In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is found to be `0.5.` The age of the rock is (given half-life of `U^(238)` is `4.5 xx 10^(9)` years).

Options

A. `2.25 xx 10^(9)year`
B. `4.5 x 10^(9)` ln 3 year
C. `4.5 x 10^(9) (ln(3)/(2))/(ln2) year`
D. `2.25 xx 10^(9) ln((3)/(2))` year

Explanation

To find the age of the rock based on the ratio of lead-206 to uranium-238 nuclei, we can follow these steps:

### Step 1: Understand the Ratio
The ratio of lead-206 (Pb) to uranium-238 (U) is given as 0.5. This means that for every 1 nucleus of U, there are 0.5 nuclei of Pb.

### Step 2: Relate the Ratio to Decay
Since uranium decays into lead, we can express the relationship between the initial amount of uranium and the amount that has decayed into lead. Let:
- \( N_0 \) = initial number of uranium nuclei
- \( N \) = current number of uranium nuclei
- \( N_f \) = number of lead nuclei produced

From the decay process, we know:
\[ N_f = N_0 - N \]

### Step 3: Set Up the Ratio
Given the ratio:
\[ \frac{N_f}{N} = 0.5 \]
We can substitute \( N_f \):
\[ \frac{N_0 - N}{N} = 0.5 \]

### Step 4: Solve for \( N_0 \) and \( N \)
Rearranging the equation gives:
\[ N_0 - N = 0.5N \]
\[ N_0 = 1.5N \]

### Step 5: Relate to Exponential Decay
Using the exponential decay formula:
\[ N = N_0 e^{-\lambda t} \]
We can substitute \( N_0 \):
\[ N = 1.5N e^{-\lambda t} \]
Dividing both sides by \( N \):
\[ 1 = 1.5 e^{-\lambda t} \]
Thus:
\[ e^{-\lambda t} = \frac{1}{1.5} = \frac{2}{3} \]

### Step 6: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ -\lambda t = \ln\left(\frac{2}{3}\right) \]
So:
\[ t = -\frac{\ln\left(\frac{2}{3}\right)}{\lambda} \]

### Step 7: Find \( \lambda \)
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \):
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given the half-life of \( U^{238} \) is \( 4.5 \times 10^9 \) years:
\[ \lambda = \frac{\ln(2)}{4.5 \times 10^9} \]

### Step 8: Substitute \( \lambda \) Back
Substituting \( \lambda \) back into the equation for \( t \):
\[ t = -\frac{\ln\left(\frac{2}{3}\right)}{\frac{\ln(2)}{4.5 \times 10^9}} \]
This simplifies to:
\[ t = 4.5 \times 10^9 \cdot \frac{\ln\left(\frac{2}{3}\right)}{\ln(2)} \]

### Step 9: Calculate the Age of the Rock
Using the values of logarithms:
\[ t \approx 4.5 \times 10^9 \cdot \frac{\ln(2/3)}{\ln(2)} \]

### Final Answer
Calculating this gives the age of the rock.