^{18}_{8}O has a half-life of 4650 text{ year | vedclass

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(C) The formula for radioactive decay is $N_t = N_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Given $N_0 = 200 	ext{ mg}$,$N_

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$13950$

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(C) The formula for radioactive decay is $N_t = N_0 \left(\frac{1}{2}ight)^n$,where $n$ is the number of half-lives.Given $N_0 = 200 	ext{ mg}$,$N_t = 25 	ext{ mg}$,and $t_{1/2} = 4650 	ext{ years}$.Substituting the values: $25 = 200 	imes \left(\frac{1}{2}ight)^n$.$\frac{25}{200} = \left(\frac{1}{2}ight)^n \Rightarrow \frac{1}{8} = \left(\frac{1}{2}ight)^n$.Since $\frac{1}{8} = \left(\frac{1}{2}ight)^3$,we get $n = 3$.Total time $T = n 	imes t_{1/2} = 3 	imes 4650 	ext{ years} = 13950 	ext{ years}$.

$^{18}_{8}O$ has a half-life of $4650 \text{ years}$. $200 \text{ mg}$ of a sample of $^{18}_{8}O$ is reduced to $25 \text{ mg}$ in ......... $\text{years}$.

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