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
Jul 28, 2016

Answer:

#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.