A piece of bone of an animal from a ruin is found to have $$^{14}$$C activity of 12 disintegrations per minute per gm of its carbon content. The $$^{14}$$C activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die? (Given half life of $$^{14}$$C is $$t_{1/2}$$ = 5760 years)

The activity of radioactive carbon-14 decreases over time due to decay. The initial activity when the animal was alive is given as $$A_0 = 16$$ disintegrations per minute per gram, and the current activity of the bone is $$A = 12$$ disintegrations per minute per gram. The half-life of carbon-14 is $$t_{1/2} = 5760$$ years.

The decay follows the exponential law: $$A = A_0 e^{-\lambda t}$$, where $$\lambda$$ is the decay constant and $$t$$ is the time elapsed since death. Substituting the given values:

$$12 = 16 e^{-\lambda t}$$

Divide both sides by 16:

$$\frac{12}{16} = e^{-\lambda t}$$

Simplify the fraction:

$$\frac{3}{4} = e^{-\lambda t}$$

Take the natural logarithm of both sides:

$$\ln\left(\frac{3}{4}\right) = \ln\left(e^{-\lambda t}\right)$$

The right side simplifies to $$-\lambda t$$:

$$\ln\left(\frac{3}{4}\right) = -\lambda t$$

Multiply both sides by -1:

$$-\ln\left(\frac{3}{4}\right) = \lambda t$$

Using the property $$-\ln(a) = \ln(1/a)$$, rewrite as:

$$\ln\left(\frac{4}{3}\right) = \lambda t$$

Therefore:

$$t = \frac{\ln\left(\frac{4}{3}\right)}{\lambda}$$

Now, find the decay constant $$\lambda$$ using the half-life formula $$t_{1/2} = \frac{\ln(2)}{\lambda}$$. Solving for $$\lambda$$:

$$\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{5760}$$

Substitute this into the equation for $$t$$:

$$t = \frac{\ln\left(\frac{4}{3}\right)}{\frac{\ln(2)}{5760}} = \ln\left(\frac{4}{3}\right) \times \frac{5760}{\ln(2)}$$

So:

$$t = 5760 \times \frac{\ln\left(\frac{4}{3}\right)}{\ln(2)}$$

Compute the numerical value. First, calculate $$\ln(4/3)$$:

$$\frac{4}{3} \approx 1.3333, \quad \ln(1.3333) \approx 0.287682$$

Next, $$\ln(2) \approx 0.693147$$:

$$\frac{\ln(4/3)}{\ln(2)} = \frac{0.287682}{0.693147} \approx 0.415037$$

Now multiply by 5760:

$$t = 5760 \times 0.415037 \approx 2390.613$$

Rounding to the nearest whole number gives approximately 2391 years.

Comparing with the options, 2391 years corresponds to option B.

Hence, the correct answer is Option B.