Carbon-14 decays with a half-life of about 5, | vedclass

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(A) The radioactive decay law states that the amount of substance remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$.Given that the r

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$2 	imes 58$

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(A) The radioactive decay law states that the amount of substance remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$.Given that the ratio of Carbon-$14$ to Carbon-$12$ is $\frac{1}{4}$ of the original amount,we have $\frac{N(t)}{N_0} = \frac{1}{4}$.Since $\frac{1}{4} = (1/2)^2$,it implies that the time elapsed $t$ is equal to two half-lives.$t = 2 	imes T_{1/2} = 2 	imes 5,800$ years.Since $1$ century $= 100$ years,$t = 2 	imes 58$ centuries.Therefore,$x = 2 	imes 58$.

Carbon-$14$ decays with a half-life of about $5,800$ years. In a sample of bone,the ratio of Carbon-$14$ to Carbon-$12$ is found to be $\frac{1}{4}$ of what it is in free air. This bone may belong to a period about $x$ centuries ago,where $x$ is nearest to:

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