The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon.

Question:

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

Decay rate of living carbon-containing matter, R = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of ${ }_{6}^{14} \mathrm{C}, T_{1 / 2}=5730$ years

The decay rate of the specimen obtained from the Mohenjodaro site:

R' = 9 decays/min

Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant, λand time, t as:

$\frac{N}{N^{\prime}}=\frac{R}{R^{\prime}}=\mathrm{e}^{-\lambda x}$

$\mathrm{e}^{-\lambda t}=\frac{9}{15}=\frac{3}{5}$

$-\lambda t=\log _{e} \frac{3}{5}=-0.5108$

$\therefore t=\frac{0.5108}{\lambda}$

But $\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{5730}$

$\therefore t=\frac{0.5108}{\frac{0.693}{5730}}=4223.5$ years

Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.