A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If the half-life of C$$^{14}$$ is 5730 years, then the age of the wooden piece placed in the museum is approximately: [This question was awarded a bonus and proper correction was made to avoid that]

The activity of carbon-14 decays exponentially over time. The activity $$ A $$ at time $$ t $$ is given by $$ A = A_0 e^{-\lambda t} $$, where $$ A_0 $$ is the initial activity, $$ \lambda $$ is the decay constant, and $$ t $$ is the time elapsed.

For the recently cut tree, the activity is 20 decays per minute, which is $$ A_0 $$. For the wooden piece in the museum, the activity is 2 decays per minute, which is $$ A $$. The ratio of the activities is:

$$ \frac{A}{A_0} = \frac{2}{20} = \frac{1}{10} $$

Substituting into the decay formula:

$$ \frac{1}{10} = e^{-\lambda t} $$

Taking the natural logarithm on both sides:

$$ \ln\left(\frac{1}{10}\right) = \ln\left(e^{-\lambda t}\right) $$

$$ \ln(1) - \ln(10) = -\lambda t $$

$$ 0 - \ln(10) = -\lambda t $$

$$ -\ln(10) = -\lambda t $$

Multiplying both sides by -1:

$$ \ln(10) = \lambda t $$

Solving for $$ t $$:

$$ t = \frac{\ln(10)}{\lambda} $$

The decay constant $$ \lambda $$ is related to the half-life $$ T_{1/2} $$ by $$ \lambda = \frac{\ln(2)}{T_{1/2}} $$. Given $$ T_{1/2} = 5730 $$ years for carbon-14:

$$ \lambda = \frac{\ln(2)}{5730} $$

Substituting this into the expression for $$ t $$:

$$ t = \frac{\ln(10)}{\frac{\ln(2)}{5730}} = \ln(10) \times \frac{5730}{\ln(2)} $$

Using approximate values $$ \ln(10) \approx 2.303 $$ and $$ \ln(2) \approx 0.693 $$:

$$ t = 2.303 \times \frac{5730}{0.693} $$

First, compute $$ \frac{5730}{0.693} $$:

$$ \frac{5730}{0.693} \approx 8268.3988 $$

Now multiply:

$$ t = 2.303 \times 8268.3988 $$

$$ 2.303 \times 8000 = 18424 $$

$$ 2.303 \times 268.3988 = 2.303 \times 200 = 460.6 \quad ; \quad 2.303 \times 68.3988 \approx 2.303 \times 68 = 156.604 \quad ; \quad 2.303 \times 0.3988 \approx 0.918 $$

Adding these:

$$ 460.6 + 156.604 = 617.204 $$

$$ 617.204 + 0.918 = 618.122 $$

Now add to the previous result:

$$ 18424 + 618.122 = 19042.122 $$

Thus, $$ t \approx 19042 $$ years.

Hence, the correct answer is Option C.