Question

If we start with 1 mg of strontium, 0.953 mg will remain after 2.0 years. What is the half life of strontium-90?

Answer 1

`t_"1/2" = "29 years"`

Explanation:

The nuclear half-life of a radioactive nuclide, `t_"1/2"`, is defined as the time needed for a sample of this nuclide to decay to half of its initial mass.

The first thing to notice here is that it took `2.0` years for only

`"1 mg " - " 0.953 mg" = "0.047 mg"`

of strontium-90 to decay. This tells you that the half-life of the nuclide is significantly longer than `2.0` years, since you need half of the original sample, the equivalent of `"0.50 mg"`, to decay in order to mark one half-life.

Your tool of choice here will be the equation

`color(blue)(|bar(ul(color(white)(a/a)A_"t" = A_0 * 1/2^ncolor(white)(a/a)|)))`

Here

`A_"t"` - the amount of the nuclide that remains undecayed after a period of time `t`
`A_0` - the initial mass of the nuclide
`n` - the number of half-lives that pass in the period of time `t`, calculated as

`color(purple)(|bar(ul(color(white)(a/a)color(black)(n = t/t_"1/2")color(white)(a/a)|)))`

Use this equation to find how many half-lives passed in `20` years

`A_"t" = A_0 * 1/2^n`

`0.953 color(red)(cancel(color(black)("mg"))) = 1color(red)(cancel(color(black)("mg"))) * 1/2^n`

`2^n = 1/0.953`

This will be equivalent to

`ln(2^n) = ln(1/0.953)`

`n * ln(2) = ln(1/0.953) implies n = ln(1/0.953)/ln(2)`

This will get you

`n = 0.06945`

This means that only `0.06945` of a half-life passed in `2.0` years. The half-life of the nuclide, `t_"1/2"`, will thus be

`n = "2.0 years"/0.06945 = color(green)(|bar(ul(color(white)(a/a)color(black)("29 years")color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the initial mass of the sample.