An archaeologist analyses wood in a prehistoric structure and finds that C^{14} (Half life = 5700, y

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13.Nuclei

easy

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

A

$5700$

B

$2850$

C

$11400$

D

$22800$

(AIIMS-2013)

Solution

$\frac{C_{14}}{C_{12}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{t / 5700}$

$\Rightarrow \frac{t}{5700}=2 \Rightarrow t=11400$ $years$

Standard 12

Physics

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