# 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?

Jun 26, 2016

${t}_{\text{1/2" = "29 years}}$

#### Explanation:

The nuclear half-life of a radioactive nuclide, ${t}_{\text{1/2}}$, 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

$\text{1 mg " - " 0.953 mg" = "0.047 mg}$

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 $\text{0.50 mg}$, to decay in order to mark one half-life.

Your tool of choice here will be the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {A}_{\text{t}} = {A}_{0} \cdot \frac{1}{2} ^ n \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${A}_{\text{t}}$ - 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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = \frac{t}{t} _ \text{1/2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

${A}_{\text{t}} = {A}_{0} \cdot \frac{1}{2} ^ n$

$0.953 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mg"))) = 1color(red)(cancel(color(black)("mg}}}} \cdot \frac{1}{2} ^ n$

${2}^{n} = \frac{1}{0.953}$

This will be equivalent to

$\ln \left({2}^{n}\right) = \ln \left(\frac{1}{0.953}\right)$

$n \cdot \ln \left(2\right) = \ln \left(\frac{1}{0.953}\right) \implies n = \ln \frac{\frac{1}{0.953}}{\ln} \left(2\right)$

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, ${t}_{\text{1/2}}$, 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.