# 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?

##### 1 Answer

#### Explanation:

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

#color(blue)(|bar(ul(color(white)(a/a) ln("A") = - lamda * t + ln("A"_0) color(white)(a/a)|)))#

Here

**initial activity** of the radioactive sample

**decay constant**

You can rearrange the above equation to solve for

#-lamda * t = ln("A") - ln("A"_0)#

#lamda = (ln("A"_0) - ln("A"))/t#

which gets you

#color(purple)(|bar(ul(color(white)(a/a)color(black)(lamda = ln("A"_0/"A")/t)color(white)(a/a)|)))#

In your case, you know that it takes **days** for the activity of the sample to decrease by *after* **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

#lamda = ln(100/60.1)/"8.57 days" = "0.05941 days"^(-1)#

Usually, the decay constant is expressed in

#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**.