The radioactive isotope X with a half-life of `10^(9)` decays to Y which is stable. A sample of rocks was found to contain both the elements X and Y in the years ratio `1: 7`. If initially, the quantity of Y in the rock was zero, then the age of the rocks is

Options

A. `2 xx 10^(9)years`
B. `3 xx 10^(9)years`
C. `6 xx 10^(9)years`
D. `7 xx 10^(9)years`

Answer

To solve the problem, we need to determine the age of the rocks based on the decay of the radioactive isotope X into the stable isotope Y. We know the following:

1. The half-life of isotope X is \(10^9\) years.
2. The ratio of X to Y in the sample is \(1:7\).
3. Initially, the quantity of Y was zero.

Let's denote:
- The initial quantity of X as \(N_0\).
- The quantity of X at time \(t\) as \(N_X\).
- The quantity of Y at time \(t\) as \(N_Y\).

### Step-by-step Solution:

**Step 1: Understand the ratio of X and Y.**
Given that the ratio of X to Y is \(1:7\), we can express this in terms of a common variable:
\[ N_X : N_Y = 1 : 7 \]
Let \(N_X = x\) and \(N_Y = 7x\).

**Step 2: Relate the quantities of X and Y.**
Since Y is formed from the decay of X, we have:
\[ N_Y = N_0 - N_X \]
Initially, \(N_Y = 0\), so:
\[ N_Y = N_0 - N_X \implies 7x = N_0 - x \]
Rearranging gives:
\[ N_0 = 8x \]

**Step 3: Apply the decay formula.**
The quantity of X remaining after time \(t\) can be expressed using the half-life formula:
\[ N_X = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
Where \(T_{1/2} = 10^9\) years. Substituting \(N_X\) and \(N_0\):
\[ x = 8x \left(\frac{1}{2}\right)^{\frac{t}{10^9}} \]

**Step 4: Simplify the equation.**
Dividing both sides by \(x\) (assuming \(x \neq 0\)):
\[ 1 = 8 \left(\frac{1}{2}\right)^{\frac{t}{10^9}} \]
This simplifies to:
\[ \left(\frac{1}{2}\right)^{\frac{t}{10^9}} = \frac{1}{8} \]

**Step 5: Solve for \(t\).**
Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), we can equate the exponents:
\[ \frac{t}{10^9} = 3 \implies t = 3 \times 10^9 \text{ years} \]

### Conclusion:
The age of the rocks is \(3 \times 10^9\) years.

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