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

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Answer 1 · 2016-12-18T18:22:02

$100 g$ $100g$

half-life of caesium: $30.2$ $30.2$ years

$90.6 = 30.2 \cdot 3$ $90.6 = 30.2 * 3$

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

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

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

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

divide by $0.125$ $0.125$ :

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

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

$= 100 g$ $=100g$

Answer 2 · 2016-12-18T18:31:47

$100 \cdot g \cdot C s$ $100 * g * Cs$

I like to use this equation:

$M (t) = M_{0} \cdot {(\frac{1}{2})}^{\frac{t}{h}}$ $M(t)=M_0*(1/2)^(t/h)$

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

$M_{0}$ $M_0$ is the initial amount of the substance in gram

$t$ $t$ is the amount of time which has elapsed

$h$ $h$ is the half life of the substance

And in our case (we are solving for the initial amount, $M_{0}$ $M_0$ :

$12.5 = M_{0} \cdot {(\frac{1}{2})}^{\frac{90.6}{30.2}}$ $12.5=M_0*(1/2)^(90.6/30.2)$

So

$M_{0} = \frac{12.5}{{(\frac{1}{2})}^{\frac{90.6}{30.2}}}$ $M_0=12.5/((1/2)^(90.6/30.2))$

and

$M_{0} = \frac{12.5}{0.125}$ $M_0=12.5/0.125$

$M_{0} = 100$ $M_0=100$