What is the half-life (in years) of a radioac | vedclass

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(C) The activity of a radioactive sample is given by $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.Given that the activity drops to $1/16$ o

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$7$

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(C) The activity of a radioactive sample is given by $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.Given that the activity drops to $1/16$ of its initial value,we have $A/A_0 = 1/16$.Since $1/16 = (1/2)^4$,we can equate the powers: $n = 4$.The number of half-lives $n$ is related to the total time $t$ and the half-life $T_{1/2}$ by the formula $n = t / T_{1/2}$.Given $t = 30$ years and $n = 4$,we have $4 = 30 / T_{1/2}$.Therefore,$T_{1/2} = 30 / 4 = 7.5$ years.

What is the half-life (in years) of a radioactive material if its activity drops to $1/16$th of its initial value in $30$ years (in $.5$)?

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