A piece of wood from a recently cut tree shows 20 decays per minute . A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute . If half life of `C^(14)` is 5730 years , then age of the wooden piece placed in the museum is approximately :

Options

A. 10439 years
B. 13094 years
C. 19039 years
D. 39049 years

Answer

To find the age of the wooden piece placed in the museum, we can use the decay formula and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Identify the initial and final decay rates
- The decay rate of the recently cut wood (n0) = 20 decays per minute.
- The decay rate of the museum wood (nt) = 2 decays per minute.

### Step 2: Use the decay formula
The relationship between the initial and final amounts in radioactive decay is given by:
\[ N_t = N_0 e^{-\lambda t} \]
Where:
- \(N_t\) = final quantity (decay rate in the museum)
- \(N_0\) = initial quantity (decay rate of the recently cut wood)
- \(\lambda\) = decay constant
- \(t\) = time elapsed

### Step 3: Rearranging the formula
We can rearrange the formula to solve for \(t\):
\[ \frac{N_t}{N_0} = e^{-\lambda t} \]
Taking the natural logarithm on both sides:
\[ \ln\left(\frac{N_t}{N_0}\right) = -\lambda t \]

### Step 4: Substitute the values
Substituting \(N_t = 2\) and \(N_0 = 20\):
\[ \ln\left(\frac{2}{20}\right) = -\lambda t \]
This simplifies to:
\[ \ln\left(\frac{1}{10}\right) = -\lambda t \]

### Step 5: Calculate the decay constant \(\lambda\)
The decay constant \(\lambda\) is related to the half-life (\(t_{1/2}\)) by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Given that the half-life of \(C^{14}\) is 5730 years:
\[ \lambda = \frac{0.693}{5730} \approx 1.21 \times 10^{-4} \text{ years}^{-1} \]

### Step 6: Substitute \(\lambda\) back into the equation
Now substituting \(\lambda\) back into the equation:
\[ \ln\left(\frac{1}{10}\right) = -\left(1.21 \times 10^{-4}\right) t \]
Calculating \(\ln\left(\frac{1}{10}\right)\):
\[ \ln\left(\frac{1}{10}\right) = -2.303 \]
Thus:
\[ -2.303 = -\left(1.21 \times 10^{-4}\right) t \]

### Step 7: Solve for \(t\)
Rearranging gives:
\[ t = \frac{2.303}{1.21 \times 10^{-4}} \approx 19043.3 \text{ years} \]

### Conclusion
The age of the wooden piece placed in the museum is approximately **19043 years**.

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