The half-life of C^{14} is 5760 years. For a | vedclass

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(D) Given: Half-life of $C^{14}$, $t_{1/2} = 5760 \ years$. Initial amount, $N_0 = 200 \ mg$. Final amount, $N_t = 25 \ mg$. Radioactive decay follows

Vedclass · Accepted Answer

$17280$

Vedclass · Answer

(D) Given: Half-life of $C^{14}$, $t_{1/2} = 5760 \ years$. Initial amount, $N_0 = 200 \ mg$. Final amount, $N_t = 25 \ mg$. Radioactive decay follows first-order kinetics. We can use the relation: $N_t = N_0 	imes (1/2)^n$, where $n$ is the number of half-lives. $25 = 200 	imes (1/2)^n$ $1/8 = (1/2)^n$ $(1/2)^3 = (1/2)^n$ Therefore, $n = 3$. Total time $t = n 	imes t_{1/2} = 3 	imes 5760 \ years = 17280 \ years$. Alternatively, as shown in the diagram: $200 \ mg$ $\xrightarrow{t_{1/2}} 100 \ mg$ $\xrightarrow{t_{1/2}} 50 \ mg$ $\xrightarrow{t_{1/2}} 25 \ mg$. Total time = $3 	imes 5760 \ years = 17280 \ years$.

The half-life of $C^{14}$ is $5760 \ years$. For a $200 \ mg$ sample of $C^{14}$, the time taken to change to $25 \ mg$ is (in $years$)

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