Q.25 An archaeological sample (remains of an animal) containing 14C isotope of Carbon is found to give 10 beta decays per minute per gram of Carbon. It is known that the natural abundance of 14C in organic matter that is in equilibrium with the atmosphere today will give 15 beta decays per minute per gram of Carbon. The half life of 14C is known to be 5730 years. The estimated age of the sample (in years) is
(A) 3010
(B) 3350
(C) 3500
(D) 3800
The correct estimated age of the archaeological sample is about 3350 years, so the right option is (B) 3350.
Detailed Solution
The question is a classic radiocarbon dating (carbon-14 dating) numerical problem.
Key Data
- Present-day activity of living organic matter: \( A_0 = 15 \) beta decays per minute per gram of carbon.
- Activity of the archaeological sample: \( A = 10 \) beta decays per minute per gram of carbon.
- Half-life of \( ^{14}\mathrm{C} \): \( T_{1/2} = 5730 \) years.
[web:9]
Decay Law
Radiocarbon dating uses the exponential decay law:
\[ A = A_0 \left( \frac{1}{2} \right)^{t / T_{1/2}} \]
Rearrange to solve for age \( t \):
\[ \frac{A}{A_0} = \left( \frac{1}{2} \right)^{t / T_{1/2}} \]
\[ t = T_{1/2} \cdot \frac{ \ln(A/A_0) }{ \ln(1/2) } \]
Calculation
Substitute values (\( A_0 = 15 \), \( A = 10 \)):
Activity ratio: \( \frac{A}{A_0} = \frac{10}{15} = \frac{2}{3} \).
\[ t = 5730 \cdot \frac{ \ln(2/3) }{ \ln(1/2) } \]
Numerical evaluation:
- \( \ln(2/3) \approx -0.4055 \)
- \( \ln(1/2) \approx -0.6931 \)
- \( t \approx 5730 \cdot 0.5847 \approx 3350 \) years
So, the estimated age ≈ 3350 years, which matches option (B).
Explanation of Each Option
| Option | Value (years) | Comment |
|---|---|---|
| (A) | 3010 | Too low. Would correspond to a smaller activity drop than 15 → 10 decays/min; decay has not been enough for 3010 years to produce the observed ratio 10/15. |
| (B) | 3350 | Correct. Matches the age computed from the exponential decay law using half-life 5730 years and activity ratio 2/3. |
| (C) | 3500 | Slightly high. Close but still implies more decay than observed (activity would be lower than 10 decays/min if age were 3500 years). |
| (D) | 3800 | Too high. Requires much greater reduction in activity than 10/15; sample would show significantly fewer beta decays per minute than measured. |
Radiocarbon dating problems are frequent in competitive exams, and understanding how to apply the decay law of carbon‑14 is crucial for scoring well. This detailed carbon 14 dating numerical example shows how to use measured beta decay rates and the known half‑life of \( ^{14}\mathrm{C} \) (5730 years) to estimate the age of an archaeological sample and correctly identify the right option in a multiple‑choice question.
