A rock is found to contain U-238 and Pb-206 in the ratio of 3 : 2. If `t""_(1//2)` of U-238 is `4.5xx10""^(-9)` years. Calculate the age rock?

Options

A. `3.71xx10""^(-9)years`
B. `3.71xx10""^(9)years`
C. `31.7xx10""^(9)years`
D. `3.71xx10""^(9)days`

Explanation

To calculate the age of the rock containing U-238 and Pb-206 in the ratio of 3:2, we can follow these steps:

### Step 1: Understand the Ratio
Given the ratio of U-238 to Pb-206 is 3:2, we can express this in terms of the number of atoms. Let the initial number of U-238 atoms be \( N_0 \) and the number of Pb-206 atoms be \( N_{Pb} \).

From the ratio:
\[ \frac{N_0}{N_{Pb}} = \frac{3}{2} \]
This implies that for every 3 atoms of U-238, there are 2 atoms of Pb-206.

### Step 2: Relate U-238 and Pb-206
Since U-238 decays to Pb-206, the number of Pb-206 atoms formed is equal to the number of U-238 atoms that have decayed. If \( N \) is the number of U-238 atoms that have decayed, then:
\[ N_{Pb} = N - N_0 \]
Where \( N \) is the total number of U-238 atoms initially present.

### Step 3: Set Up the Equation
From the ratio, we can express \( N_{Pb} \) in terms of \( N_0 \):
\[ N_{Pb} = \frac{2}{3} N_0 \]
Substituting into the decay equation:
\[ N - N_0 = \frac{2}{3} N_0 \]
This leads us to:
\[ N = N_0 + \frac{2}{3} N_0 = \frac{5}{3} N_0 \]

### Step 4: Use the Decay Formula
The decay of U-238 can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
Where \( \lambda \) is the decay constant, which can be calculated from the half-life \( t_{1/2} \):
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Given \( t_{1/2} = 4.5 \times 10^{-9} \) years, we can calculate \( \lambda \):
\[ \lambda = \frac{0.693}{4.5 \times 10^{-9}} \approx 1.54 \times 10^{8} \text{ years}^{-1} \]

### Step 5: Substitute into the Decay Equation
Now substituting \( N \) and \( N_0 \) into the decay equation:
\[ \frac{5}{3} N_0 = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{5}{3} = e^{-\lambda t} \]

### Step 6: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{5}{3}\right) = -\lambda t \]
Now substituting \( \lambda \):
\[ t = -\frac{\ln\left(\frac{5}{3}\right)}{\lambda} \]

### Step 7: Calculate the Age of the Rock
Substituting the value of \( \lambda \):
\[ t = -\frac{\ln\left(\frac{5}{3}\right)}{1.54 \times 10^{8}} \]
Calculating \( \ln\left(\frac{5}{3}\right) \):
\[ \ln\left(\frac{5}{3}\right) \approx 0.5108 \]
Thus:
\[ t \approx -\frac{0.5108}{1.54 \times 10^{8}} \approx 3.32 \times 10^{-9} \text{ years} \]

### Final Answer
The age of the rock is approximately \( 3.32 \times 10^{-9} \) years.

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