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(C) The radioactive decay follows first-order kinetics.Given:Final activity $(A_t)$ = $0.01 \mu Ci$Time $(t)$ = $24 \ hours$Half-life $(t_{1/2})$ = $6

Vedclass · Accepted Answer

$0.16$

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(C) The radioactive decay follows first-order kinetics.Given:Final activity $(A_t)$ = $0.01 \mu Ci$Time $(t)$ = $24 \ hours$Half-life $(t_{1/2})$ = $6 \ hours$Number of half-lives $(n)$ = $t / t_{1/2} = 24 / 6 = 4$.The relation between initial activity $(A_0)$ and final activity $(A_t)$ is given by:$A_t = A_0 	imes (1/2)^n$$0.01 = A_0 	imes (1/2)^4$$0.01 = A_0 	imes (1/16)$$A_0 = 0.01 	imes 16 = 0.16 \mu Ci$.Therefore,the maximum initial activity that can be administered is $0.16 \mu Ci$.

$A$ radioactive substance is required to have an activity of $0.01 \mu Ci$ in the human body after $24 \ hours$. The half-life of the radioactive substance is $6 \ hours$. What is the maximum initial activity that can be administered (in $\mu Ci$)?

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