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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 $_{6}^{14} C$ present with the stable carbon isotope $_{6}^{12} C$. 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 (in $years$) of the Indus-Valley civilisation
$3842.6$
$5263.4$
$4223.5$
$4826.5$
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 $_{6}^{14} C, T_{1/2}=5730$ years
The decay rate of the specimen obtained from the Mohenjodaro site
$R ^{\prime}=9$ decays/min Let $N ^{\prime}$ be the number of radioactive atoms present in the specimen during the Mohenjodaroperiod.
Therefore, we can relate the decay constant, $\lambda$ and time, $t$ as
$\frac{N^{\prime}}{N}=\frac{R^{\prime}}{R}=e^{-\lambda t}$
$e^{-\lambda t}=\frac{9}{15}=\frac{3}{5}$
$-\lambda t=\log _{e} \frac{3}{5}=-0.5108$
$\therefore t=\frac{0.5108}{2}$
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 civilization $4223.5$ years.
