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Exercises · 13.19

Q.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}\text{C} present with the stable carbon isotope 612C^{12}_{6}\text{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}\text{C}, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 614C^{14}_{6}\text{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.

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Use A=A0e−λtA = A_0 e^{-\lambda t} with A0=15A_0 = 15/min/g (living tissue), A=9A = 9/min/g (specimen), and λ=ln⁡2/5730 yr\lambda = \ln 2 / 5730\ \text{yr}. Solving for tt gives an age of about 4223 years, consistent with the known era of the Indus Valley Civilisation.

The activity of a decaying sample follows:

A=A0 e−λtA = A_0\, e^{-\lambda t}

Given A0=15A_0 = 15 decays/min/g (the equilibrium activity while alive) and A=9A = 9 decays/min/g (measured today):

AA0=e−λt=915=0.6\frac{A}{A_0} = e^{-\lambda t} = \frac{9}{15} = 0.6

Taking the natural log:

−λt=ln⁡(0.6)=−0.51083-\lambda t = \ln(0.6) = -0.51083

λt=0.51083\lambda t = 0.51083

The decay constant from the given half-life:

λ=ln⁡2T1/2=0.6931475730 yr=1.2095×10−4 yr−1\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693147}{5730\ \text{yr}} = 1.2095 \times 10^{-4}\ \text{yr}^{-1}

Solving for tt: …

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