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Q3.14:
The half-life for radioactive decay of (^{14}C) is 5730 years. An archaeological artifact containing wood had only 80% of the (^{14}C) found in a living tree. Estimate the age of the sample.

Solution :

Step 1: Establishing the Kinetic Framework

Radioactive decay follows first-order kinetics. The rate constant ($k$) for a first-order reaction is related to the half-life ($t_{1/2}$) by the expression derived from the integrated rate law [Per the First-Order Integrated Rate Law]:

$k = \frac{0.693}{t_{1/2}}$

Given $t_{1/2} = 5730 \text{ years}$, we calculate $k$:

$k = \frac{0.693}{5730 \text{ years}} \approx 1.209 \times 10^{-4} \text{ year}^{-1}$

Step 2: Defining the Concentration Parameters

The integrated rate law for first-order reactions is expressed as:

$t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$

Where:

  • $[A]_0$ = Initial concentration of $^{14}C$ in a living tree (represented as $100\%$).
  • $[A]_t$ = Concentration of $^{14}C$ at time $t$ (given as $80\%$ of the initial).
  • $t$ = Age of the artifact.

Step 3: Calculating the Age of the Sample

Substituting the values into the rate equation:

$t = \frac{2.303}{1.209 \times 10^{-4} \text{ year}^{-1}} \log \left( \frac{100}{80} \right)$

$t = \frac{2.303}{1.209 \times 10^{-4}} \log (1.25)$

[Using logarithmic values: $\log(1.25) \approx 0.0969$]

$t = \frac{2.303}{1.209 \times 10^{-4}} \times 0.0969$

$t \approx 1.9048 \times 10^4 \times 0.0969$

$t \approx 1845 \text{ years}$

Graphical Representation of Decay

100% 0 Time (Years) Decay curve of 14C 1845 yrs 80%

Final Solution: The estimated age of the archaeological artifact is approximately 1845 years.


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