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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: Identifying Parameters and Constants

Radioactive decay follows first-order kinetics [Per the Integrated Rate Law for first-order reactions].

  • Half-life ($t_{1/2}$) = $5730 \text{ years}$
  • Remaining concentration fraction ($\frac{[A]_t}{[A]_0}$) = $80\% = 0.80$
  • Decay constant ($k$): Calculated via the relation $k = \frac{\ln(2)}{t_{1/2}}$

Step 2: Calculating the Decay Constant ($k$)

Using the definition of half-life for a first-order process:

$k = \frac{0.693}{5730 \text{ years}}$

$k \approx 1.2094 \times 10^{-4} \text{ year}^{-1}$ [Mathematical evaluation of $\frac{0.693}{5730}$]

Step 3: Applying the First-Order Integrated Rate Equation

The age of the sample ($t$) is determined by the formula [Derived from the Integrated Rate Law]:

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

Substituting the known values:

$t = \frac{2.303}{1.2094 \times 10^{-4}} \log\left(\frac{1}{0.80}\right)$

Step 4: Algebraic Computation

First, calculate the logarithmic component:

$\log(1.25) \approx 0.0969$ [Logarithmic property: $\log(1) - \log(0.8)$]

Now, solve for $t$:

$t = (1.9042 \times 10^4) \times 0.0969$

$t \approx 1845 \text{ years}$

Visual Representation of Decay

[A] t Decay Curve: N(t) = N₀e⁻ᵏᵗ
Parameter Value
Decay Constant ($k$) $1.2094 \times 10^{-4} \text{ yr}^{-1}$
Ratio ($[A]_0 / [A]_t$) $1.25$
Calculated Age ($t$) $\approx 1845 \text{ years}$

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


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