Q37 / 120 · VITEEE 2010 · Chemistry
The activity of an old piece of wood is just $25 \%$ of the fresh piece of wood. If $t_{1 / 2}$ of $-14$ is 6000 yr, the age of piece of wood is
- A$6000 \mathrm{yr}$
- B$3000 \mathrm{yr}$
- C$9000 \mathrm{yr}$
- Correct answerD$12000 \mathrm{yr}$
Solution
As we know that,
$$
\begin{aligned}
k &=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6000}=1.155 \times 10^{-4} \\
t &=\frac{2.303}{k} \log \frac{N_{0}}{N}=\frac{2.303}{1.155 \times 10^{-4}} \log \frac{100}{25} \\
&=12000 \mathrm{yr}(\text { age of piece of wood })
\end{aligned}
$$
Hence, the correct answer is option: 'D'.