the equation

$\textcolor{b l u e}{A = {A}_{0} \cdot \frac{1}{2} ^ n} \text{ }$, where

$n$ - the number of half-lives that pass in a given period of time

$\textcolor{b l u e}{n = \text{period of time"/"half-life}}$

So, you want to know what fraction of an initial sample of $\text{^131"I}$ remains after $\text{24.21 days}$.

How many half-lives do you get in that period of time, knowing that one half-life is equal to $\text{8.07 days}$?

$n = \left(24.21 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{days"))))/(8.07color(red)(cancel(color(black)("days}}}}\right) = 3$

This means that you have

$A = {A}_{0} \cdot \frac{1}{2} ^ 3$

$A = {A}_{0} \cdot \frac{1}{8}$

Therefore, your initial sample of $\text{^131"I}$ will be reduced to $\frac{1}{8} \text{th}$ of its initial value after $\text{24.21 days}$.