The half-life of plutonium-239 is 24,100 years. Of an original mass of 100g, how much plutonium-239 remains after 96,440 years?

Nov 10, 2015

$6.25 g$

Explanation:

Note: I edited the question to use the half-life of plutonium-239 as the correct value of $24 , 100$ years, not the value in the original question of $24 , 110$ years, which I assume was a typo.

Use the equation

${M}_{r} = {M}_{i} \times {\left(\frac{1}{2}\right)}^{n}$

Where $n$ is the number of half-lives
${M}_{r}$ is the mass remaining after $n$ half-lives
and ${M}_{i}$ is the initial mass of the sample

to find $n$, the number of half-lives, divide the total time ($96 , 400$) by the time of the half-life ($24 , 100$)

$n = \frac{96 , 400}{24 , 100} = 4$

So:
${M}_{r} = 100 g \times {\left(\frac{1}{2}\right)}^{4} = 6.25 g$

The mass remaining is $6.25 g$

Nov 10, 2015

$\text{6.25 g}$

Explanation:

The nuclear half-life of a nuclide tells you how much time must pass before half of an initial sample of said nuclide undergoes radioactive decay.

Simply put, an initial sample of radioactive isotope is halved with every passing of a half-life.

This means that for a ${A}_{0}$ sample of a radioactive isotope, you can say that you'll be left with

• ${A}_{0} \cdot \frac{1}{2} \to$ 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.

Notice that you can express the amount of the sample that remains after $n$ half-lives like this

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

Now, notice that the time given to you is actually a multiple of the half-life

$n = \frac{96440}{24110} = 4$

This means that four half-lives of plutonium-239 will pass in $\text{96,440}$, and so the remaining sample after this period of time will be

$m = \text{100 g" * 1/2^4 = "6.25 g}$

I'll leave the answer rounded to three sig figs, despite the fact that you only have one sig fig for the mass of the sample.