Question

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"`?

Answer 1

`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.