# Question #7f902

Oct 27, 2017

$\text{4797 years}$

#### Explanation:

The idea here is that the amount of radium-226 you have in your sample will be halved with every passing half-life.

This is the case because a radioactive nuclide's nuclear half-life tells you the amount of time needed for half of an initial sample to undergo radioactive decay.

Mathematically, you can write this as

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

Here

• ${A}_{t}$ is the amount of the radioactive nuclide that remains undecayed after a period of time $t$
• ${A}_{0}$ is the initial amount of the radioactive nuclide
• $\textcolor{red}{n}$ is the number of half-lives that pass in the period of time $t$

Now, you know that $\frac{1}{8} \text{th}$ of an initial sample of radium-226 remains undecayed. This is equivalent to saying that you have

${A}_{t} = {A}_{0} \cdot \frac{1}{8}$

Since you know that

$8 = 2 \cdot 2 \cdot 2 = {2}^{3}$

you can rewrite this as

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{3}}$

You can thus say that in order for the initial sample to be reduced to $\frac{1}{8} \text{th}$ of its initial value, $\textcolor{red}{n} = \textcolor{red}{3}$ half-lives must pass.

Since you know that radium-226 has a half-life of $1599$ years, you can say that a total of

$3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{half-lives"))) * "1599 years"/(1color(red)(cancel(color(black)("half-life")))) = color(darkgreen)(ul(color(black)("4797 years}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the half-life of the nuclide.

Keep in mind that you have three sig figs for the initial mass of the sample, but you're not using this value in your calculations.