How long does it take a radioactive element to decay to #10%# of its original quantity in a first-order process if its half-life is #"43.80 s"#?

#t = "145.5 s"#

(For the record, I did this without looking at the answer. It's much more important you know HOW to do it.)

The first method is a very general approach as in radioactive decay problems (which are first-order processes!). The second method is from the kinetics unit in general chemistry.

METHOD 1

This can be done simply by knowing the definition of a half-life:

The time it takes for the concentration of a sample to halve.

Since we know that

#t_"1/2" = "43.80 s"#,

consider the following recursive train of thought for first-order half-lives only.

• After one half-lives, #[A] = 1/2[A]_0#.
• After two half-lives, #[A] = 1/4[A]_0#.
• After three half-lives, #[A] = 1/8[A]_0#.
• After four half-lives, #[A] = 1/16[A]_0#.

and so on. Thus, we can write the current concentration as a function of the starting concentration:

#[A] = (1/2)^n [A]_0#

where #n# is the number of half-lives that passed.

The number of half-lives passed, #n#, is simply the total time divided by the known half-life.

Thus, #n = t//t_"1/2"#, and we have the current concentration as a function of the starting concentration, half-life, and total time passed:

#barul|stackrel(" ")(" "[A] = (1/2)^(t//t_"1/2") [A]_0" ")|#

Given that #10%# of the sample is left after decay, we know that

#[A] = 0.1[A]_0#,

and thus,

#(1/2)^(t//t_"1/2") = 0.1#.

To solve for #t#, take the #ln# of both sides.

#ln(1/2)^(t//t_"1/2") = ln0.1#

#t/(t_"1/2")ln(1/2) = ln0.1#

Therefore, it will take this long for #10%# of the sample to remain:

#color(blue)(t) = t_"1/2"cdotln0.1/ln(1/2)#

#= "43.80 s" cdot (ln 0.1)/(ln (1/2))#

#=# #color(blue)("145.5 s")#

METHOD 2

The more usual method uses the half-life of a first-order reaction (refer to your textbook):

#t_"1/2" = (ln2)/k#

And thus, the rate constant is given by:

#k = (ln2)/t_"1/2"#

Via the first-order integrated rate law (refer to your textbook),

#ln[A] = -kt + ln[A]_0#,

one can then solve for the time passed simply by knowing that #[A] = 0.1[A]_0# as mentioned earlier.

#ln(0.1[A]_0) - ln[A]_0 = -kt#

#ln((0.1cancel([A]_0))/(cancel([A]_0))) = -kt#

#t = ln0.1/(-k)#

#= (ln0.1)/(- (ln2)/(t_"1/2"))#

#= t_"1/2" cdot (ln0.1)/(ln (1/2))#

which is of the same form as in method 1. We would also get #"145.5 s"# in this way.