The half-life of { }_{84}^{209} Po is 103 tex | vedclass

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(D) The initial amount of the sample is $N_0 = 100 	ext{ g}$.The final amount of the sample is $N = 3.125 	ext{ g}$.The half-life $T_{1/2} = 103 	e

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$515 	ext{ years}$

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(D) The initial amount of the sample is $N_0 = 100 	ext{ g}$.The final amount of the sample is $N = 3.125 	ext{ g}$.The half-life $T_{1/2} = 103 	ext{ years}$.Using the radioactive decay formula: $N = N_0 \left( \frac{1}{2} ight)^n$, where $n$ is the number of half-lives.$3.125 = 100 \left( \frac{1}{2} ight)^n$$\left( \frac{1}{2} ight)^n = \frac{3.125}{100} = \frac{1}{32}$Since $\frac{1}{32} = \left( \frac{1}{2} ight)^5$, we have $n = 5$.The total time taken is $t = n 	imes T_{1/2} = 5 	imes 103 	ext{ years} = 515 	ext{ years}$.

The half-life of ${ }_{84}^{209} Po$ is $103 \text{ years}$. The time it takes for a $100 \text{ g}$ sample of ${ }_{84}^{209} Po$ to decay to $3.125 \text{ g}$ is

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