Question

An element X of a half-life of \(1.4 \times 10^9\) years decays to form another stable element Y. A sample is taken from a rock that contains both X and Y in the ratio 1 : 7. If at the time of formation of the rock, Y was not present in the sample, then the age of the rock in years is

• A. \(4.2 \times 10^{9}\)
• B. \(1.4 \times 10^{9}\)
• C. \(0.35 \times 10^{9}\)
• D. \(2.8 \times 10^{9}\)

Hint:

Relate remaining fraction to half-lives to find age of radioactive samples.

Answer 1

The Correct Option is A

Step 1: Use decay formula
\[ N_0 = N + N_d \] Where \(N_0\) initial nuclei, \(N\) remaining, \(N_d\) decayed. Given ratio \(X:Y = 1:7\), so \(N : N_d = 1 : 7\)
Step 2: Calculate fraction remaining
\[ \frac{N}{N_0} = \frac{1}{1+7} = \frac{1}{8} \] Step 3: Use decay law
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T}} \] where \(T = 1.4 \times 10^9\) years (half-life), solve for \(t\): \[ \left(\frac{1}{2}\right)^{\frac{t}{1.4 \times 10^9}} = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \] \[ \frac{t}{1.4 \times 10^9} = 3 \implies t = 4.2 \times 10^9\, \text{years} \] Step 4: Conclusion
The age of the rock is \(4.2 \times 10^9\) years.