# Question 5e9e6

Jul 19, 2016

$\text{2.25 years}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {A}_{t} = {A}_{0} \cdot \frac{1}{2} ^ n \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${A}_{t}$ - the amount undecayed after a period of time $t$
${A}_{0}$ - the initial mass of the radioactive isotope
$n$ - the number of half-lives that pass in a period of time $t$

In your case, you know that it took $9$ years for a sample of a given radioactive isotope to decay from $\text{448 g}$, the initial mass of the sample, to $\text{28 g}$, the mass that remains undecayed.

Your goal here will be to use the above equation to find the value of $n$, the number of half-lives that pass in $9$ years. Plug in your values to find

$28 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) = 448 color(red)(cancel(color(black)("g}}}} \cdot \frac{1}{2} ^ n$

Rearrange to find

${2}^{n} = \frac{448}{28} = 16$

Since $16$ can be written as a power of $2$

$16 = 2 \cdot 2 \cdot 2 \cdot 2 = {2}^{4}$

you will have

${2}^{n} = {2}^{4}$

This implies that

$n = 4$

So, you know that it takes $9$ years for $4$ half-lives to pass, which means that one half-life, ${t}_{\text{1/2}}$, is

t_"1/2" = "9 years"/4 = color(green)(|bar(ul(color(white)(a/a)color(black)("2.25 years")color(white)(a/a)|)))#