Question

An ancient artifact retains 75% of its original carbon-14. Given $ T_{1/2} = 5730 \, \text{years} $, and $ \ln(0.5) = -0.7 $, $ \ln(0.75) = -0.3 $, find the age of the sample.

• A. 2300 years
• B. 2456 years
• C. 2546 years
• D. 3456 years

Hint:

Use \( t = \frac{-\ln(N/N_0)}{\lambda} \) and \( \lambda = \frac{0.693}{T_{1/2}} \) for radioactive decay problems.

Answer 1

The Correct Option is B

Decay law: \[ N = N_0 e^{-\lambda t} \Rightarrow \ln\left(\frac{N}{N_0}\right) = -\lambda t \Rightarrow t = \frac{-\ln(N/N_0)}{\lambda} \] \[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{5730} \] Now: \[ \ln(0.75) = -0.3 \Rightarrow t = \frac{0.3}{\lambda} = \frac{0.3 \cdot 5730}{0.693} \approx 2456\, \text{years} \]