A radioisotope X  with a half life 1.4 × 109  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

Number of Nuclei at t = 0Number of Nuclei at time t As per questionN0-xx=177N0-7x=x or x=78N0∴ Remaining nuclei of isotope X=N0-x=N0-78N0=18N0=123N0So three half lives would have been passed∴  t=nT1/2=3×1.4×109 years =4.2×109 years  Hence, the age of the rock is 4.2×109 years.