Half lives of two isotopes X and Y of a material are known to be `2xx10^(9)` years and `4xx10^(9)` years respectively if a planet was formed with equal number of these isotopes, then the current age of planet, given that currently the material has `20%` of X and `80%` of Y by number, will be:

Options

A. `8xx10^(9)` years
B. `6xx10^(9)` years
C. `4xx10^(9)` years
D. `2xx10^(9)` years

Answer

To solve the problem, we need to determine the current age of the planet based on the given half-lives of isotopes X and Y, their initial conditions, and their current proportions.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
At the formation of the planet, we assume there were equal numbers of isotopes X and Y. Let's denote the initial number of nuclei of both isotopes as \(N_0\). Therefore, we have:
\[ N_{0X} = N_{0Y} = \frac{N_0}{2} \]

2. **Current Proportions**:
The problem states that currently, the material has 20% of X and 80% of Y. If the total number of nuclei at present is \(N\), then:
\[ N_X = 0.2N \quad \text{and} \quad N_Y = 0.8N \]

3. **Decay Formula**:
The number of remaining nuclei of a radioactive isotope at time \(t\) is given by the formula:
\[ N = N_0 e^{-\lambda t} \]
where \(\lambda\) is the decay constant related to the half-life (\(T_{1/2}\)) by:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]

4. **Calculate Decay Constants**:
For isotopes X and Y, we have:
- Half-life of X, \(T_{1/2X} = 2 \times 10^9\) years
- Half-life of Y, \(T_{1/2Y} = 4 \times 10^9\) years

Therefore, the decay constants are:
\[ \lambda_X = \frac{\ln 2}{2 \times 10^9} \quad \text{and} \quad \lambda_Y = \frac{\ln 2}{4 \times 10^9} \]

5. **Set Up Equations**:
Using the decay formula, we can express the current number of nuclei for both isotopes:
\[ N_X = \frac{N_0}{2} e^{-\lambda_X t} = 0.2N \]
\[ N_Y = \frac{N_0}{2} e^{-\lambda_Y t} = 0.8N \]

6. **Express Total Nuclei**:
Since \(N = N_X + N_Y\), we can express \(N\) in terms of \(N_0\):
\[ N = N_0 e^{-\lambda_X t} + N_0 e^{-\lambda_Y t} \]

7. **Divide the Equations**:
Dividing the equation for \(N_Y\) by the equation for \(N_X\):
\[ \frac{N_Y}{N_X} = \frac{0.8N}{0.2N} = 4 \]
This leads to:
\[ \frac{e^{-\lambda_Y t}}{e^{-\lambda_X t}} = 4 \quad \Rightarrow \quad e^{(\lambda_X - \lambda_Y)t} = 4 \]

8. **Take Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ (\lambda_X - \lambda_Y)t = \ln 4 \]

9. **Substituting for \(\lambda\)**:
Substitute \(\lambda_X\) and \(\lambda_Y\) into the equation:
\[ \left(\frac{\ln 2}{2 \times 10^9} - \frac{\ln 2}{4 \times 10^9}\right)t = \ln 4 \]

10. **Simplify and Solve for \(t\)**:
This simplifies to:
\[ \left(\frac{\ln 2}{4 \times 10^9}\right)t = \ln 4 \quad \Rightarrow \quad t = \frac{4 \times 10^9 \ln 4}{\ln 2} \]
Since \(\ln 4 = 2 \ln 2\):
\[ t = \frac{4 \times 10^9 \cdot 2 \ln 2}{\ln 2} = 8 \times 10^9 \text{ years} \]

### Final Answer:
The current age of the planet is \(8 \times 10^9\) years.