Question

The activity of an old piece of wood is just \(25%\) of the fresh piece of wood. If \(t_{1/2}\) of \(C\text{-}14\) is \(6000\,yr\), the age of piece of wood is

• A. \(6000\,yr\)
• B. \(3000\,yr\)
• C. \(9000\,yr\)
• D. \(12000\,yr\)

Hint:

If activity becomes \(25%\), that means two half-lives passed because \((1/2)^2 = 1/4\).

Answer 1

The Correct Option is D

Step 1: Use radioactive decay law.
Activity is proportional to number of atoms:
\[ \frac{A}{A_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}} \] Step 2: Given activity ratio.
\[ \frac{A}{A_0} = 0.25 = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \] Step 3: Compare powers.
So
\[ \left(\frac{1}{2}\right)^{t/t_{1/2}} = \left(\frac{1}{2}\right)^2 \Rightarrow \frac{t}{t_{1/2}} = 2 \] Step 4: Calculate time.
\[ t = 2 \times 6000 = 12000\,yr \] Final Answer: \[ \boxed{12000\,yr} \]