In a sample of rock, the ratio of number of `""^(206)Pb` to `""^(238)U` nuclei is found to be 0.5. The age of the rock is `(18//n)xx10^(9)(ln((3)/(2)))/(ln2)` year (Assume that all the Pb nuclides in he rock was produced due to the decay of Uranium nuclides and `T_(1//2)(""^(238)U)=4.5xx10^(9)` year). Find n.

Answer

To solve the problem, we need to find the value of \( n \) in the given age of the rock formula. The ratio of the number of \( ^{206}Pb \) to \( ^{238}U \) nuclei is given as 0.5, and the half-life of \( ^{238}U \) is \( 4.5 \times 10^9 \) years.

### Step-by-step Solution:

1. **Understanding the Ratio**:
The ratio of the number of \( ^{206}Pb \) to \( ^{238}U \) is given as:
\[ \frac{N_{Pb}}{N_{U}} = 0.5 \]
This implies that for every 1 nucleus of \( ^{238}U \), there are 0.5 nuclei of \( ^{206}Pb \).

2. **Relating \( N_{Pb} \) and \( N_{U} \)**:
If we let \( N_{U} = N_0 \) (the initial number of \( ^{238}U \) nuclei), then:
\[ N_{Pb} = 0.5 N_{U} = 0.5 N_0 \]

3. **Decay of Uranium**:
The number of \( ^{238}U \) nuclei remaining after time \( t \) can be expressed using the decay formula:
\[ N_{U} = N_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

4. **Finding the Decay Constant**:
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Substituting the half-life of \( ^{238}U \):
\[ \lambda = \frac{\ln(2)}{4.5 \times 10^9} \]

5. **Setting Up the Equation**:
From the ratio, we can set up the equation:
\[ 0.5 N_0 = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ 0.5 = e^{-\lambda t} \]

6. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln(0.5) = -\lambda t \]
Rearranging gives:
\[ t = -\frac{\ln(0.5)}{\lambda} \]

7. **Substituting for \( \lambda \)**:
Substitute \( \lambda \):
\[ t = -\frac{\ln(0.5)}{\frac{\ln(2)}{4.5 \times 10^9}} = \frac{4.5 \times 10^9 \ln(0.5)}{\ln(2)} \]

8. **Calculating \( t \)**:
We know that \( \ln(0.5) = -\ln(2) \), thus:
\[ t = \frac{4.5 \times 10^9 (-\ln(2))}{\ln(2)} = 4.5 \times 10^9 \times 1 = 4.5 \times 10^9 \text{ years} \]

9. **Finding \( n \)**:
The age of the rock is given as:
\[ \frac{18}{n} \times 10^9 \frac{\ln(3/2)}{\ln(2)} \]
Setting this equal to our calculated \( t \):
\[ \frac{18}{n} \times 10^9 \frac{\ln(3/2)}{\ln(2)} = 4.5 \times 10^9 \]
Dividing both sides by \( 10^9 \):
\[ \frac{18}{n} \frac{\ln(3/2)}{\ln(2)} = 4.5 \]
Rearranging gives:
\[ n = \frac{18 \ln(3/2)}{4.5 \ln(2)} \]

10. **Calculating \( n \)**:
Simplifying:
\[ n = 4 \frac{\ln(3/2)}{\ln(2)} \]

### Final Answer:
To find the numerical value of \( n \), we can compute:
\[ n \approx 4 \times \frac{0.4055}{0.6931} \approx 4 \times 0.5849 \approx 2.3396 \approx 2.34 \]
Thus, \( n \) is approximately \( 2.34 \).