Question e3336

Dec 27, 2017

$\text{15.5 mg}$

Explanation:

The half-life of a radioactive nuclide, ${t}_{\text{1/2}}$, is simply a measure of the time needed for half of a given sample to undergo radioactive decay.

So if you take ${A}_{0}$ to be the initial mass of carbon-14, you can say that this sample will be reduced to

• ${A}_{0} \cdot \frac{1}{2} = {A}_{0} / 2 = {A}_{0} / {2}^{\textcolor{red}{1}} \to$ after $\textcolor{red}{1}$ half-life
• ${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 = {A}_{0} / {2}^{\textcolor{red}{2}} \to$ after $\textcolor{red}{2}$ half-lives
• ${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 = {A}_{0} / {2}^{\textcolor{red}{3}} \to$ after $\textcolor{red}{3}$ half-lives
$\vdots$

and so on.

Now, let's say that ${A}_{t}$ represents the mass of carbon-14 that remains undecayed after a period of time $t$. You can say that ${A}_{t}$ will be equal to

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

with

$\textcolor{red}{n} = \frac{t}{t} _ \text{1/2}$

Here $\textcolor{red}{n}$ represents the number of half-lives that pass in the given period of time $t$.

In your case, the sample has an initial mass of $\text{52 mg}$ and the half-life of carbon-14 is equal to $5730$years, so you can say that after a period of time $t$ passes, the sample will be reduced to

A_t = "52 mg" * (1/2)^(t/"5730 years")

To find the amount of carbon-14 that remains undecayed after $\text{10,000}$ years, simply plug in this value into the above equation.

You will have

color(red)(n) = ("10,000" color(red)(cancel(color(black)("years"))))/(5730color(red)(cancel(color(black)("years"))))

and

A_t = "52 mg" * (1/2)^("10,000"/5730) = color(darkgreen)(ul(color(black)("15.5 mg")))#

The answer is rounded to the nearest tenth.