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 ${}_{6}^{14}C$ ${}_{6}^{14}C$ present with the stable carbon istope ${}_{6}^{12}C .$ 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 ${}_{6}^{14}C,$ and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${}_{6}^{14}C$ 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.

Solution

Given,
Normal activity, R₀ = 15 decays/min,
Present activity R = 9 decays/min,
Half life, T = 5730 years,
Age t =?

As, activity is proportional to the number of radioactive atoms, therefore,

\frac{N}{N_{0}} = \frac{R}{R_{0}} = \frac{9}{15}

But,

\frac{N}{N_{0}} = e^{- λt}

∴

e^{- λt} = \frac{9}{15} = \frac{3}{5}

e^{+ λt} = \frac{5}{3}

\Rightarrow

λt \log_{e} e = \log_{e} \frac{5}{3} = 2.3026 \log 1.6667

\Rightarrow

λt = 2.3026 \times 0.2218 = 0.5109 ∴ t = \frac{0.5109}{λ}

But,

λ = \frac{0.693}{T} = \frac{0.693}{5730} {Yr}^{- 1}

Therefore,

t = \frac{0.5109}{0.693 / 5730} = \frac{0.5109 \times 5730}{0.693}

t = 4224.3 years .

is the approximate age.

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