Question

How many half-lives have elapsed when 25% of the parent nuclide is left?

Answer 1

`"2 half-lives"`

Explanation:

The thing to remember about a radioactive nuclide's nuclear half-life, `t_"1/2"`, is that it represents the time needed for half, hence the term half-life, of an initial sample of said nuclide to undergo radioactive decay.

In other words, the mass of a radioactive nuclide, regardless of its initial value, will always be halved after `1` half-life passes.

So, if you start with `A_0`, you can say that you will be left with

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

and so on.

In your case, you know that

`25% = 25/100 = 1/4`

of the initial sample is left after a time `t` passes, which means that you will have

`A_t = A_0 * 1/4 = A_0/4 = A_0/2^color(red)(2)`

As you can see, this is exactly what you would expect to get after `color(red)(2)` half-lives pass, so

`t = color(red)(2) * t_"1/2"`

and

`A_ (color(red)(2) xx t_"1/2") = A_0/2^color(red)(2) ->` the initial sample is down to `25%` of its initial value after `color(red)(2)` half-lives