Question

Titanium-44 is radioactive isotope such that every 63 years, its mass decreases by half. For a sample of titanium-44 with an initial mass of 100 grams, what is a function that will give the mass of the sample remaining after any amount of time?

Answer 1

`A_t = "100 g" * (1/2)^(t/"63 years")`

Explanation:

The first thing to notice here is that the problem is giving you the half-life of titanium-44.

As you know, the half-life of a radioactive nuclide, `t_"1/2"`, tells you the amount of time needed for half of an initial sample of said nuclide to undergo radioactive decay.

In this case, you know that every `63` years, the mass of titanium-44 present in a sample is halved.

`t_"1/2" = "63 years"`

Now, if you take `A_0` to be the initial mass of titanium-44, you can say that you will have

• `A_0 * 1/2 = A_0 * (1/2)^color(red)(1)->` after `color(red)(1)` half-life
• `A_0/2 * 1/2 = A_0 * 1/4 = A_0 * (1/2)^color(red)(2) ->` after `color(red)(2)` half-lives
• `A_0/4 * 1/2 = A_0 * 1/8 = A_0 * (1/2)^color(red)(3) ->` after `color(red)(3)` half-lives
• `A_0/8 * 1/2 = A_0 * 1/16 = A_0 * (1/2)^color(red)(4) ->` after `color(red)(4)` half-lives
  `vdots`

and so on. You can thus say that `A_t`, which will be the amount of titanium-44 that remains undecayed after a period of time `t`, will be equal to

`A_t = A_0 * (1/2)^color(red)(n)`

Here `color(red)(n)` represents the number of half-lives that pass in the given period of time `t`. Since you know that

`"number of half-lives" = "total time"/"one half-life"`

you can express the number of half-lives that pass in the period of time `t` by using the half-life of the nuclide

`color(red)(n) = t/t_"1.2"`

Plug this into the equation to get

`A_t = A_0 * (1/2)^(t/t_"1/2")`

Since you know that

`A_0 = "100 g"`

you can rewrite the equation as

`color(darkgreen)(ul(color(black)(A_t = "100 g" * (1/2)^(t/"63 years"))))`

This equation will allow you to find the mass of titanium-44 that remains undecayed after a given period of time `t` passes.

Notice that if `t` is a multiple of the half-life `t_"1/2"`, let's say

`t = 3 xx t_"1/2"`

you have

`color(red)(n) = (3 * color(red)(cancel(color(black)(t_"1/2"))))/color(red)(cancel(color(black)(t_"1/2"))) = color(red)(3)`

and

`A_t = "100 g" * (1/2)^color(red)(3)`

`A_t = "100 g" * 1/8 = "12.5 g"`

This means that after

`3 * "63 years" = "189 years"`

pass, the initial `"100-g"` sample of titanium-44 will be reduced to `"12.5 g"`.