# Question #95320

Dec 15, 2015

$\text{10.8 days}$

#### Explanation:

Here's a quick way to solve this problem - divide the mass of the initial sample by the mass of the final sample, then find a number that when used as an exponent for $2$ gives you this ratio.

Once you have that number, multiply it by the half-life of the isotope and you have your answer.

In this case,

$\left(180 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(11.25color(red)(cancel(color(black)("g}}}}\right) = 16$

${2}^{x} = 16 \implies x = 4$

Therefore,

$t = 4 \cdot \text{2.69 days" = "10.76 days}$

*Now, let me explain what's going on here. *

As you know, an isotope's nuclear half-life tells you the time needed for a sample of this isotope to decay to half of its initial value.

So, if your initial sample is ${A}_{0}$, you can say that you'll be left with

• ${A}_{0} \cdot \frac{1}{2} = {A}_{0} / 2 \to$ after one half-life
• ${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 \to$ after two half-lives
• ${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 \to$ after three half-lives
• ${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} / 16 \to$ after four half-lives
$\vdots$

and so on.

This of course allows you to find a relationship between the initial sample, ${A}_{0}$, and the remaining sample, $A$, by using the number of half-lives that pass

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

$n$ - how many half-lives pass in a given period of time

In your case, you know that the initial sample of $\text{^198"Au}$ had a mass of $\text{180 g}$, and that after a period of time, the sample is down to $\text{11.25 g}$.

Plug this into the above equation to get

$11.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) = 180 color(red)(cancel(color(black)("g}}}} \cdot \frac{1}{2} ^ n$

This is equivalent to

$\frac{1}{2} ^ n = \frac{11.25}{180}$

$\frac{1}{2} ^ n = \frac{1}{16} \implies {2}^{n} = 16 \iff n = 4$

Since $n$ is equal to

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

you can say that

$4 = \text{period of time"/"2.69 days}$

Therefore, the initial sample will be reduced to $\text{11.25 g}$ in

$\text{period of time" = 4 xx "2.69 days" = "10.76 days}$

You should round this off to two sig figs, the number of sig figs you have for the initial sample, but I'll leave it rounded to three sig figs

$t = \textcolor{g r e e n}{\text{10.8 days}}$