# A sample of a radioactive isotope is found to have lost 39.9% of its original activity after 8.57 days. What is the decay constant of this isotope?

Jul 28, 2016

$6.88 \cdot {10}^{- 6} {\text{s}}^{- 1}$

#### Explanation:

As you know, radioactive decay is a first-order reaction described by the integrated rate law

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \ln \left({\text{A") = - lamda * t + ln("A}}_{0}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${\text{A}}_{0}$ - the initial activity of the radioactive sample
$\text{A}$ - the activity of the radioactive sample after a period of time $t$
$l a m \mathrm{da}$ - the decay constant

You can rearrange the above equation to solve for $l a m \mathrm{da}$

$- l a m \mathrm{da} \cdot t = \ln \left({\text{A") - ln("A}}_{0}\right)$

$l a m \mathrm{da} = \frac{\ln \left(\text{A"_0) - ln("A}\right)}{t}$

which gets you

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{l a m \mathrm{da} = \ln \frac{\text{A"_0/"A}}{t}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you know that it takes $8.57$ days for the activity of the sample to decrease by 39.9% of its original value. This is equivalent to saying that after $8.57$ days, the sample is left with an activity that is

"A" = 100% - 39.9% = 60.1%

of its original value. You can now plug in your values to find the value of the decay constant

$l a m \mathrm{da} = \ln \frac{\frac{100}{60.1}}{\text{8.57 days" = "0.05941 days}} ^ \left(- 1\right)$

Usually, the decay constant is expressed in ${\text{s}}^{- 1}$. Since the problem doesn't specify which units to use for $l a m \mathrm{da}$, you can try your hand at converting this value to ${\text{s}}^{- 1}$

0.05941 1/color(red)(cancel(color(black)("days"))) * (1color(red)(cancel(color(black)("day"))))/(24 color(red)(cancel(color(black)("hrs")))) * (1color(red)(cancel(color(black)("hr"))))/(60color(red)(cancel(color(black)("min")))) * (1color(red)(cancel(color(black)("min"))))/"60 s" = color(green)(|bar(ul(color(white)(a/a)color(black)(6.88 * 10^(-6)"s"^(-1))color(white)(a/a)|)))

The answer is rounded to three sig figs.