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(D) The activity $A$ of a radioactive sample at time $t$ is given by the formula $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A_0$

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$1/8$

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(D) The activity $A$ of a radioactive sample at time $t$ is given by the formula $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A_0$ is the initial activity and $T_{1/2}$ is the half-life.Given: $T_{1/2} = 3 \ hours$ and $t = 9 \ hours$.The number of half-lives elapsed is $n = \frac{t}{T_{1/2}} = \frac{9}{3} = 3$.Substituting these values into the formula:$\frac{A}{A_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.Thus,the activity becomes $1/8$ of its initial value.

If the half-life of a radioactive element is $3 \ hours$,what fraction of its initial activity remains after $9 \ hours$?

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