A sample of carbon from an ancient frame gives 7 counts of `""^(14)C` per minute per gram of carbon. If freshly cut wood gives 15.3 counts of `""^(14)C` per minute per gram, what is the age of the frame ? (Half-life period of `""^(14)C=5770` years)

Explanation

To find the age of the ancient frame using the given data about the counts of \(^{14}C\), we can use the radioactive decay formula. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have two counts of \(^{14}C\):
- Current count from the ancient frame: \(N_t = 7\) counts/minute/gram
- Initial count from freshly cut wood: \(N_0 = 15.3\) counts/minute/gram

We also know the half-life of \(^{14}C\) is \(T_{1/2} = 5770\) years.

### Step 2: Use the Decay Formula
The decay of a radioactive substance can be described by the equation:
\[ N_t = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \(N_t\) is the remaining quantity (current count)
- \(N_0\) is the initial quantity (initial count)
- \(t\) is the time elapsed (age of the frame)
- \(T_{1/2}\) is the half-life of the substance

### Step 3: Rearranging the Formula
We need to rearrange the formula to solve for \(t\):
\[ \frac{N_t}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{N_t}{N_0}\right) = \frac{t}{T_{1/2}} \ln\left(\frac{1}{2}\right) \]
Now, solving for \(t\):
\[ t = T_{1/2} \cdot \frac{\ln\left(\frac{N_t}{N_0}\right)}{\ln\left(\frac{1}{2}\right)} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ t = 5770 \cdot \frac{\ln\left(\frac{7}{15.3}\right)}{\ln\left(\frac{1}{2}\right)} \]

### Step 5: Calculate the Natural Logarithms
Calculate \( \frac{7}{15.3} \):
\[ \frac{7}{15.3} \approx 0.456 \]
Now, calculate the natural logarithm:
\[ \ln(0.456) \approx -0.784 \]
And for \( \ln\left(\frac{1}{2}\right) \):
\[ \ln\left(\frac{1}{2}\right) \approx -0.693 \]

### Step 6: Substitute Back to Find \(t\)
Now substituting back into the equation:
\[ t = 5770 \cdot \frac{-0.784}{-0.693} \approx 5770 \cdot 1.13 \approx 6521 \text{ years} \]

### Conclusion
The age of the ancient frame is approximately **6521 years**.

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