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Standard XII

Biology

Palaeozoic Era

An archaeolog...

Question

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:

57000 years

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2850 years

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11,400 years

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22,800 years

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Solution

The correct option is C 11,400 years
As we know, Age = $\frac{2.303 t_{1 / 2}}{0.693} l o g (\frac{A_{0}}{A})$
$= \frac{2.303 \times 5770}{0.693} l o g (\frac{1}{0.25}) = 11, 400$ years

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An archaeologist analyses the wood in a prehistoric structure and finds that C14 (Half life = 5700 years) to C12 is only one-fourth of that found in the cells of buried plants.The age of the wood is about:

The correct option is C 11,400 yearsAs we know, Age = 2.303t1/20.693log(A0A)=2.303×57700.693log(10.25)=11,400 years

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:

The correct option is C 11,400 years
As we know, Age = $\frac{2.303 t_{1 / 2}}{0.693} l o g (\frac{A_{0}}{A})$
$= \frac{2.303 \times 5770}{0.693} l o g (\frac{1}{0.25}) = 11, 400$ years