The half-life of a radioactive sample,where t | vedclass

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(B) The activity of a radioactive sample follows the law $A = A_0 \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.Given,initial ac

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(B) The activity of a radioactive sample follows the law $A = A_0 \left( \frac{1}{2} ight)^n$,where $n$ is the number of half-lives.Given,initial activity $A_0 = 8 	ext{ counts}$ and final activity $A = 1 	ext{ count}$.Substituting the values: $1 = 8 \left( \frac{1}{2} ight)^n$.This simplifies to $\left( \frac{1}{2} ight)^n = \frac{1}{8}$.Since $\frac{1}{8} = \left( \frac{1}{2} ight)^3$,we have $n = 3$.The total time $t = 3 	ext{ hours}$.The number of half-lives is given by $n = \frac{t}{T_{1/2}}$,where $T_{1/2}$ is the half-life.Therefore,$3 = \frac{3}{T_{1/2}}$,which gives $T_{1/2} = 1 	ext{ hour}$.

The half-life of a radioactive sample,where the initial activity of the material was $8 \text{ counts}$ and after $3 \text{ hours}$ it becomes $1 \text{ count}$,is ............... $hours$.

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