A piece of wood was found to have `""^(14)C//^(12)C` ratio 0.7 times that in a living plant. Calculate the period (in years) when the plant died. (`t_((1)/(2)) " for " C^(14)= 5760yr`)

Answer

To solve the problem of determining the time since the plant died based on the given ratio of carbon-14 to carbon-12, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The ratio of \(^{14}C\) to \(^{12}C\) in the wood is 0.7 times that in living plants.
- The half-life of \(^{14}C\) is given as \(t_{1/2} = 5760\) years.

2. **Define the Variables:**
- Let \(N_0\) be the initial amount of \(^{14}C\) in a living plant.
- Let \(N_t\) be the amount of \(^{14}C\) in the wood.
- According to the problem, \(N_t = 0.7 N_0\).

3. **Use the Decay Formula:**
The decay of a radioactive substance can be described by the formula:
\[ N_t = N_0 e^{-\lambda t} \]
where \(\lambda\) is the decay constant and \(t\) is the time elapsed.

4. **Calculate the Decay Constant (\(\lambda\)):**
The decay constant \(\lambda\) can be calculated using the half-life:
\[ \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \text{ years}^{-1} \]

5. **Rearranging the Decay Formula:**
Substitute \(N_t\) and \(\lambda\) into the decay formula:
\[ 0.7 N_0 = N_0 e^{-\lambda t} \]
Dividing both sides by \(N_0\) gives:
\[ 0.7 = e^{-\lambda t} \]

6. **Taking the Natural Logarithm:**
Taking the natural logarithm of both sides:
\[ \ln(0.7) = -\lambda t \]
Rearranging gives:
\[ t = -\frac{\ln(0.7)}{\lambda} \]

7. **Substituting \(\lambda\) into the Equation:**
Substitute \(\lambda\) into the equation:
\[ t = -\frac{\ln(0.7)}{\frac{0.693}{5760}} = -\frac{5760 \ln(0.7)}{0.693} \]

8. **Calculating \(\ln(0.7)\):**
Using a calculator, we find:
\[ \ln(0.7) \approx -0.3567 \]

9. **Final Calculation:**
Substitute \(\ln(0.7)\) back into the equation:
\[ t = -\frac{5760 \times (-0.3567)}{0.693} \approx \frac{5760 \times 0.3567}{0.693} \approx 2966 \text{ years} \]

### Final Answer:
The period when the plant died is approximately **2966 years**.