The initial activity of a certain radioactive | vedclass

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(C) The activity of a radioactive sample follows the law $A = A_0 (1/2)^{t/T_{1/2}}$,where $A$ is the final activity,$A_0$ is the initial activity,$t$

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$4$

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(C) The activity of a radioactive sample follows the law $A = A_0 (1/2)^{t/T_{1/2}}$,where $A$ is the final activity,$A_0$ is the initial activity,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.Given $A_0 = 16000 \ counts \ min^{-1}$,$A = 2000 \ counts \ min^{-1}$,and $t = 12 \ h$.Substituting these values into the equation: $2000 = 16000 	imes (1/2)^{12/T_{1/2}}$.Dividing both sides by $16000$: $2000/16000 = (1/2)^{12/T_{1/2}}$.$1/8 = (1/2)^{12/T_{1/2}}$.Since $1/8 = (1/2)^3$,we have $(1/2)^3 = (1/2)^{12/T_{1/2}}$.Equating the exponents: $3 = 12/T_{1/2}$.Solving for $T_{1/2}$: $T_{1/2} = 12/3 = 4 \ h$.

The initial activity of a certain radioactive isotope was measured as $16000 \ counts \ min^{-1}$. Given that the only activity measured was due to this isotope and that its activity after $12 \ h$ was $2000 \ counts \ min^{-1}$,its half-life,in hours,is nearest to

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