# Question #c035b

Nov 30, 2015

$\text{26 g}$

#### Explanation:

You don't actually need to use the formula given to you, all you need to know here is that the initial sample of plutonium-239 will be halved with every passing of a half-life.

This means that you can write

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

$n$ - the number of half-lives that pass in a given period of time
$A$ - the mass of the sample that remains undecayed
${A}_{0}$ - the initial mass of the sample

So, your goal here is to find the value of $n$ by using the fact that

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

In your case, the half-life of plutonium-239 is said to be $2.4 \cdot {10}^{4}$ years. This means that after $9500$ years, $n$ will be

$n = \left(9500 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{years"))))/(2.4 * 10^4 color(red)(cancel(color(black)("years}}}}\right) = \frac{19}{48}$

If the initial sample had a mass of $34.2$ grams, then you'll be left with

$A = \text{34.2 g" * 1/2^(19/48) = "25.994 g}$

Rounded to two sig figs, the answer will be

$A = \textcolor{g r e e n}{\text{26 g}}$