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NCERT Class XII Chapter
Nuclei
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Question : 8 of 31
Marks: +1, -0
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  614C\,{}^{14}_{6}\mathrm{C} present with the stable carbon isotope 612C.{}^{12}_{6}\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  614C,\,{}^{14}_{6}\mathrm{C}, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 614C^{14}_{6}\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 civilization.
Solution:  
In order to estimate age, let us first find the activity ratio in form of time ‘t’.
Given normal activity, R0R_0 = 15 decays min−1\mathrm{min}^{-1}
Present activity, R = 9 decays min−1\mathrm{min}^{-1},
T1/2T_{1/2} = 5730 years
Since activity is proportional to the number of radioactive atoms, therefore,
RR0\frac{R}{R_0} = NN0\frac{N}{N_0} = N0e−λtN0\frac{N_0 e^{-\lambda t}}{N_0} = e−λte^{-\lambda t} or 115\frac{1}{15} = e−λte^{-\lambda t} or eλte^{\lambda t} = 159\frac{15}{9}
Taking natural logarithm ,
log⁡eeλt\log_e e^{\lambda t} = log⁡e159\log_e \frac{15}{9} or λt log⁡e\log_e e = 2.303 log⁡1053\log_{10} \frac{5}{3} = 2.303 × 0.2218
or t = 0.5109λ\frac{0.5109}{\lambda} ... (i) [Since log⁡e\log_e e = 1]
Now we know, half life, T1/2T_{1/2} = 0.693λ\frac{0.693}{\lambda}
∴ t = 0.51090.693T1/2\frac{0.5109}{\frac{0.693}{T_{1/2}}} = 0.51090.693×T1/2\frac{0.5109}{0.693} \times T_{1/2} = 0.5109×57300.693\frac{0.5109 \times 5730}{0.693} years = 4224 years
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