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

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13.Nuclei

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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 buried plants. The age of the wood is about ........$years$

A

$5700$

B

$2850 $

C

$11400$

D

$22800 $

Solution

(c) $\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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