An archaeologist analyses the wood in a prehi | vedclass

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(C) The ratio of radioactive isotope ${C^{14}}$ to stable isotope ${C^{12}}$ decreases over time due to radioactive decay.Given that the current ratio

Vedclass · Accepted Answer

$11400$

Vedclass · Answer

(C) The ratio of radioactive isotope ${C^{14}}$ to stable isotope ${C^{12}}$ decreases over time due to radioactive decay.Given that the current ratio is $\frac{1}{4}$ of the initial ratio,we use the radioactive decay formula: $N(t) = N_0 \left( \frac{1}{2} ight)^{\frac{t}{T_{1/2}}}$.Here,$\frac{N(t)}{N_0} = \frac{1}{4}$ and $T_{1/2} = 5700 \, years$.Substituting the values: $\frac{1}{4} = \left( \frac{1}{2} ight)^{\frac{t}{5700}}$.Since $\frac{1}{4} = \left( \frac{1}{2} ight)^2$,we equate the exponents: $2 = \frac{t}{5700}$.Therefore,$t = 2 	imes 5700 = 11400 \, years$.

An archaeologist analyses the wood in a prehistoric structure and finds that the ratio of ${C^{14}}$ (half-life $= 5700 \, years$) to ${C^{12}}$ is only one-fourth of that found in the cells of living plants. The age of the wood is about ........ $years$.

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