A radioactive isotope X with half-life `1.5xx10^(9)` yr decays into a stable nucleus Y .A rock sample contains both elements X and Y in the ratio 1 : 15. They age of the rock is

Options

A. `3.0xx10^(9)yr`
B. `4.5xx10^(9)yr`
C. `6.0xx10^(9)yr`
D. `9.0xx10^(9)yr`

Explanation

To find the age of the rock sample containing the radioactive isotope X and its stable decay product Y, we can follow these steps:

### Step 1: Understand the Ratio
The rock sample contains elements X and Y in the ratio 1:15. This means that for every 1 atom of X, there are 15 atoms of Y. Therefore, the total number of atoms in the sample can be expressed as:
- Number of atoms of X (N_X) = 1
- Number of atoms of Y (N_Y) = 15
- Total atoms (N_total) = N_X + N_Y = 1 + 15 = 16

### Step 2: Relate the Number of Atoms
Initially, when the rock was formed, it contained only the radioactive isotope X. Over time, as X decays into Y, the number of atoms of X decreases while the number of atoms of Y increases.

Let N_0 be the initial number of atoms of X. After some time t, the number of atoms of X remaining (N) can be expressed using the decay formula:
\[ N = N_0 e^{-\lambda t} \]
Where:
- \( \lambda \) is the decay constant.
- \( t \) is the time elapsed.

### Step 3: Calculate Decay Constant
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
Given that the half-life \( T_{1/2} \) of isotope X is \( 1.5 \times 10^9 \) years, we can calculate \( \lambda \):
\[ \lambda = \frac{0.693}{1.5 \times 10^9 \text{ years}} \]

### Step 4: Set Up the Equation
At the present time, the number of atoms of X is 1, and the number of atoms of Y is 15. Since Y is produced from the decay of X, we can express the relationship as:
\[ N_Y = N_0 - N \]
Where \( N_Y = 15 \) and \( N = 1 \). Therefore, we have:
\[ 15 = N_0 - 1 \]
This gives:
\[ N_0 = 16 \]

### Step 5: Substitute into the Decay Formula
Now we substitute \( N_0 \) into the decay formula:
\[ 1 = 16 e^{-\lambda t} \]
Dividing both sides by 16 gives:
\[ \frac{1}{16} = e^{-\lambda t} \]

### Step 6: Take Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{16}\right) = -\lambda t \]
Since \( \frac{1}{16} = 2^{-4} \), we can rewrite this as:
\[ \ln(2^{-4}) = -\lambda t \]
This simplifies to:
\[ -4 \ln(2) = -\lambda t \]
Thus:
\[ t = \frac{4 \ln(2)}{\lambda} \]

### Step 7: Substitute \( \lambda \)
Now substituting \( \lambda \):
\[ t = \frac{4 \ln(2)}{\frac{0.693}{1.5 \times 10^9}} \]
Calculating this gives:
\[ t = 4 \times 1.5 \times 10^9 \text{ years} \]
\[ t = 6.0 \times 10^9 \text{ years} \]

### Final Answer
The age of the rock is \( 6.0 \times 10^9 \) years.

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