Question

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?

Answer 1

`6.88 * 10^(-6)"s"^(-1)`

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

`"A"_0` - the initial activity of the radioactive sample
`"A"` - the activity of the radioactive sample after a period of time `t`
`lamda` - the decay constant

You can rearrange the above equation to solve for `lamda`

`-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 `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

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

Usually, the decay constant is expressed in `"s"^(-1)`. Since the problem doesn't specify which units to use for `lamda`, you can try your hand at converting this value to `"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.