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 $$1.02$$:$$1$$.$$)$$

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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$$−λ$$1tN0e$$−λ$$2t=e$$λ$$2$$−λ$$1t$$  ………$$(ii)Equating$$ Eqs. $$(i) and (i), we have           eλ$$2$$−λ$$1t=1001.02$$ or  λ$$2$$−λ$$1t=ln$$⁡(100)−ln⁡(1.02)∴ $$t=ln$$⁡(100)−ln⁡$$(1.02)12$$×$$109$$−$$14$$×$$109Substituting$$ the values, we have               $$t=1.834$$×$$1010yr$$

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