# If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

Dec 18, 2016

$100 g$

#### Explanation:

half-life of caesium: $30.2$ years

$90.6 = 30.2 \cdot 3$

remaining mass after $90.6$ years = $12.5 g$
remaining mass after $3$ half-lives = $12.5 g$

1 half-life = original mass $\cdot {0.5}^{1}$

3 half-lives = original mass $\cdot {0.5}^{3}$

original mass $\cdot 0.125$ = $12.5 g$

divide by $0.125$:

original mass = $\left(\frac{12.5}{0.125}\right) g$

$= \frac{12500}{125} g$

$= 100 g$

Dec 18, 2016

$100 \cdot g \cdot C s$

#### Explanation:

I like to use this equation:

$M \left(t\right) = {M}_{0} \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{h}}$

Where the mass of the substance that remains is represented as a function of t, the time which has elapsed in years, and:

${M}_{0}$ is the initial amount of the substance in gram

$t$ is the amount of time which has elapsed

$h$ is the half life of the substance

And in our case (we are solving for the initial amount, ${M}_{0}$:

$12.5 = {M}_{0} \cdot {\left(\frac{1}{2}\right)}^{\frac{90.6}{30.2}}$

So

${M}_{0} = \frac{12.5}{{\left(\frac{1}{2}\right)}^{\frac{90.6}{30.2}}}$

and

${M}_{0} = \frac{12.5}{0.125}$

${M}_{0} = 100$