A bone fragment found in a cave contains $0.21$ times as much $_{6}^{14} C$ as in equal amount of carbon in air when the organism containing bone died. Find the approximate age of fragment $t_{1 / 2}$ of $_{6}^{14} C$ = $5730$ years

1.15 \times 10^{4}

y

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1.3 \times 10^{4}

y

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1.24 \times 10^{4}

y

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1.4 \times 10^{4}

y

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Solution

The correct option is B $1.3 \times 10^{4}$ $y$
$λ$ = $\frac{0.693}{t_{1 / 2}}$ = $\frac{0.693}{5730}$ = $1.209 \times 10^{- 4} y^{- 1}$
$t$ = $\frac{2.303 (.6794)}{1.209 \times 10^{- 4}}$ = $1.3 \times 10^{- 4}$ $y$

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Q. A bone fragment found in a cave contains 0.2 times as much

_{6}^{14}

C as an equal amount of carbon in the air when the organism containing bone died.The approximate age of fragment is (t

\frac{1}{2}

_{6}^{14}

C = 5730 years )

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive present with the stable carbon isotope . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.