In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be 100:1. The mean lives of the two isotopes are 4×109 yr and 2×109 yr, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal proportion, the age of the rock p×1010yr, where the value of p is ... Take, ratio of the atomic weights of the two isotopes is 1.02:1.)

At the time of observation (t = t),      m1m2=1001            (given)Further, it is given that A1A2=1.021Number of atoms N=mA∴ N1N2=m1m2×A2A1=1001.02   …….(i)Let N0 be the number of atoms of both the isotopes at the time of formation, then             N1N2=N0e−λ1tN0e−λ2t=eλ2−λ1t  ………(ii)Equating Eqs. (i) and (i), we have           eλ2−λ1t=1001.02 or  λ2−λ1t=ln⁡(100)−ln⁡(1.02)∴ t=ln⁡(100)−ln⁡(1.02)12×109−14×109Substituting the values, we have               t=1.834×1010yr