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

Answer

To find the current age of the planet formed with equal numbers of isotopes X and Y, we can follow these steps:

### Step 1: Define Initial Conditions
Let the initial number of nuclei of both isotopes X and Y be \( N_0 \). Since the planet was formed with equal numbers of isotopes, we have:
- Initial number of nuclei of isotope X, \( N_{X0} = \frac{N_0}{2} \)
- Initial number of nuclei of isotope Y, \( N_{Y0} = \frac{N_0}{2} \)

### Step 2: Define Current Conditions
According to the problem, currently, the material has:
- 20% of isotope X
- 80% of isotope Y

Let the current total number of nuclei be \( N \). Thus:
- Current number of nuclei of isotope X, \( N_X = 0.2N \)
- Current number of nuclei of isotope Y, \( N_Y = 0.8N \)

### Step 3: Use the Decay Formula
The number of remaining nuclei after time \( t \) can be expressed using the decay formula:
\[ N = N_0 e^{-\lambda t} \]
Where \( \lambda \) is the decay constant related to half-life \( T_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

### Step 4: Write Equations for Each Isotope
For isotope X:
\[ N_X = N_{X0} e^{-\lambda_X t} \]
Substituting the values:
\[ 0.2N = \frac{N_0}{2} e^{-\lambda_X t} \]

For isotope Y:
\[ N_Y = N_{Y0} e^{-\lambda_Y t} \]
Substituting the values:
\[ 0.8N = \frac{N_0}{2} e^{-\lambda_Y t} \]

### Step 5: Relate Decay Constants to Half-Lives
Given:
- Half-life of X, \( T_{1/2X} = 2 \times 10^9 \) years
- Half-life of Y, \( T_{1/2Y} = 4 \times 10^9 \) years

Calculate the decay constants:
\[ \lambda_X = \frac{\ln(2)}{2 \times 10^9} \]
\[ \lambda_Y = \frac{\ln(2)}{4 \times 10^9} \]

### Step 6: Solve the Equations
Now, we can set up the equations:
1. From isotope X:
\[ 0.2N = \frac{N_0}{2} e^{-\lambda_X t} \]
Rearranging gives:
\[ e^{-\lambda_X t} = \frac{0.4N}{N_0} \]

2. From isotope Y:
\[ 0.8N = \frac{N_0}{2} e^{-\lambda_Y t} \]
Rearranging gives:
\[ e^{-\lambda_Y t} = \frac{1.6N}{N_0} \]

### Step 7: Divide the Equations
Dividing the two equations:
\[ \frac{e^{-\lambda_Y t}}{e^{-\lambda_X t}} = \frac{1.6N/N_0}{0.4N/N_0} \]
This simplifies to:
\[ e^{-(\lambda_Y - \lambda_X)t} = 4 \]

Taking the natural logarithm of both sides:
\[ -(\lambda_Y - \lambda_X)t = \ln(4) \]
Thus:
\[ t = \frac{\ln(4)}{\lambda_Y - \lambda_X} \]

### Step 8: Substitute Values
Substituting the values of \( \lambda_X \) and \( \lambda_Y \):
\[ t = \frac{\ln(4)}{\frac{\ln(2)}{4 \times 10^9} - \frac{\ln(2)}{2 \times 10^9}} \]
This simplifies to:
\[ t = \frac{\ln(4)}{\frac{\ln(2)}{4 \times 10^9} - \frac{2\ln(2)}{4 \times 10^9}} \]
\[ t = \frac{\ln(4)}{-\frac{\ln(2)}{4 \times 10^9}} \]
\[ t = -4 \times 10^9 \cdot \frac{\ln(4)}{\ln(2)} \]

### Step 9: Calculate Age
Since \( \ln(4) = 2\ln(2) \):
\[ t = -4 \times 10^9 \cdot 2 = -8 \times 10^9 \text{ years} \]

Since we are looking for the age in positive terms, the current age of the planet is approximately:
\[ t \approx 8 \times 10^9 \text{ years} \]

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

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