Q. A radio isotope 'X' with a half life $1.4 \times 10^9$ years decays to 'Y' which is stable. A sample of the rock from a cave was found to contain 'X' and 'Y' in the ratio 1 : 7. The age of the rock is :
A) $1.96 \times 10^9$ years
B) $3.92 \times 10^9$ years
C) $4.20 \times 10^9$ years
D) $8.40 \times 10^9$ years
Explanation
[Solution] X remaining = 1/8 of the original, i.e. 3 half-lives = $4.2 \times 10^9$ years. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.