# The half-life for radioactive decay (a first-order process) of plutonium-239 is 24,000 years. How many years would it take for one mole of this radioactive material to decay so that just one atom remains?

Jun 15, 2016

$1.9 \cdot {10}^{6} \text{years}$

#### Explanation:

Your strategy here will be to use Avogadro's number to calculate the number of atoms of plutonium-239 that you're starting with.

One you know that, use the equation that allows you to calculate the amount of a radioactive nuclide that remains undecayed, ${\text{A}}_{t}$, in terms of the initial amount of the nuclide, ${\text{A}}_{0}$, and the number of half-lives, $n$, that pass in a given period of time $t$.

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

Here you can say that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = \frac{t}{t} _ \text{1/2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where ${t}_{\text{1/2}}$ is the half-life of the nuclide.

So, you know that Avogadro's number acts as a conversion factor between the number of moles of a element and the number of atoms it contains

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

Since you're dealing with one mole of plutonium-239, you can say that the initial amount of this isotope will be

$\text{A"_ 0 = 6.022 * 10^(23)"atoms}$

The amount that remains undecayed is

$\text{A"_t = "1 atom}$

Now, rearrange the above equation to solve for $n$

${\text{A"_t/"A}}_{0} = \frac{1}{2} ^ n$

${2}^{n} = {\text{A"_0/"A}}_{t}$

This will be equivalent to

$\ln \left({2}^{n}\right) = \ln \left({\text{A"_0/"A}}_{t}\right)$

$n \cdot \ln \left(2\right) = \ln \frac{{\text{A"_0/"A"_t) implies n = ln("A"_0/"A}}_{t}}{\ln} \left(2\right)$

Plug in your values to get

$n = \frac{1}{\ln} \left(2\right) \cdot \ln \left(\left(6.022 \cdot {10}^{23} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atoms"))))/(1color(red)(cancel(color(black)("atom}}}}\right)\right) = 78.99$

This means that it takes $78.99$ half-lives for your sample of plutonium-239 to decay from one mole to one atom.

Since the half-life of the nuclide is equal to $\text{24,000 years}$, it follows that you have

$n = \frac{t}{t} _ \text{1/2" implies t = n * t_"1/2}$

t = 78.99 * "24,000 years" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.9 * 10^6"years")color(white)(a/a)|)))

The answer is rounded to two sig figs, the number of sig figs you have for the half-life of the nuclide.

You can thus say that it will take $1.9$ million years for one mole of plutonium-239 to decay to one atom.