# Question #d4494

Jan 3, 2018

Here's what I got.

#### Explanation:

The thing to remember about an isotope's nuclear half-life, ${t}_{\text{1/2}}$, is that it represents the time needed for half of an initial sample of that isotope to undergo radioactive decay.

In other words, the half-life tells you how much must pass in order for your sample to be halved.

If you take ${A}_{t}$ to be the amount of a radioactive isotope that remains undecayed after a time $t$ passes and ${A}_{0}$ to be the initial amount of that isotope, you can say that you will have

• ${A}_{t} = {A}_{0} \cdot \frac{1}{2} = {A}_{0} / 2 = {A}_{0} / {2}^{\textcolor{red}{1}} \to$ after $\textcolor{red}{1}$ half-life
• ${A}_{t} = {A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 = {A}_{0} / {2}^{\textcolor{red}{2}} \to$ after $\textcolor{red}{2}$ half-lives
• ${A}_{t} = {A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 = {A}_{0} / {2}^{\textcolor{red}{3}} \to$ after $\textcolor{red}{3}$ half-lives
$\vdots$

and so on. So with every passing half-life, you get to divide the initial amount by $2$. This means that if $\textcolor{red}{n}$ half-lives pass in a given period of tiem $t$, you will divide the initial amount by $2$ a total of $\textcolor{red}{n}$ times.

${A}_{t} = {A}_{0} / \left({\underbrace{2 \cdot 2 \cdot \ldots \cdot 2}}_{\textcolor{b l a c k}{\textcolor{red}{n} \textcolor{w h i t e}{.} \text{times}}}\right) = {A}_{0} / {2}^{\textcolor{red}{n}}$

This is equivalent to

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

with

$\textcolor{red}{n} = \frac{t}{t} _ \text{1/2}$

In your case, you start with $\text{200 g}$ of this radioactive isotope and end up with $\text{25 g}$. This means that you have

$25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) = 200color(red)(cancel(color(black)("g}}}} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

Divide both sides by $200$ to get

$\frac{25}{200} = \frac{1}{2} ^ \textcolor{red}{n}$

This is equivalent to

$\frac{1}{4} = \frac{1}{2} ^ \textcolor{red}{n}$

$\frac{1}{2} ^ 2 = \frac{1}{2} ^ \textcolor{red}{n} \implies \textcolor{red}{n} = 2$

So you can say that in order for your sample to be reduced from $\text{200 g}$ to $\text{25 g}$, two half-lives must pass.

This means that you have

$t = 2 \cdot {t}_{\text{1/2}}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{t = 2 \cdot \text{8.4 days" = "16.8 days}}}}$

I'll leave the answer rounded to three sig figs, but a more accurate answer would be

$t = \text{20 days}$

because you have only one significant figure for the initial mass of the sample.