# Question b3120

Jun 8, 2017

Here's what I got.

#### Explanation:

The idea here is that a radioactive isotope's nuclear half-life, ${t}_{\text{1/2}}$, tells you the time needed for half of the atoms present in a sample of said isotope to undergo radioactive decay.

This means that if you start with ${A}_{0}$ radioactive isotope, you can say that you will end up with

• ${A}_{0} \cdot \frac{1}{2} = {A}_{0} / {2}^{\textcolor{red}{1}} \to$ 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
• ${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} / 16 = {A}_{0} / {2}^{\textcolor{red}{4}} \to$ after $\textcolor{red}{4}$ half-lives
$\vdots$

and so on. You can thus say that the number of radioactive isotopes that remain undecayed after a period of time $t$ is equal to

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

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

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

In your case, you know that you start with $100$ radioactive isotopes and that $88$ have already decayed to their daughter nuclides. This, of course, implies that you're left with $12$ radioactive isotopes after an unknown time $t$ passes.

You can thus say that

$12 = 100 \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

Rearrange to get

$\frac{12}{100} = {\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}$

This will be equivalent to

$\ln \left(\frac{12}{100}\right) = \ln \left[{\left(\frac{1}{2}\right)}^{\textcolor{red}{n}}\right]$

which gets you

$\textcolor{red}{n} = \ln \frac{\frac{12}{100}}{\ln} \left(\frac{1}{2}\right) = 3.06$

So, you know that $3.06$ half-lives must pass in order for the number of radioactive isotopes to go from $100$ to $12$.

You now have

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

and since you know that

${t}_{\text{1/2" = "1000 years}}$

you can say that the rock is

"age of rock" = 3.06 * "1000 years" = color(darkgreen)(ul(color(black)("3060 years")))#

I'll leave the answer rounded to three sig figs, but keep in mind that your values do not justify this value.