# Question cc5db

Apr 8, 2017

Here's what I got.

#### Explanation:

The nuclear half-life, ${t}_{\text{1/2}}$, tells you the amount of time needed in order for an initial sample of a radioactive nuclide to decay to half of its original mass.

If you take ${B}_{0}$ to be the initial mass of the radioactive substance, you can say that you will have

• ${B}_{0} \cdot \frac{1}{2} = {B}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{1}} \to$ after $\textcolor{red}{1}$ half-life
• ${B}_{0} / 2 \cdot \frac{1}{2} = {B}_{0} \cdot \frac{1}{4} = {B}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{2}} \to$ after $\textcolor{red}{2}$ half-lives
• ${B}_{0} / 4 \cdot \frac{1}{2} = {B}_{0} \cdot \frac{1}{8} = {B}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{3}} \to$ after $\textcolor{red}{3}$ half-lives
• ${B}_{0} / 8 \cdot \frac{1}{2} = {B}_{0} \cdot \frac{1}{16} = {B}_{0} \cdot {\left(\frac{1}{2}\right)}^{\textcolor{red}{4}} \to$ after $\textcolor{red}{4}$ half-lives
$\vdots$

and so on. Now, if you take ${B}_{t}$ to be the amount that remains undecayed after a given period of time $t$, you can say that ${B}_{t}$ will be equal to

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

Here $\textcolor{red}{n}$ represents the number of half-lives that pass in the given period of time $t$. You can express the number of half-lives that pass in the given period of time $t$ by using the half-life of the substance

color(red)(n) = t/t_"1/2" color(white)((acolor(black)( larr " the total time that passes")/(color(black)( larr " one half-life")aaaaaaaaaaa)

${t}_{\text{1/2" = "40 years" " }}$ and $\text{ " B_0 = "3 g}$

so

$\textcolor{red}{n} = \frac{t}{\text{40 years}}$

and the exponential equation will look like this

color(darkgreen)(bar(|ul(color(black)(color(white)(a/a)B_t = "3 g" * (1/2)^(t/"40 years")color(white)(a/a)))|))

To determine how much $B$ is left after $8$ years, simply calculate the number of half-lives that will pass in this amount of time

$\left(8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{years"))))/(40color(red)(cancel(color(black)("years}}}}\right) = \frac{1}{5}$

and plug the result into the equation

B_"8 years" = "3 g" * (1/2)^(1/5) = color(darkgreen)(ul(color(black)("2.61 g B")))

Finally, to determine the number of years needed for the initial sample to be reduced to $\text{0.5 g}$, rearrange the equation as

0.5 color(red)(cancel(color(black)("g"))) = 3 color(red)(cancel(color(black)("g"))) * (1/2)^(t/"40 years")

${\left(\frac{1}{2}\right)}^{\frac{t}{\text{40 years}}} = \frac{0.5}{3}$

Take the natural log from both sides

$\ln \left[{\left(\frac{1}{2}\right)}^{\frac{t}{\text{40 years}}}\right] = \ln \left(\frac{1}{6}\right)$

This will be equivalent to

$\frac{t}{\text{40 years}} \cdot \ln \left(\frac{1}{2}\right) = \ln \left(\frac{1}{6}\right)$

You will end up with

t = ln(1/6)/ln(1/2) * "40 years" = color(darkgreen)(ul(color(black)("103 years")))#

I'll leave the values rounded to three sig figs.