In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be `100:1`. The mean lives of the two isotopes are `4xx10^9` years and `2xx10^9` years, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal propotional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is `1.02:1`.

Explanation

To calculate the age of the rock based on the given information about the two radioactive isotopes, we can follow these steps:

### Step 1: Define the Given Information
- Mass ratio of isotopes: \( \frac{m_1}{m_2} = 100:1 \)
- Mean lives of isotopes:
- \( \tau_1 = 4 \times 10^9 \) years (for isotope 1)
- \( \tau_2 = 2 \times 10^9 \) years (for isotope 2)
- Ratio of atomic weights: \( \frac{A_1}{A_2} = 1.02:1 \)

### Step 2: Calculate the Decay Constants
The decay constant \( \lambda \) is related to the mean life \( \tau \) by the formula:
\[ \lambda = \frac{1}{\tau} \]
Thus, we can calculate:
- \( \lambda_1 = \frac{1}{4 \times 10^9} \)
- \( \lambda_2 = \frac{1}{2 \times 10^9} \)

### Step 3: Set Up the Number of Atoms
Let \( n \) be the initial number of atoms of both isotopes, then:
- Number of atoms of isotope 1: \( n_1 = n \)
- Number of atoms of isotope 2: \( n_2 = n \)

After time \( t \):
- \( n_1 = n e^{-\lambda_1 t} \)
- \( n_2 = n e^{-\lambda_2 t} \)

### Step 4: Formulate the Ratio of Remaining Atoms
From the definitions above, we can write:
\[ \frac{n_1}{n_2} = \frac{n e^{-\lambda_1 t}}{n e^{-\lambda_2 t}} = e^{-(\lambda_1 - \lambda_2)t} \]
This simplifies to:
\[ \frac{n_1}{n_2} = e^{(\lambda_2 - \lambda_1)t} \]

### Step 5: Relate the Mass Ratio to the Number of Atoms
Using the mass ratio and atomic weights, we have:
\[ \frac{m_1}{m_2} = \frac{100}{1} = 100 \]
And since the number of atoms is related to mass and atomic weight:
\[ \frac{n_1}{n_2} = \frac{m_1/A_1}{m_2/A_2} = \frac{m_1}{m_2} \cdot \frac{A_2}{A_1} \]
Substituting the known values:
\[ \frac{n_1}{n_2} = 100 \cdot \frac{1}{1.02} \]

### Step 6: Set Up the Equation
Equating the two expressions for \( \frac{n_1}{n_2} \):
\[ e^{(\lambda_2 - \lambda_1)t} = 100 \cdot \frac{1}{1.02} \]

### Step 7: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ (\lambda_2 - \lambda_1)t = \ln\left(100 \cdot \frac{1}{1.02}\right) \]

### Step 8: Solve for \( t \)
Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ \lambda_2 - \lambda_1 = \frac{1}{2 \times 10^9} - \frac{1}{4 \times 10^9} = \frac{2 - 1}{4 \times 10^9} = \frac{1}{4 \times 10^9} \]
Thus,
\[ t = \frac{\ln\left(100 \cdot \frac{1}{1.02}\right)}{\frac{1}{4 \times 10^9}} \]

### Step 9: Calculate \( t \)
Calculating \( \ln(100) \) and \( \ln(1.02) \):
\[ \ln(100) = 4.6052, \quad \ln(1.02) \approx 0.0198 \]
So,
\[ t = \frac{4.6052 - 0.0198}{\frac{1}{4 \times 10^9}} = \frac{4.5854}{\frac{1}{4 \times 10^9}} = 4.5854 \times 4 \times 10^9 \approx 1.834 \times 10^{10} \text{ years} \]

### Final Answer
The age of the rock is approximately \( 1.834 \times 10^{10} \) years.

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