The `._1^3H` isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 12.33 yr.It can be used to measure the age of objects up to about 100 yr. It is produced in the upper atmosphere by cosmic rays and brought to Earth by rain. As an application , determine approximately the age of a bottle of wine whose `._1^3H` radiation is about `1/10` that present in new wine .

Answer

To determine the age of a bottle of wine based on its tritium content, we can use the concept of radioactive decay and the half-life of tritium. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- The half-life of tritium (\(T_{1/2}\)) is 12.33 years.
- The current activity (\(A\)) of the wine is \( \frac{1}{10} \) of the activity of new wine (\(A_0\)).

### Step 2: Use the Decay Formula
The radioactive decay can be described by the formula:
\[ A = A_0 e^{-\lambda t} \]
Where:
- \(A\) is the current activity,
- \(A_0\) is the initial activity,
- \(\lambda\) is the decay constant,
- \(t\) is the time elapsed.

### Step 3: Relate the Decay Constant to Half-Life
The decay constant (\(\lambda\)) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Substituting the half-life of tritium:
\[ \lambda = \frac{\ln(2)}{12.33 \text{ years}} \]

### Step 4: Express the Current Activity in Terms of Initial Activity
Given that the current activity is \( \frac{1}{10} \) of the initial activity:
\[ A = \frac{A_0}{10} \]

### Step 5: Substitute into the Decay Formula
Substituting \(A\) into the decay formula gives:
\[ \frac{A_0}{10} = A_0 e^{-\lambda t} \]
Dividing both sides by \(A_0\) (assuming \(A_0 \neq 0\)):
\[ \frac{1}{10} = e^{-\lambda t} \]

### Step 6: Take the Natural Logarithm of Both Sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{10}\right) = -\lambda t \]

### Step 7: Substitute \(\lambda\) and Solve for \(t\)
Substituting \(\lambda = \frac{\ln(2)}{12.33}\):
\[ \ln\left(\frac{1}{10}\right) = -\left(\frac{\ln(2)}{12.33}\right) t \]
Now, solving for \(t\):
\[ t = -\frac{12.33 \cdot \ln\left(\frac{1}{10}\right)}{\ln(2)} \]

### Step 8: Calculate the Values
Calculating \(\ln\left(\frac{1}{10}\right)\):
\[ \ln\left(\frac{1}{10}\right) = -\ln(10) \approx -2.3026 \]
Now substituting this value:
\[ t = -\frac{12.33 \cdot (-2.3026)}{\ln(2)} \]
Calculating \(\ln(2) \approx 0.6931\):
\[ t \approx \frac{12.33 \cdot 2.3026}{0.6931} \]
Calculating the final value:
\[ t \approx \frac{28.426}{0.6931} \approx 41 years \]

### Conclusion
The approximate age of the bottle of wine is **41 years**.