# Question 3890e

Nov 23, 2015

$\text{1.7 g}$

#### Explanation:

Your go-to equation when it comes to calculating the amount of a radioactive sample that remains after the passing of a certain period of time looks like this

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

$A$ - the amount of the radioactive substance that remains after the passing of $n$ half-lives
${A}_{0}$ - the initial mass of the sample
$n$ - the number of half-lives that passed

As you know, the nuclear half-life of a radioactive substance is defined as the time needed for half of the atoms of that sample to undergo radioactive decay.

In other words, the nuclear half-life tells you how much time must before before a sample of a radioactive substance is reduced to half of its original mass.

In your case, the half-life of plutonium is known to $2.4 \cdot {10}^{4}$ years. To determine the value of $n$, the number of half-lives that pass in a given period of time, divide the respective period of time by the half-life

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

$n = \left(5.0 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{years"))))/(2.4 * 10^4color(red)(cancel(color(black)("years}}}}\right) = \frac{25}{12} \cdot {10}^{- 1} = \frac{5}{24}$

This means that the amount of plutonium that remains undecayed after that much time will be

A = "2.0 g" * 1/2^(5/24) = color(green)("1.7 g")#