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

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

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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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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A
$3296 \text{ years}$
B
$103 \sqrt{2} \text{ years}$
C
$1648 \text{ years}$
D
$515 \text{ years}$

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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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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