Uranium `._(92)U^(238)` decayed to `._(82)Pb^(206)`. They decay process is `._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206))`
`t_(1//2)` of `U^(238) = 4.5 xx 10^(9)` years
A sample of rock south America contains equal number of atoms of `U^(238)` and `Pb^(206)`. The age of rock will be

Options

A. `4.5 xx 10^(9)` years
B. `9 xx 10^(9)` years
C. `13.5 xx 10^(9)` years
D. `2.25 xx 10^(9)` years

Explanation

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

### Step 1: Understand the decay process
Uranium-238 decays to lead-206 through a series of alpha and beta decays. The question states that the rock contains equal numbers of uranium-238 and lead-206 atoms.

### Step 2: Set up the decay equation
Let \( x \) be the number of alpha decays and \( y \) be the number of beta decays. The decay process can be represented as:
\[ _{92}^{238}\text{U} \xrightarrow{(x \text{ alpha}, y \text{ beta})} _{82}^{206}\text{Pb} \]
The total number of decay events (alpha and beta) will lead to a decrease in the number of uranium atoms and an increase in lead atoms.

### Step 3: Relate the number of atoms
Since the rock contains equal numbers of uranium and lead:
\[ N_{\text{Pb}} = N_{\text{U}} + (x + y) \]
Given that \( N_{\text{Pb}} = N_{\text{U}} \), we can write:
\[ N_{\text{U}} + (x + y) = N_{\text{U}} \]
This implies that \( x + y = N_{\text{U}} \).

### Step 4: Use the half-life formula
The age of the rock can be calculated using the formula:
\[ T = \frac{t_{1/2}}{\ln(2)} \cdot \ln\left(1 + \frac{N_{\text{Pb}}}{N_{\text{U}}}\right) \]
Since \( N_{\text{Pb}} = N_{\text{U}} \), we have:
\[ T = \frac{t_{1/2}}{\ln(2)} \cdot \ln(1 + 1) = \frac{t_{1/2}}{\ln(2)} \cdot \ln(2) \]
This simplifies to:
\[ T = t_{1/2} \]

### Step 5: Substitute the half-life value
The half-life of uranium-238 is given as \( t_{1/2} = 4.5 \times 10^9 \) years. Therefore, the age of the rock is:
\[ T = 4.5 \times 10^9 \text{ years} \]

### Final Answer
The age of the rock is approximately \( 4.5 \times 10^9 \) years.

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